Find maximum score by performing given operations

Given two arrays a[] and b[][2] with both having size N. N operations have to be performed, and initially counter will be 0. Additionally, there is a M bonus score for a hitting streak of b[i][0] and the bonus score received will be b[i][1], the task is to find the maximum score by performing either of the below operations optimally for all i from 1 to N.

• Operation 1: increases the counter’s value by 1 and receives a[i] score
• Operation 2: resets the counter’s value to 0, without receiving any score.

Examples:

Input: a[] = {2, 7, 1, 8, 2, 8}, b[][2] = {{2, 10}, {3, 1}, {5, 5}}
Output: 48
Explanation: 

• For i = 1, Change the counter’s value from 0 to 1 and receive a 2 score.
• For i = 2, Change the counter’s value from 1 to 2 and receive a 7 score and get 10 scores as a streak bonus as well.
• For i = 3, Change the counter’s value from 2 to 0.
• For i = 4, Change the counter’s value from 0 to 1 and receive 8 scores.
• For i = 5, Change the counter’s value from 1 to 2 and receive 2 scores and get a 10 score as a streak bonus.
• For i = 6, Change the counter’s value from 2 to 3 and receive 8 scores. and get 1 score as a streak bonus.

Total score : 2 + 7 + 10 + 8 + 2 +10 + 8 +1 = 48.

Input:  a[] = {1000000000, 1000000000, 1000000000}, b[][2] = {{1, 1000000000}, {3, 1000000000}}
Output: 5000000000

Naive approach: The problem can be solved based on The basis of the following idea:

Basic way to solve this problem is to generate all 2^N combinations by recursive brute force.

Time Complexity: O(2^N)
Auxiliary Space: O(1)

Efficient Approach:  The above approach can be optimized based on the following idea:

Dynamic programming along with hashing can be used to solve to this problem.

• dp[i][j] represents maximum score till i’th operation with current counter j.
• recurrence relation: dp[i][j] = max(dp[i  – 1][j + 1] + a[i] + hash[j + 1], dp[i – 1][0])

It can be observed that there are N * N states but the recursive function is called exponential times. That means that some states are called repeatedly. So the idea is to store the value of states. This can be done using recursive structure intact and just store the value in a HashMap and whenever the function is called, return the value store without computing .

Follow the steps below to solve the problem:

• Creating a hashmap array for mapping values of b[i][0] to b[i][1] named hash[].
• Create a recursive function that takes two parameters i representing i’th operation and j representing counter at i’th operation.
• Call the recursive function for both increasing the current counter by 1 and resetting the counter.
• Check the base case if all N operations are over then return 0.
• Create a 2d array of dp[5001][5001] initially filled with -1.
• If the answer for a particular state is computed then save it in dp[i][j].
• If the answer for a particular state is already computed then just return dp[i][j].

Below is the implementation of the above approach.

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// To avoid integer overflow
#define int long long

// dp table initialized with - 1
int dp[5001][5001];

// Recursive function to count maximum score
// by performing following operations
int recur(int i, int j, int a[], int hash[], int N)
{

    // Base case
    if (i == N) {
        return 0;
    }

    // If answer for current state is already
    // calculated then just return dp[i][j]
    if (dp[i][j] != -1)
        return dp[i][j];

    // Calling recursive function for
    // performing operation 1
    int ans = recur(i + 1, j + 1, a, hash, N) + a[i]
              + hash[j + 1];

    // Calling recursive function for
    // performing operation 2
    ans = max(ans, recur(i + 1, 0, a, hash, N) + 0LL);

    // Save and return dp value
    return dp[i][j] = ans;
}

// Function to count maximum score
// by performing following operations
void findMaximumScore(int a[], int b[][2], int N, int M)
{

    // Filling dp table with -1
    memset(dp, -1, sizeof(dp));

    // Creating Hash table
    int hash[N + 1] = { 0 };

    // Mapping hash table with values
    for (int i = 0; i < M; i++) {
        hash[b[i][0]] += b[i][1];
    }

    cout << recur(0, 0, a, hash, N) << endl;
}

// Driver Code
int32_t main()
{

    // Input 1
    int a[] = { 2, 7, 1, 8, 2, 8 },
        b[][2] = { { 2, 10 }, { 3, 1 }, { 5, 5 } };
    int N = sizeof(a) / sizeof(a[0]);
    int M = sizeof(b) / sizeof(b[0]);

    // Function Call
    findMaximumScore(a, b, N, M);

    // Input 2
    int a1[] = { 1000000000, 1000000000, 1000000000 },
        b1[][2] = { { 1, 1000000000 }, { 3, 1000000000 } };
    int N1 = sizeof(a1) / sizeof(a1[0]);
    int M1 = sizeof(b1) / sizeof(b1[0]);

    // Function Call
    findMaximumScore(a1, b1, N1, M1);
    return 0;
}

Java

// Java code for the above approach
import java.util.Arrays;

class Main
{

    // To avoid integer overflow
    static long MAX = 1000000007;

    // dp table initialized with - 1
    static long[][] dp = new long[5001][5001];

    // Recursive function to count maximum score
    // by performing following operations
    static long recur(int i, int j, int[] a, int[] hash, int N)
    {

        // Base case
        if (i == N) {
            return 0;
        }

        // If answer for current state is already
        // calculated then just return dp[i][j]
        if (dp[i][j] != -1) {
            return dp[i][j];
        }

        // Calling recursive function for
        // performing operation 1
        long ans = recur(i + 1, j + 1, a, hash, N) + a[i]
                + hash[j + 1];

        // Calling recursive function for
        // performing operation 2
        ans = Math.max(ans, recur(i + 1, 0, a, hash, N) + 0);

        // Save and return dp value
        return dp[i][j] = ans;
    }

    // Function to count maximum score
    // by performing following operations
    static void findMaximumScore(int[] a, int[][] b, int N, int M)
    {

        // Filling dp table with -1
        for (long[] row : dp) {
            Arrays.fill(row, -1);
        }

        // Creating Hash table
        int[] hash = new int[N + 1];

        // Mapping hash table with values
        for (int i = 0; i < M; i++) {
            hash[b[i][0]] += b[i][1];
        }

        System.out.println(recur(0, 0, a, hash, N));
    }

    // Driver Code
    public static void main(String[] args) {
        // Input 1
        int[] a = { 2, 7, 1, 8, 2, 8 };
        int[][] b = { { 2, 10 }, { 3, 1 }, { 5, 5 } };
        int N = a.length;
        int M = b.length;

        // Function Call
        findMaximumScore(a, b, N, M);

        // Input 2
        int[] a1 = { 1000000000, 1000000000, 1000000000 };
        int[][] b1 = { { 1, 1000000000 }, { 3, 1000000000 } };
        int N1 = a1.length;
        int M1 = b1.length;

        // Function Call
        findMaximumScore(a1, b1, N1, M1);
    }
}

// This code is contributed by Potta Lokesh

Python3

#Python code to implement the approach

# dp table initialized with - 1
dp=[[-1 for i in range(5001)] for j in range(5001)]

# Recursive function to count maximum score
# by performing following operations
def recur(i,j,a,hash,N):
    # Base case
    if(i==N):
        return 0

    # If answer for current state is already
    # calculated then just return dp[i][j]
    if(dp[i][j]!=-1):
        return dp[i][j]

    # Calling recursive function for
    # performing operation 1
    ans=recur(i+1,j+1,a,hash,N)+a[i]+hash[j+1]

    # Calling recursive function for
    # performing operation 2
    ans=max(ans,recur(i+1,0,a,hash,N))

    # Save and return dp value
    dp[i][j]=ans
    return dp[i][j]

# Function to count maximum score
# by performing following operations
def findMaximumScore(a,b,N,M):

    # Filling dp table with -1
    for i in range(len(dp)):
        for j in range(len(dp[0])):
            dp[i][j]=-1

    # Creating Hash table
    hash=[0]*(N+1)

    # Mapping hash table with values
    for i in range(M):
        hash[b[i][0]]+=b[i][1]

    print(recur(0,0,a,hash,N))

# Driver Code

# Input 1
a=[2, 7, 1, 8, 2, 8]
b=[[2,10],[3,1],[5,5]]
N=len(a)
M=len(b)

# Function call
findMaximumScore(a,b,N,M)

# Input 2
a1=[1000000000, 1000000000, 1000000000]
b1=[[1,1000000000],[3,1000000000]]
N1=len(a1)
M1=len(b1)

# Function call
findMaximumScore(a1,b1,N1,M1)

# This code is contributed by Pushpesh Raj.

C#

using System;
using System.Linq;

class GFG
{

  // To avoid integer overflow
  static long MAX = 1000000007;

  // dp table initialized with - 1
  static long[, ] dp = new long[5001, 5001];

  // Recursive function to count maximum score
  // by performing following operations
  static long recur(int i, int j, int[] a, int[] hash,
                    int N)
  {

    // Base case
    if (i == N) {
      return 0;
    }

    // If answer for current state is already
    // calculated then just return dp[i][j]
    if (dp[i, j] != -1) {
      return dp[i, j];
    }

    // Calling recursive function for
    // performing operation 1
    long ans = recur(i + 1, j + 1, a, hash, N) + a[i]
      + hash[j + 1];

    // Calling recursive function for
    // performing operation 2
    ans = Math.Max(ans,
                   recur(i + 1, 0, a, hash, N) + 0);

    // Save and return dp value
    return dp[i, j] = ans;
  }

  // Function to count maximum score
  // by performing following operations
  static void findMaximumScore(int[] a, int[, ] b, int N,
                               int M)
  {

    // Filling dp table with -1
    for (int i = 0; i < 5001; i++) {
      for (int j = 0; j < 5001; j++) {
        dp[i, j] = -1;
      }
    }

    // Creating Hash table
    int[] hash = new int[N + 1];

    // Mapping hash table with values
    for (int i = 0; i < M; i++) {
      hash[b[i, 0]] += b[i, 1];
    }

    Console.WriteLine(recur(0, 0, a, hash, N));
  }

  // Driver Code
  public static void Main(string[] args)
  {
    // Input 1
    int[] a = { 2, 7, 1, 8, 2, 8 };
    int[, ] b = { { 2, 10 }, { 3, 1 }, { 5, 5 } };
    int N = a.Length;
    int M = b.GetLength(0);

    // Function Call
    findMaximumScore(a, b, N, M);

    // Input 2
    int[] a1 = { 1000000000, 1000000000, 1000000000 };
    int[, ] b1
      = { { 1, 1000000000 }, { 3, 1000000000 } };
    int N1 = a1.Length;
    int M1 = b1.GetLength(0);

    // Function Call
    findMaximumScore(a1, b1, N1, M1);
  }
}

// This code is contributed by ik_9

Javascript

// Javascript code to implement the approach

// dp table initialized with - 1
let dp=new Array(5001);
for(let i=0; i<5001; i++)
    dp[i]=new Array(5001);

// Recursive function to count maximum score
// by performing following operations
function recur(i, j, a, hash, N)
{

    // Base case
    if (i == N) {
        return 0;
    }

    // If answer for current state is already
    // calculated then just return dp[i][j]
    if (dp[i][j] != -1)
        return dp[i][j];

    // Calling recursive function for
    // performing operation 1
    let ans = recur(i + 1, j + 1, a, hash, N) + a[i]
              + hash[j + 1];

    // Calling recursive function for
    // performing operation 2
    ans = Math.max(ans, recur(i + 1, 0, a, hash, N) + 0);

    // Save and return dp value
    return dp[i][j] = ans;
}

// Function to count maximum score
// by performing following operations
function findMaximumScore(a, b, N, M)
{

    // Filling dp table with -1
    for(let i=0; i<5001; i++)
        for(let j=0; j<5001; j++)
            dp[i][j]=-1;

    // Creating Hash table
    let hash = new Array(N+1).fill(0);

    // Mapping hash table with values
    for (let i = 0; i < M; i++) {
        hash[b[i][0]] += b[i][1];
    }

    document.write(recur(0, 0, a, hash, N))
}

// Driver Code
    // Input 1
    let a = [ 2, 7, 1, 8, 2, 8 ],
        b = [[ 2, 10 ], [ 3, 1 ], [ 5, 5 ]];
    let N = a.length;
    let M = b.length;

    // Function Call
    findMaximumScore(a, b, N, M);

    document.write("<br>");

    // Input 2
    let a1 = [ 1000000000, 1000000000, 1000000000 ],
        b1 = [ [ 1, 1000000000 ], [ 3, 1000000000 ] ];
    let N1 = a1.length;
    let M1 = b1.length;

    // Function Call
    findMaximumScore(a1, b1, N1, M1);

Output

48
5000000000

Time Complexity: O(N²)
Auxiliary Space: O(N²)

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.

Implementation :

C++

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

// To avoid integer overflow
#define int long long

// dp table
int dp[5001][5001];

// Function to count maximum score
// by performing following operations
void findMaximumScore(int a[], int b[][2], int N, int M)
{
    // Creating Hash table
    int hash[N + 1] = { 0 };

    // Mapping hash table with values
    for (int i = 0; i < M; i++) {
        hash[b[i][0]] += b[i][1];
    }

    // Base case initialization
    for (int i = 0; i <= N; i++) {
        dp[N][i] = 0;
    }

    // Tabulation
    for (int i = N - 1; i >= 0; i--) {
        for (int j = 0; j <= i; j++) {
            // Operation 1
            int ans = dp[i + 1][j + 1] + a[i] + hash[j + 1];
            // Operation 2
            ans = max(ans, dp[i + 1][0]);
            dp[i][j] = ans;
        }
    }

    cout << dp[0][0] << endl;
}

// Driver Code
int32_t main()
{

    // Input 1
    int a[] = { 2, 7, 1, 8, 2, 8 },
        b[][2] = { { 2, 10 }, { 3, 1 }, { 5, 5 } };
    int N = sizeof(a) / sizeof(a[0]);
    int M = sizeof(b) / sizeof(b[0]);

    // Function Call
    findMaximumScore(a, b, N, M);

    // Input 2
    int a1[] = { 1000000000, 1000000000, 1000000000 },
        b1[][2] = { { 1, 1000000000 }, { 3, 1000000000 } };
    int N1 = sizeof(a1) / sizeof(a1[0]);
    int M1 = sizeof(b1) / sizeof(b1[0]);

    // Function Call
    findMaximumScore(a1, b1, N1, M1);
    return 0;
}

Output

48
5000000000

Time Complexity: O(N^2)
Auxiliary Space: O(N^2)

Efficient approach : Space optimization

In previous approach the dp[i][j] is depend upon the current and previous row of 2D matrix. So to optimize space we use two vectors curr and next that keep track of current and next row of DP.

Implementation :

C++

// C++ code for the above approach

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

// To avoid integer overflow
#define int long long

// Function to count maximum score
// by performing following operations
void findMaximumScore(int a[], int b[][2], int N, int M)
{
    // Creating Hash table
    int hash[N + 1] = { 0 };

    // to store the values of next row of DP
    vector<int> next(N + 1, 0);

    // Mapping hash table with values
    for (int i = 0; i < M; i++) {
        hash[b[i][0]] += b[i][1];
    }

    // iterating over DP to get the current
    // value from previous computations
    for (int i = N - 1; i >= 0; i--) {

        vector<int> curr(N + 1);
        for (int j = 0; j <= i; j++) {
            // Operation 1
            int ans = next[j + 1] + a[i] + hash[j + 1];
            // Operation 2
            ans = max(ans, next[0]);
            curr[j] = ans;
        }

        // assigning values to iterate further
        next = curr;
    }

    // print final answer
    cout << next[0] << endl;
}

// Driver Code
int32_t main()
{

    // Input 1
    int a[] = { 2, 7, 1, 8, 2, 8 },
        b[][2] = { { 2, 10 }, { 3, 1 }, { 5, 5 } };
    int N = sizeof(a) / sizeof(a[0]);
    int M = sizeof(b) / sizeof(b[0]);

    // Function Call
    findMaximumScore(a, b, N, M);

    // Input 2
    int a1[] = { 1000000000, 1000000000, 1000000000 },
        b1[][2] = { { 1, 1000000000 }, { 3, 1000000000 } };
    int N1 = sizeof(a1) / sizeof(a1[0]);
    int M1 = sizeof(b1) / sizeof(b1[0]);

    // Function Call
    findMaximumScore(a1, b1, N1, M1);
    return 0;
}

Output

48
5000000000

Time Complexity: O(N^2)
Auxiliary Space: O(N)