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Introduction to Dynamic Programming – Data Structures and Algorithm Tutorials

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  • Difficulty Level : Easy
  • Last Updated : 01 Feb, 2023
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Dynamic Programming (DP) is defined as a technique that solves some particular type of problems in Polynomial Time. Dynamic Programming solutions are faster than the exponential brute method and can be easily proved their correctness.

Dynamic Programming is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of subproblems so that we do not have to re-compute them when needed later. This simple optimization reduces time complexities from exponential to polynomial.

Introduction to Dynamic Programming - Data Structures and Algorithm Tutorials

Introduction to Dynamic Programming – Data Structures and Algorithm Tutorials

Characteristics of Dynamic Programming Algorithm:

  • In general, dynamic programming (DP) is one of the most powerful techniques for solving a certain class of problems. 
  • There is an elegant way to formulate the approach and a very simple thinking process, and the coding part is very easy. 
  • Essentially, it is a simple idea, after solving a problem with a given input, save the result as a reference for future use, so you won’t have to re-solve it.. briefly ‘Remember your Past’ :). 
  • It is a big hint for DP if the given problem can be broken up into smaller sub-problems, and these smaller subproblems can be divided into still smaller ones, and in this process, you see some overlapping subproblems. 
  • Additionally, the optimal solutions to the subproblems contribute to the optimal solution of the given problem (referred to as the Optimal Substructure Property).
  •  The solutions to the subproblems are stored in a table or array (memoization) or in a bottom-up manner (tabulation) to avoid redundant computation.
  • The solution to the problem can be constructed from the solutions to the subproblems.
  • Dynamic programming can be implemented using a recursive algorithm, where the solutions to subproblems are found recursively, or using an iterative algorithm, where the solutions are found by working through the subproblems in a specific order.

Dynamic programming works on following principles: 

  • Characterize structure of optimal solution, i.e. build a mathematical model of the solution.
  • Recursively define the value of the optimal solution. 
  • Using bottom-up approach, compute the value of the optimal solution for each possible subproblems.
  •  Construct optimal solution for the original problem using information computed in the previous step.

Applications:

Dynamic programming is used to solve optimization problems. It is used to solve many real-life problems such as,

(i) Make a change problem

 (ii) Knapsack problem

(iii) Optimal binary search tree

What is the difference between a Dynamic programming algorithm and recursion?

  • In dynamic programming, problems are solved by breaking them down into smaller ones to solve the larger ones, while recursion is when a function is called and executed by itself. While dynamic programming can function without making use of recursion techniques, since the purpose of dynamic programming is to optimize and accelerate the process, programmers usually make use of recursion techniques to accelerate and turn the process efficiently.
  • When a function can execute a specific task by calling itself, receive the name of the recursive function. In order to perform and accomplish the work, this function calls itself when it has to be executed.
  • Using dynamic programming, you can break a problem into smaller parts, called subproblems, to solve it. Dynamic programming involves solving the problem for the first time, then using memoization to store the solutions.
  • Therefore, the main difference between the two techniques is their intended use; recursion is used to automate a function, whereas dynamic programming is an optimization technique used to solve problems.
  • Recursive functions recognize when they are needed, execute themselves, then stop working. When the function identifies the moment it is needed, it calls itself and is executed; this is called a recursive case. As a result, the function must stop once the task is completed, known as the base case.
  • By establishing states, dynamic programming recognizes the problem and divides it into sub-problems in order to solve the whole scene. After solving these sub-problems, or variables, the programmer must establish a mathematical relationship between them. Last but not least, these solutions and results are stored as algorithms, so they can be accessed in the future without having to solve the whole problem again.

Techniques to solve Dynamic Programming Problems:

1. Top-Down(Memoization):

Break down the given problem in order to begin solving it. If you see that the problem has already been solved, return the saved answer. If it hasn’t been solved, solve it and save it. This is usually easy to think of and very intuitive, This is referred to as Memoization.

2. Bottom-Up(Dynamic Programming):

Analyze the problem and see in what order the subproblems are solved, and work your way up from the trivial subproblem to the given problem. This process ensures that the subproblems are solved before the main problem. This is referred to as Dynamic Programming.

Types of the approach of dynamic programming algorithm

Types of the approach of dynamic programming algorithm

Tabulation(Dynamic Programming) vs Memoization:

There are two different ways to store the values so that the values of a sub-problem can be reused. Here, will discuss two patterns of solving dynamic programming (DP) problems: 

  • Tabulation: Bottom Up
  • Memoization: Top Down

Before getting to the definitions of the above two terms consider the following statements:

  • Version 1: I will study the theory of DP from GeeksforGeeks, then I will practice some problems on classic DP and hence I will master DP.
  • Version 2: To Master DP, I would have to practice Dynamic problems and practice problems – Firstly, I would have to study some theories of DP from GeeksforGeeks

Both versions say the same thing, the difference simply lies in the way of conveying the message and that’s exactly what Bottom-Up and Top-Down DP do. Version 1 can be related to Bottom-Up DP and Version-2 can be related to Top-Down DP.

     Tabulation       Memoization   
State State transition relation is difficult to think State Transition relation is easy to think
Code Code gets complicated when a lot of 
conditions are required
Code is easy and less complicated
 
Speed Fast, as we directly access previous states from the table Slow due to a lot of recursive calls and return statements
Subproblem solving If all subproblems must be solved at least once, a bottom-up dynamic programming algorithm usually outperforms a top-down memoized algorithm by a constant factor If some subproblems in the subproblem space need not be solved at all, the memoized solution has the advantage of solving only those subproblems that are definitely required 
Table entries In the Tabulated version, starting from the first entry, all entries are filled one by one Unlike the Tabulated version, all entries of the lookup table are not necessarily filled in Memoized version. The table is filled on demand.
Approach Generally, tabulation(dynamic programming) is an iterative approach On the other hand, memoization is a recursive approach.

How to solve a Dynamic Programming Problem?

To dynamically solve a problem, we need to check two necessary conditions: 

  • Overlapping Subproblems: When the solutions to the same subproblems are needed repetitively for solving the actual problem. The problem is said to have overlapping subproblems property.
     
N-th Fibonacci Series as Overlapping Subproblems

N-th Fibonacci Series as Overlapping Subproblems

  • Optimal Substructure Property: If the optimal solution of the given problem can be obtained by using optimal solutions of its subproblems then the problem is said to have Optimal Substructure Property.

Steps to solve a Dynamic programming problem:

  1. Identify if it is a Dynamic programming problem.
  2. Decide a state expression with the Least parameters.
  3. Formulate state and transition relationships.
  4. Do tabulation (or memorization).

1) How to classify a problem as a Dynamic Programming algorithm Problem?

  • Typically, all the problems that require maximizing or minimizing certain quantities or counting problems that say to count the arrangements under certain conditions or certain probability problems can be solved by using Dynamic Programming.
  • All dynamic programming problems satisfy the overlapping subproblems property and most of the classic Dynamic programming problems also satisfy the optimal substructure property. Once we observe these properties in a given problem be sure that it can be solved using Dynamic Programming.

2) Deciding the state:

Problems with dynamic programming are mostly concerned with the state and its transition. The most fundamental phase must be carried out with extreme care because the state transition depends on the state definition you select.

State:

A state is a collection of characteristics that can be used to specifically describe a given position or standing in a given challenge. To minimise state space, this set of parameters has to be as compact as feasible.

3) Formulating a relation among the states:

The hardest part of a Dynamic Programming challenge is this step, which calls for a lot of intuition, observation, and training.

Example:

Given 3 numbers {1, 3, 5}, the task is to tell the total number of ways we can form a number N using the sum of the given three numbers. (allowing repetitions and different arrangements).

The total number of ways to form 6 is: 8
1+1+1+1+1+1
1+1+1+3
1+1+3+1
1+3+1+1
3+1+1+1
3+3
1+5
5+1

Following are the steps to solve the problem:

  • We choose a state for the given problem.
  • N will be used as the determining factor for the state because it can be used to identify any subproblem.
  • The DP state will resemble state(N), where the state(N) is the total number of arrangements required to create N using the elements 1, 3, and 5. Identify the relationship of the transition between any two states.
  • We must now calculate the state (N).

3.1) How to Compute the state?

As we can only use 1, 3, or 5 to form a given number N. Let us assume that we know the result for N = 1, 2, 3, 4, 5, 6 

Let us say we know the result for:
state (n = 1), state (n = 2), state (n = 3) ……… state (n = 6) 
Now, we wish to know the result of the state (n = 7). See, we can only add 1, 3, and 5. Now we can get a sum total of 7 in the following 3 ways:

1) Adding 1 to all possible combinations of state (n = 6) 
Eg: [ (1+1+1+1+1+1) + 1] 
[ (1+1+1+3) + 1] 
[ (1+1+3+1) + 1] 
[ (1+3+1+1) + 1] 
[ (3+1+1+1) + 1] 
[ (3+3) + 1] 
[ (1+5) + 1] 
[ (5+1) + 1]

2) Adding 3 to all possible combinations of state (n = 4);
[(1+1+1+1) + 3] 
[(1+3) + 3] 
[(3+1) + 3]

3) Adding 5 to all possible combinations of state(n = 2) 
[ (1+1) + 5]

(Note how it sufficient to add only on the right-side – all the add-from-left-side cases are covered, either in the same state, or another, e.g. [ 1+(1+1+1+3)]  is not needed in state (n=6) because it’s covered by state (n = 4) [(1+1+1+1) + 3])

Now, think carefully and satisfy yourself that the above three cases are covering all possible ways to form a sum total of 7;
Therefore, we can say that result for 
state(7) = state (6) + state (4) + state (2) 
OR
state(7) = state (7-1) + state (7-3) + state (7-5)
In general, 
state(n) = state(n-1) + state(n-3) + state(n-5)

Below is the implementation for the above approach:

C++




// Returns the number of arrangements to
// form 'n'
int solve(int n)
{
// base case
if (n < 0)
    return 0;
if (n == 0)
    return 1;
 
return solve(n-1) + solve(n-3) + solve(n-5);
}


Java




import java.io.*;
 
class GFG {
 
    // Returns the number of arrangements to
    // form 'n'.
    public static int solve(int n)
    {
        // base case
        if (n < 0)
            return 0;
        if (n == 0)
            return 1;
 
        return solve(n - 1) + solve(n - 3) + solve(n - 5);
    }
    public static void main(String[] args) {}
}


Python3




# Python program to Returns the number of arrangements to form 'n'
def solve(n):
  # Base case
    if(n < 0):
        return 0
    if(n == 0):
        return 1
    return solve(n-1)+solve(n-3)+solve(n-5)
 
  # This code is contributed by ishankhandelwals.


C#




using System;
 
public class GFG {
 
    // Returns the number of arrangements to
    // form 'n'
    public static int solve(int n)
    {
        // base case
        if (n < 0)
            return 0;
        if (n == 0)
            return 1;
 
        return solve(n - 1) + solve(n - 3) + solve(n - 5);
    }
 
    static public void Main() {}
}
 
// This code is contributed by akashish__


Javascript




//JS code that returns the number of arrangements to
// form 'n'
function solve(n)
{
 
// base case
if (n < 0)
    return 0;
if (n == 0)
    return 1;
return solve(n-1) + solve(n-3) + solve(n-5);
}
 
// This code is contributed by ishankhandelwals.


Time Complexity: O(3n), As at every stage we need to take three decisions and the height of the tree will be of the order of n.
Auxiliary Space: O(n), The extra space is used due to the recursion call stack.

The above code seems exponential as it is calculating the same state again and again. So, we just need to add memoization.

4) Adding memoization or tabulation for the state

The simplest portion of a solution based on dynamic programming is this. Simply storing the state solution will allow us to access it from memory the next time that state is needed.

Adding memoization to the above code:

C++




// Initialize to -1
int dp[MAXN];
 
// This function returns the number of
// arrangements to form 'n'
int solve(int n)
{
// base case
if (n < 0)
    return 0;
if (n == 0)
    return 1;
 
// Checking if already calculated
if (dp[n]!=-1)
    return dp[n];
 
// Storing the result and returning
return dp[n] = solve(n-1) + solve(n-3)
                          + solve(n-5);
}


Java




// Initialize to -1
int dp = new int[MAXN];
 
// This function returns the number of
// arrangements to form 'n'
int solve(int n)
{
    // base case
    if (n < 0)
        return 0;
    if (n == 0)
        return 1;
 
    // Checking if already calculated
    if (dp[n] != -1)
        return dp[n];
 
    // Storing the result and returning
    return dp[n]
        = solve(n - 1) + solve(n - 3) + solve(n - 5);
}


Python3




# Initialize to -1
dp = []
 
# This function returns the number of
# arrangements to form 'n'
def solve(n):
    # base case
    if n < 0:
        return 0
    if n == 0:
        return 1
 
# Checking if already calculated
    if dp[n] != -1:
        return dp[n]
 
# Storing the result and returning
    dp[n] = solve(n-1) + solve(n-3) + solve(n-5)
    return dp[n]
 
  # This code is contributed by ishankhandelwals.


C#




using System;
 
public class GFG
{
   
      // Initialize to -1
    public static int dp = new int[MAXN];
 
    // This function returns the number of
    // arrangements to form 'n'
    public static int solve(int n)
    {
    // base case
    if (n < 0)
        return 0;
    if (n == 0)
        return 1;
 
    // Checking if already calculated
    if (dp[n]!=-1)
        return dp[n];
 
    // Storing the result and returning
    return dp[n] = solve(n-1) + solve(n-3)
                              + solve(n-5);
    }
 
    static public void Main (){
 
        // Code
    }
}
 
// This code is contributed by akashish__


Javascript




// Initialize to -1
let dp = [];
for (let i = 0; i < MAXN; i++) {
    dp.push(0);
}
 
// This function returns the number of
// arrangements to form 'n'
function solve(n) {
    // base case
    if (n < 0)
        return 0;
    if (n == 0)
        return 1;
 
    // Checking if already calculated
    if (dp[n] != -1)
        return dp[n];
 
    // Storing the result and returning
    return dp[n] = solve(n - 1) + solve(n - 3)
        + solve(n - 5);
}
// contributed by akashish__


Time Complexity: O(n), As we just need to make 3n function calls and there will be no repetitive calculations as we are returning previously calculated results.
Auxiliary Space: O(n), The extra space is used due to the recursion call stack.

How to solve Dynamic Programming problems through Example?

Problem: Let’s find the Fibonacci sequence up to the nth term. A Fibonacci series is the sequence of numbers in which each number is the sum of the two preceding ones. For example, 0, 1, 1, 2, 3, and so on. Here, each number is the sum of the two preceding numbers.

Naive Approach: The basic way to find the nth Fibonacci number is to use recursion.

Below is the implementation for the above approach:

C++




// C++ code for the above approach:
#include <iostream>
using namespace std;
 
// Function to find nth fibonacci number
int fib(int n)
{
    if (n <= 1) {
        return n;
    }
    int x = fib(n - 1);
    int y = fib(n - 2);
 
    return x + y;
}
 
// Drivers code
int main()
{
    int n = 5;
 
    // Function Call
    cout << fib(n);
    return 0;
}


Java




// Java code for the above approach:
import java.io.*;
 
class GFG {
    // Function to find nth fibonacci number
    public static int fib(int n)
    {
        if (n <= 1) {
            return n;
        }
        int x = fib(n - 1);
        int y = fib(n - 2);
 
        return x + y;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 5;
 
        // Function Call
        System.out.print(fib(n));
    }
}
 
// This code is contributed by Rohit Pradhan


Python3




# Function to find nth fibonacci number
def fib(n):
    if (n <= 1):
        return n
    x = fib(n - 1)
    y = fib(n - 2)
 
    return x + y
 
n = 5;
 
# Function Call
print(fib(n))
 
#contributed by akashish__


C#




using System;
 
public class GFG {
 
    // Function to find nth fibonacci number
    public static int fib(int n)
    {
        if (n <= 1) {
            return n;
        }
        int x = fib(n - 1);
        int y = fib(n - 2);
 
        return x + y;
    }
 
    static public void Main()
    {
 
        int n = 5;
 
        // Function Call
        Console.WriteLine(fib(n));
    }
}
// contributed by akashish__


Javascript




// Function to find nth fibonacci number
function fib(n)
{
    if (n <= 1) {
        return n;
    }
    let x = fib(n - 1);
    let y = fib(n - 2);
 
    return x + y;
}
 
// Drivers code
let n = 5;
 
// Function Call
console.log(fib(n));
 
// This code is contributed by akashish__


Output

5

Complexity Analysis: 

Time Complexity: O(2n)

  • Here, for every n, we are required to make a recursive call to fib(n – 1) and fib(n – 2). For fib(n – 1), we will again make the recursive call to fib(n – 2) and fib(n – 3). Similarly, for fib(n – 2), recursive calls are made on fib(n – 3) and fib(n – 4) until we reach the base case.
  • During each recursive call, we perform constant work(k) (adding previous outputs to obtain the current output). We perform 2nK work at every level (where n = 0, 1, 2, …). Since n is the number of calls needed to reach 1, we are performing 2n-1k at the final level. Total work can be calculated as:
  • If we draw the recursion tree of the Fibonacci recursion then we found the maximum height of the tree will be n and hence the space complexity of the Fibonacci recursion will be O(n).

Efficient approach: As it is a very terrible complexity(Exponential), thus we need to optimize it with an efficient method. (Memoization)

Let’s look at the example below for finding the 5th Fibonacci number.

Representation of 5th Fibonacci number

Observations:

  • The entire program repeats recursive calls. As in the above figure, for calculating fib(4), we need the value of fib(3) (first recursive call over fib(3)), and for calculating fib(5), we again need the value of fib(3)(second similar recursive call over fib(3)). 
  • Both of these recursive calls are shown above in the outlining circle. 
  • Similarly, there are many others for which we are repeating the recursive calls. 
  • Recursion generally involves repeated recursive calls, which increases the program’s time complexity. 
  • By storing the output of previously encountered values (preferably in arrays, as these can be traversed and extracted most efficiently), we can overcome this problem. The next time we make a recursive call over these values, we will use their already stored outputs instead of calculating them all over again. 
  • In this way, we can improve the performance of our code. Memoization is the process of storing each recursive call’s output for later use, preventing the code from calculating it again. 

Way to memoize: To achieve this in our example we will simply take an answer array initialized to -1. As we make a recursive call, we will first check if the value stored in the answer array corresponding to that position is -1. The value -1 indicates that we haven’t calculated it yet and have to recursively compute it. The output must be stored in the answer array so that, next time, if the same value is encountered, it can be directly used from the answer array.   

Now in this process of memoization, considering the above Fibonacci numbers example, it can be observed that the total number of unique calls will be at most (n + 1) only.

Below is the implementation for the above approach:

C++




// C++ code for the above approach:
#include <iostream>
using namespace std;
 
// Helper Function
int fibo_helper(int n, int* ans)
{
 
    // Base case
    if (n <= 1) {
        return n;
    }
 
    // To check if output already exists
    if (ans[n] != -1) {
        return ans[n];
    }
 
    // Calculate output
    int x = fibo_helper(n - 1, ans);
    int y = fibo_helper(n - 2, ans);
 
    // Saving the output for future use
    ans[n] = x + y;
 
    // Returning the final output
    return ans[n];
}
 
int fibo(int n)
{
    int* ans = new int[n + 1];
 
    // Initializing with -1
    for (int i = 0; i <= n; i++) {
        ans[i] = -1;
    }
    fibo_helper(n, ans);
}
 
// Drivers code
int main()
{
    int n = 5;
 
    // Function Call
    cout << fibo(n);
    return 0;
}


Java




// Java code for the above approach:
import java.io.*;
 
class GFG {
    // Helper Function
    public static int fibo_helper(int n, int ans[])
    {
 
        // Base case
        if (n <= 1) {
            return n;
        }
 
        // To check if output already exists
        if (ans[n] != -1) {
            return ans[n];
        }
 
        // Calculate output
        int x = fibo_helper(n - 1, ans);
        int y = fibo_helper(n - 2, ans);
 
        // Saving the output for future use
        ans[n] = x + y;
 
        // Returning the final output
        return ans[n];
    }
 
    public static int fibo(int n)
    {
        int ans[] = new int[n + 1];
 
        // Initializing with -1
        for (int i = 0; i <= n; i++) {
            ans[i] = -1;
        }
        return fibo_helper(n, ans);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 5;
 
        // Function Call
        System.out.print(fibo(n));
    }
}
 
// This code is contributed by Rohit Pradhan


C#




using System;
 
public class GFG {
 
    // Helper Function
    public static int fibo_helper(int n, int[] ans)
    {
 
        // Base case
        if (n <= 1) {
            return n;
        }
 
        // To check if output already exists
        if (ans[n] != -1) {
            return ans[n];
        }
 
        // Calculate output
        int x = fibo_helper(n - 1, ans);
        int y = fibo_helper(n - 2, ans);
 
        // Saving the output for future use
        ans[n] = x + y;
 
        // Returning the final output
        return ans[n];
    }
 
    public static int fibo(int n)
    {
        int[] ans = new int[n + 1];
 
        // Initializing with -1
        for (int i = 0; i <= n; i++) {
            ans[i] = -1;
        }
        return fibo_helper(n, ans);
    }
 
    static public void Main()
    {
 
        // Code
 
        int n = 5;
 
        // Function Call
        Console.WriteLine(fibo(n));
    }
}
// contributed by akashish__


Python3




# Helper Function
def fibo_helper(n, ans):
  # Base case
  if (n <= 1):
    return n
 
  # To check if output already exists
  if (ans[n] is not -1):
    return ans[n]
 
  # Calculate output
  x = fibo_helper(n - 1, ans)
  y = fibo_helper(n - 2, ans)
 
  # Saving the output for future use
  ans[n] = x + y
 
  # Returning the final output
  return ans[n]
 
 
def fibo(n):
  ans = [-1]*(n+1)
 
  # Initializing with -1
  #for (i = 0; i <= n; i++) {
  for i in range(0,n+1):
    ans[i] = -1
     
  return fibo_helper(n, ans)
 
 
# Code
n = 5
 
# Function Call
print(fibo(n))
# contributed by akashish__


Javascript




<script>
 
// Javascript code for the above approach:
 
// Helper Function
function fibo_helper(n, ans) {
    // Base case
    if (n <= 1) {
        return n;
    }
 
    // To check if output already exists
    if (ans[n] != -1) {
        return ans[n];
    }
 
    // Calculate output
    let x = fibo_helper(n - 1, ans);
    let y = fibo_helper(n - 2, ans);
 
    // Saving the output for future use
    ans[n] = x + y;
 
    // Returning the final output
    return ans[n];
}
 
function fibo(n) {
    let ans = [];
 
    // Initializing with -1
    for (let i = 0; i <= n; i++) {
        ans.push(-1);
    }
    return fibo_helper(n, ans);
}
 
// Drivers code
let n = 5;
 
// Function Call
console.log(fibo(n));
 
// contributed by akashish__
 
</script>


Output

5

Complexity analysis:

  • Time complexity: O(n)
  • Auxiliary Space: O(n)

Optimized approach: Following a bottom-up approach to reach the desired index. This approach of converting recursion into iteration is known as Dynamic programming(DP).

Observations:

  • Finally, what we do is recursively call each response index field and calculate its value using previously saved outputs. 
  • Recursive calls terminate via the base case, which means we are already aware of the answers which should be stored in the base case indexes. 
  • In the case of Fibonacci numbers, these indices are 0 and 1 as f(ib0) = 0 and f(ib1) = 1. So we can directly assign these two values ​​into our answer array and then use them to calculate f(ib2), which is f(ib1) + f(ib0), and so on for each subsequent index. 
  • This can easily be done iteratively by running a loop from i = (2 to n). Finally, we get our answer at the 5th index of the array because we already know that the ith index contains the answer to the ith value.
  • Simply, we first try to find out the dependence of the current value on previous values ​​and then use them to calculate our new value. Now, we are looking for those values which do not depend on other values, which means they are independent(base case values, since these, are the smallest problems
    which we are already aware of).

Below is the implementation for the above approach:

C++




// C++ code for the above approach:
#include <iostream>
using namespace std;
 
// Function for calculating the nth
// Fibonacci number
int fibo(int n)
{
    int* ans = new int[n + 1];
 
    // Storing the independent values in the
    // answer array
    ans[0] = 0;
    ans[1] = 1;
 
    // Using the bottom-up approach
    for (int i = 2; i <= n; i++) {
        ans[i] = ans[i - 1] + ans[i - 2];
    }
 
    // Returning the final index
    return ans[n];
}
 
// Drivers code
int main()
{
    int n = 5;
 
    // Function Call
    cout << fibo(n);
    return 0;
}


Java




// Java code for the above approach:
import java.io.*;
 
class GFG {
    // Function for calculating the nth
    // Fibonacci number
    public static int fibo(int n)
    {
        int ans[] = new int[n + 1];
 
        // Storing the independent values in the
        // answer array
        ans[0] = 0;
        ans[1] = 1;
 
        // Using the bottom-up approach
        for (int i = 2; i <= n; i++) {
            ans[i] = ans[i - 1] + ans[i - 2];
        }
 
        // Returning the final index
        return ans[n];
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 5;
 
        // Function Call
        System.out.print(fibo(n));
    }
}
 
// This code is contributed by Rohit Pradhan


Python3




# Python3 code for the above approach:
 
# Function for calculating the nth
# Fibonacci number
def fibo(n):
  ans = [None] * (n + 1)
 
  # Storing the independent values in the
  # answer array
  ans[0] = 0
  ans[1] = 1
 
  # Using the bottom-up approach
  for i in range(2,n+1):
    ans[i] = ans[i - 1] + ans[i - 2]
 
  # Returning the final index
  return ans[n]
 
# Drivers code
n = 5
 
# Function Call
print(fibo(n))
#contributed by akashish__


C#




// C# code for the above approach:
 
using System;
 
public class GFG {
 
    // Function for calculating the nth
    // Fibonacci number
    public static int fibo(int n)
    {
        int[] ans = new int[n + 1];
 
        // Storing the independent values in the
        // answer array
        ans[0] = 0;
        ans[1] = 1;
 
        // Using the bottom-up approach
        for (int i = 2; i <= n; i++) {
            ans[i] = ans[i - 1] + ans[i - 2];
        }
 
        // Returning the final index
        return ans[n];
    }
 
    static public void Main()
    {
 
        // Code
        int n = 5;
 
        // Function Call
        Console.Write(fibo(n));
    }
}
 
// This code is contributed by lokeshmvs21.


Javascript




<script>
// Javascript code for the above approach:
// Function for calculating the nth
// Fibonacci number
function fibo(n)
{
    let ans = new Array(n + 1).fill(0);
 
    // Storing the independent values in the
    // answer array
    ans[0] = 0;
    ans[1] = 1;
 
    // Using the bottom-up approach
    for (let i = 2; i <= n; i++) {
        ans[i] = ans[i - 1] + ans[i - 2];
    }
 
    // Returning the final index
    return ans[n];
}
 
// Drivers code
let n = 5;
 
// Function Call
console.log(fibo(n));
 
// contributed by akashish__
</script>


Output

5

Complexity analysis: 

  • Time complexity: O(n)
  • Auxiliary Space: O(n)

Optimization of above method

  • in above code we can see that the current state of any fibonacci number depend only on prev two number
  • so using this observation we can conclude that we did not need to store the whole table of size n but instead of that we can only store the prev two values
  • so this way we can optimize the space complexity in the above code O(n) to O(1)

C++




// C++ code for the above approach:
#include <iostream>
using namespace std;
 
// Function for calculating the nth Fibonacci number
int fibo(int n)
{
    int prevPrev, prev, curr;
 
    // Storing the independent values
    prevPrev = 0;
    prev = 1;
    curr = 1;
 
    // Using the bottom-up approach
    for (int i = 2; i <= n; i++) {
        curr = prev + prevPrev;
        prevPrev = prev;
        prev = curr;
    }
 
    // Returning the final answer
    return curr;
}
 
// Drivers code
int main()
{
    int n = 5;
 
    // Function Call
    cout << fibo(n);
    return 0;
}


Output

5

Greedy approach vs Dynamic programming

Feature

Greedy method Dynamic programming

Feasibility 

In a greedy Algorithm, we make whatever choice seems best at the moment in the hope that it will lead to global optimal solution. In Dynamic Programming we make decision at each step considering current problem and solution to previously solved sub problem to calculate optimal solution .

Optimality

In Greedy Method, sometimes there is no such guarantee of getting Optimal Solution. It is guaranteed that Dynamic Programming will generate an optimal solution as it generally considers all possible cases and then choose the best.

Recursion

A greedy method follows the problem solving heuristic of making the locally optimal choice at each stage. A Dynamic programming is an algorithmic technique which is usually based on a recurrent formula that uses some previously calculated states.

Memoization                                   

It is more efficient in terms of memory as it never look back or revise previous choices It requires Dynamic Programming table for Memoization and it increases it’s memory complexity.

    Time        complexity                    

Greedy methods are generally faster. For example, Dijkstra’s shortest path algorithm takes O(ELogV + VLogV) time. Dynamic Programming is generally slower. For example, Bellman Ford algorithm takes O(VE) time.

Fashion

The greedy method computes its solution by making its choices in a serial forward fashion, never looking back or revising previous choices. Dynamic programming computes its solution bottom up or top down by synthesizing them from smaller optimal sub solutions.

Example

Fractional knapsack . 
 
0/1 knapsack problem 

Some commonly asked problems in Dynamic programming:

S. No.

Problem

Practice link

1

Min Cost Path 

solve

2

Subset Sum Problem

solve

3

Knapsack problem

solve

4

Coin Change 

solve

5

Edit Distance 

solve

6

Cutting a Rod

solve

7

Subset Sum Problem

solve

8

Longest Common Subsequence

solve

9

Matrix chain multiplication

solve

10

Count Distinct Subsequences

solve

11

Prefix Sum of Matrix (Or 2D Array)

solve 

12

Check if it is possible to transform one string into another

solve

FAQs about Dynamic Programming Algorithm:

1) Is dynamic programming just recursion?

Dynamic programming and recursion are things completely different. While dynamic programming can use recursion techniques, recursion itself doesn’t have anything similar to dynamic programming. . Dynamic programming involves breaking down a problem into smaller subproblems, storing the solutions to these subproblems to avoid redundant computation, and using these solutions to construct the overall solution. Recursion, on the other hand, is a technique for solving problems by breaking them down into smaller subproblems and solving them recursively.

2) How does dynamic programming works?

Dynamic Programming (DP) is a technique that solves some particular type of problems in Polynomial Time. Dynamic Programming solutions are faster than the exponential brute method and can be easily proved their correctness. Dynamic programming works by breaking down a problem into smaller subproblems, solving each subproblem independently, and using the solutions to these subproblems to construct the overall solution. The solutions to the subproblems are stored in a table or array (memoization) or in a bottom-up manner (tabulation) to avoid redundant computation.

3) How greedy algorithms are similar to Dynamic programming?

Greedy Algorithms are similar to dynamic programming in the sense that they are both tools for optimization. Both dynamic programming and greedy algorithms are used for optimization problems. However, while dynamic programming breaks down a problem into smaller subproblems and solves them independently, greedy algorithms make a locally optimal choice at each step with the hope of finding a globally optimal solution.

4) What are the basics of Dynamic programming?

You can solve subproblems faster by using dynamic programming, which is nothing more than recursion and memoization, thereby reducing the complexity of your code and making it faster. Following are the basic points:

  • Breaking down a problem into smaller subproblems.
  • Solving each subproblem independently.
  • Storing the solutions to subproblems to avoid redundant computation.
  • Using the solutions to the subproblems to construct the overall solution.
  • Using the principle of optimality to ensure that the solution is optimal.

5) What are the advantages of Dynamic programming?

Dynamic programming has the advantage of being able to find both a local and a global optimal solution. Additionally, practical experience can be exploited to benefit from dynamic programming’s better efficiency. However, there isn’t a single, accepted paradigm for dynamic programming, and other conditions could show up as the problem is being solved. Dynamic programming algorithms are guaranteed to find the optimal solution among a set of possible solutions, provided that the problem satisfies the principle of optimality. The solutions to subproblems can be stored in a table, which can be reused for similar problems. Dynamic programming can be applied to a wide range of problems, including optimization, sequence alignment, and resource allocation.

Conclusion:

In conclusion, dynamic programming is a powerful problem-solving technique that is used for optimization problems. Dynamic programming is a superior form of recursion that overcomes its limitations. It involves breaking down a problem into smaller subproblems, solving each subproblem independently, and using the solutions to these subproblems to construct the overall solution. The key characteristics of a dynamic programming algorithm include overlapping subproblems, optimal substructure, memoization or tabulation, and the use of either iterative or recursive methods.

Dynamic programming has several advantages over other problem-solving techniques, including efficiency, simplicity, flexibility, optimality, clarity, and code reusability. It is not just recursion, although it can be implemented using a recursive algorithm. Dynamic programming is different from greedy algorithms in that it breaks down the problem into smaller subproblems, solves each subproblem independently, and uses the solutions to the subproblems to construct the overall solution.

The basics of dynamic programming include breaking down a problem into smaller subproblems, solving each subproblem independently, storing the solutions to subproblems to avoid redundant computation, using the solutions to the subproblems to construct the overall solution and using the principle of optimality to ensure that the solution is optimal.  However, DP can occasionally be challenging to comprehend, making it a well-liked option for coding interviews. Understanding how DP functions work can be useful to everyone, whether they are a professional or a student getting ready for the placements. 

Overall, dynamic programming is a valuable tool for solving complex optimization problems and can lead to more efficient and effective solutions.

Some important links:


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