Maximum sum such that no two elements are adjacent | Set 2

Given an array of positive numbers, find the maximum sum of a subsequence with the constraint that no 2 numbers in the sequence should be adjacent in the array. So 3 2 7 10 should return 13 (sum of 3 and 10) or 3 2 5 10 7 should return 15 (sum of 3, 5, and 7).

Examples:

Input :  arr[] = {3, 5, 3} 
Output : 6 
Explanation : 
Selecting indexes 0 and 2 will maximise the sum 
i.e 3+3 = 6

Input : arr[] = {2, 5, 2}
Output : 5 

We have already discussed the efficient approach of solving this problem in the previous article.
However, we can also solve this problem using the Dynamic Programming approach.
Dynamic Programming Approach: Let’s decide the states of ‘dp’. Let dp[i] be the largest possible sum for the sub-array starting from index ‘i’ and ending at index ‘N-1’. Now, we have to find a recurrence relation between this state and a lower-order state.
In this case for an index ‘i’, we will have two choices.  

1) Choose the current index:
   In this case, the relation will be dp[i] = arr[i] + dp[i+2]
2) Skip the current index:
   Relation will be dp[i] = dp[i+1]

We will choose the path that maximizes our result. 
Thus, the final relation will be:  

dp[i] = max(dp[i+2]+arr[i], dp[i+1])

Below is the implementation of the above approach:  

C++

// C++ program to implement above approach

#include <bits/stdc++.h>
#define maxLen 10
using namespace std;

// variable to store states of dp
int dp[maxLen];

// variable to check if a given state
// has been solved
bool v[maxLen];

// Function to find the maximum sum subsequence
// such that no two elements are adjacent
int maxSum(int arr[], int i, int n)
{
    // Base case
    if (i >= n)
        return 0;

    // To check if a state has
    // been solved
    if (v[i])
        return dp[i];
    v[i] = 1;

    // Required recurrence relation
    dp[i] = max(maxSum(arr, i + 1, n),
                arr[i] + maxSum(arr, i + 2, n));

    // Returning the value
    return dp[i];
}

// Driver code
int main()
{
    int arr[] = { 12, 9, 7, 33 };

    int n = sizeof(arr) / sizeof(int);

    cout << maxSum(arr, 0, n);

    return 0;
}

Java

// Java program to implement above approach
class GFG
{

static int maxLen = 10;

// variable to store states of dp
static int dp[] = new int[maxLen];

// variable to check if a given state
// has been solved
static boolean v[] = new boolean[maxLen];

// Function to find the maximum sum subsequence
// such that no two elements are adjacent
static int maxSum(int arr[], int i, int n)
{
    // Base case
    if (i >= n)
        return 0;

    // To check if a state has
    // been solved
    if (v[i])
        return dp[i];
    v[i] = true;

    // Required recurrence relation
    dp[i] = Math.max(maxSum(arr, i + 1, n),
                arr[i] + maxSum(arr, i + 2, n));

    // Returning the value
    return dp[i];
}

// Driver code
public static void main(String args[])
{
    int arr[] = { 12, 9, 7, 33 };
    int n = arr.length;
    System.out.println( maxSum(arr, 0, n));
}
}

// This code is contributed by Arnab Kundu

Python3

# Python 3 program to implement above approach
maxLen = 10

# variable to store states of dp
dp = [0 for i in range(maxLen)]

# variable to check if a given state
# has been solved
v = [0 for i in range(maxLen)]

# Function to find the maximum sum subsequence
# such that no two elements are adjacent
def maxSum(arr, i, n):
    # Base case
    if (i >= n):
        return 0

    # To check if a state has
    # been solved
    if (v[i]):
        return dp[i]
    v[i] = 1

    # Required recurrence relation
    dp[i] = max(maxSum(arr, i + 1, n),
            arr[i] + maxSum(arr, i + 2, n))

    # Returning the value
    return dp[i]

# Driver code
if __name__ == '__main__':
    arr = [12, 9, 7, 33]

    n = len(arr)
    print(maxSum(arr, 0, n))

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to implement above approach
using System;

class GFG
{

static int maxLen = 10;

// variable to store states of dp
static int[] dp = new int[maxLen];

// variable to check if a given state
// has been solved
static bool[] v = new bool[maxLen];

// Function to find the maximum sum subsequence
// such that no two elements are adjacent
static int maxSum(int[] arr, int i, int n)
{
    // Base case
    if (i >= n)
        return 0;

    // To check if a state has
    // been solved
    if (v[i])
        return dp[i];
    v[i] = true;

    // Required recurrence relation
    dp[i] = Math.Max(maxSum(arr, i + 1, n),
                arr[i] + maxSum(arr, i + 2, n));

    // Returning the value
    return dp[i];
}

// Driver code
public static void Main()
{
    int[] arr = { 12, 9, 7, 33 };
    int n = arr.Length;
    Console.Write( maxSum(arr, 0, n));
}
}

// This code is contributed by ChitraNayal

PHP

<?php
// PHP program to implement above approach

$maxLen = 10;

// variable to store states of dp
$dp = array_fill(0, $GLOBALS['maxLen'], 0);

// variable to check if a given state
// has been solved
$v = array_fill(0, $GLOBALS['maxLen'], 0);

// Function to find the maximum sum subsequence
// such that no two elements are adjacent
function maxSum($arr, $i, $n)
{
    // Base case
    if ($i >= $n)
        return 0;

    // To check if a state has
    // been solved
    if ($GLOBALS['v'][$i])
        return $GLOBALS['dp'][$i];

    $GLOBALS['v'][$i] = 1;

    // Required recurrence relation
    $GLOBALS['dp'][$i] = max(maxSum($arr, $i + 1, $n),
                $arr[$i] + maxSum($arr, $i + 2, $n));

    // Returning the value
    return $GLOBALS['dp'][$i];
}

    // Driver code
    $arr = array( 12, 9, 7, 33 );

    $n = count($arr);

    echo maxSum($arr, 0, $n);

    // This code is contributed by AnkitRai01
?>

Javascript

<script>

// Javascript program to implement above approach
var maxLen = 10;

// variable to store states of dp
var dp = Array(maxLen);

// variable to check if a given state
// has been solved
var v = Array(maxLen);

// Function to find the maximum sum subsequence
// such that no two elements are adjacent
function maxSum(arr, i, n)
{
    // Base case
    if (i >= n)
        return 0;

    // To check if a state has
    // been solved
    if (v[i])
        return dp[i];
    v[i] = 1;

    // Required recurrence relation
    dp[i] = Math.max(maxSum(arr, i + 1, n),
                arr[i] + maxSum(arr, i + 2, n));

    // Returning the value
    return dp[i];
}

// Driver code
var arr = [12, 9, 7, 33 ];
var n = arr.length;
document.write( maxSum(arr, 0, n));

</script>

Output

45

Time Complexity : O(n)
Auxiliary Space : O(10)

Efficient approach : Using DP Tabulation method ( Iterative approach )

The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.

Steps to solve this problem :

• Create a DP of size N+1 to store the solution of the subproblems.
• Initialize the DP  with base cases
• Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
• Return the final solution stored in dp[n-1].

Implementation :

C++

#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum sum subsequence
// such that no two elements are adjacent
int maxSum(int arr[], int n)
{
    // Base case
    if (n == 0)
        return 0;
    if (n == 1)
        return arr[0];
    if (n == 2)
        return max(arr[0], arr[1]);

    // DP table to store solutions to subproblems
    int dp[n];

    // Initializing base cases
    dp[0] = arr[0];
    dp[1] = max(arr[0], arr[1]);

    // Filling up the DP table using recurrence relation
    for (int i = 2; i < n; i++) {
        dp[i] = max(dp[i - 1], arr[i] + dp[i - 2]);
    }

    // Returning the final answer
    return dp[n - 1];
}

// Driver code
int main()
{
    int arr[] = { 12, 9, 7, 33 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << maxSum(arr, n) << endl;
    return 0;
}

Java

import java.util.*;

public class Main {
    // Function to find the maximum sum subsequence
    // such that no two elements are adjacent
    public static int maxSum(int[] arr, int n)
    {
        // Base case
        if (n == 0)
            return 0;
        if (n == 1)
            return arr[0];
        if (n == 2)
            return Math.max(arr[0], arr[1]);

        // DP table to store solutions to subproblems
        int[] dp = new int[n];

        // Initializing base cases
        dp[0] = arr[0];
        dp[1] = Math.max(arr[0], arr[1]);

        // Filling up the DP table using recurrence relation
        for (int i = 2; i < n; i++) {
            dp[i] = Math.max(dp[i - 1], arr[i] + dp[i - 2]);
        }

        // Returning the final answer
        return dp[n - 1];
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 12, 9, 7, 33 };
        int n = arr.length;
        System.out.println(maxSum(arr, n));
    }
}

Python3

def maxSum(arr, n):
    # Base case
    if n == 0:
        return 0
    if n == 1:
        return arr[0]
    if n == 2:
        return max(arr[0], arr[1])

    # DP table to store solutions to subproblems
    dp = [0] * n

    # Initializing base cases
    dp[0] = arr[0]
    dp[1] = max(arr[0], arr[1])

    # Filling up the DP table using recurrence relation
    for i in range(2, n):
        dp[i] = max(dp[i - 1], arr[i] + dp[i - 2])

    # Returning the final answer
    return dp[n - 1]


arr = [12, 9, 7, 33]
n = len(arr)
print(maxSum(arr, n))

C#

using System;

class GFG {
    // Function to find the maximum sum subsequence
    // such that no two elements are adjacent
    static int maxSum(int[] arr, int n)
    {
        // Base case
        if (n == 0)
            return 0;
        if (n == 1)
            return arr[0];
        if (n == 2)
            return Math.Max(arr[0], arr[1]);

        // DP table to store solutions to subproblems
        int[] dp = new int[n];

        // Initializing base cases
        dp[0] = arr[0];
        dp[1] = Math.Max(arr[0], arr[1]);

        // Filling up the DP table using recurrence relation
        for (int i = 2; i < n; i++) {
            dp[i] = Math.Max(dp[i - 1], arr[i] + dp[i - 2]);
        }

        // Returning the final answer
        return dp[n - 1];
    }

    // Driver code
    static void Main()
    {
        int[] arr = { 12, 9, 7, 33 };
        int n = arr.Length;
        Console.WriteLine(maxSum(arr, n));
    }
}

Output

45

Time Complexity : O(n)
Auxiliary Space : O(n)

Another Approach(Using Greedy Strategy):

1. Initialize include as the first element and exclude as 0.
2. For each element in the array, update include as exclude + current element (considering the current element).
3. Update exclude as the maximum of include and exclude (excluding the current element).
4. Repeat steps 2 and 3 for all elements in the array.
5. Finally, return the maximum of include and exclude as the maximum sum.

Below is the implementation of the above approach:  

C++

#include <iostream>
#include <algorithm>
using namespace std;

int maxSumNoAdjacent(int arr[], int n) {
    if (n == 0)
        return 0;
    if (n == 1)
        return arr[0];

    int include = arr[0];
    int exclude = 0;

    for (int i = 1; i < n; i++) {
        int prevInclude = include;
        include = exclude + arr[i];
        exclude = max(prevInclude, exclude);
    }

    return max(include, exclude);
}

int main() {
    int arr[] = {12, 9, 7, 33};
    int n = sizeof(arr) / sizeof(arr[0]);

    int maxSum = maxSumNoAdjacent(arr, n);
    cout<< maxSum << endl;

    return 0;
}

Output

45

Time Complexity :O(n), where n is the size of the input array. This is because the code iterates through the array once in the for loop, performing a constant number of operations for each element.
Auxiliary Space :  O(1), meaning it uses a constant amount of extra space. The space usage remains constant throughout the execution, regardless of the size of the input array. This is because the code only uses a few variables (include, exclude, prevInclude) to keep track of the maximum sum, and their space requirements do not depend on the size of the array.