Maximize Sum possible from an Array by the given moves - GeeksforGeeks

07 May, 2021
Given three integers N, M and K and an array a[] consisting of N integers, where M and K denotes the total number of possible moves and the number of possible moves(shift by an index) on the left of the current element in an array respectively, the task is to maximize the sum possible by traversing the array utilizing all the available moves.
Examples:
Input: N = 5, M = 4, K = 0, a[] = {1, 5, 4, 3, 2} Output: 15 Explanation: Since no moves towards left is possible, therefore, the only possible path is a[0] -> a[1] -> a[2] -> a[3] -> a[4]. Therefore, the sum calculated is 15.Input: N = 5, M = 4, K = 1, a[]= {1, 5, 4, 3, 2} Output: 19 Explanation: The maximum sum can be obtained in the path a[0] -> a[1] -> a[2] -> a[1] -> a[2] Therefore, the maximum possible sum = 19
Approach: The above problem can be solved using Dynamic Programming. Follow the steps below to solve the problem:
Initialize a dp[][] matrix such that dp[i][j] stores the maximum sum possible up to ith index by using j left moves.
It can be observed that left move is possible only if i ≥ 1 and k > 0 and a right move is possible if i < n – 1.
Check the conditions and update the maximum of the sums possible from the above two moves and store in dp[i][j].
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;const int k = 1;const int m = 4; // Function to find the maximum sum possible// by given moves from the arrayint maxValue(int a[], int n, int pos,             int moves, int left,             int dp[][k + 1]){    // Checking for boundary    if (moves == 0 || (pos > n - 1 || pos < 0))        return 0;     // If previously computed subproblem occurs    if (dp[pos][left] != -1)        return dp[pos][left];     int value = 0;     // If element can be moved left    if (left > 0 && pos >= 1)         // Calculate maximum possible sum        // by moving left from current index        value = max(value, a[pos] +                    maxValue(a, n, pos - 1, moves - 1,                             left - 1, dp));     // If element can be moved right    if (pos <= n - 1)         // Calculate maximum possible sum        // by moving right from current index        // and update the maximum sum        value = max(value, a[pos] +                    maxValue(a, n, pos + 1,                             moves - 1, left, dp));     // Store the maximum sum    return dp[pos][left] = value;} // Driver Codeint main(){    int n = 5;    int a[] = { 1, 5, 4, 3, 2 };     int dp[n + 1][k + 1];    memset(dp, -1, sizeof(dp));    cout << (a[0] + maxValue(a, n, 1, m, k, dp))         << endl;} // This code is contributed by sapnasingh4991
// Java Program to implement// the above approachimport java.io.*;import java.util.*; public class GFG {     // Function to find the maximum sum possible    // by given moves from the array    public static int maxValue(int a[], int n, int pos,                               int moves, int left,                               int dp[][])    {        // Checking for boundary        if (moves == 0 || (pos > n - 1 || pos < 0))            return 0;         // If previously computed subproblem occurs        if (dp[pos][left] != -1)            return dp[pos][left];         int value = 0;         // If element can be moved left        if (left > 0 && pos >= 1)             // Calculate maximum possible sum            // by moving left from current index            value = Math.max(                value, a[pos] + maxValue(a, n, pos - 1,                                         moves - 1, left - 1, dp));         // If element can be moved right        if (pos <= n - 1)             // Calculate maximum possible sum            // by moving right from current index            // and update the maximum sum            value = Math.max(                value, a[pos]                           + maxValue(a, n, pos + 1,                                      moves - 1, left, dp));         // Store the maximum sum        return dp[pos][left] = value;    }     // Driver Code    public static void main(String args[])    {        int n = 5;        int a[] = { 1, 5, 4, 3, 2 };        int k = 1;        int m = 4;         int dp[][] = new int[n + 1][k + 1];        for (int i[] : dp)            Arrays.fill(i, -1);         System.out.println(            (a[0] + maxValue(a, n, 1, m, k, dp)));    }}
# Python3 program to implement# the above approach # Function to find the maximum sum possible# by given moves from the arraydef maxValue(a, n, pos, moves, left, dp):     # Checking for boundary    if(moves == 0 or (pos > n - 1 or pos < 0)):        return 0     # If previously computed subproblem occurs    if(dp[pos][left] != -1):        return dp[pos][left]     value = 0     # If element can be moved left    if(left > 0 and pos >= 1):         # Calculate maximum possible sum        # by moving left from current index        value = max(value, a[pos] +                    maxValue(a, n, pos - 1,                                 moves - 1,                                  left - 1, dp))     # If element can be moved right    if(pos <= n - 1):         # Calculate maximum possible sum        # by moving right from current index        # and update the maximum sum        value = max(value, a[pos] +                    maxValue(a, n, pos + 1,                                 moves - 1,                                 left, dp))                                      # Store the maximum sum    dp[pos][left] = value     return dp[pos][left] # Driver Coden = 5a = [ 1, 5, 4, 3, 2 ]k = 1m = 4 dp = [[-1 for x in range(k + 1)]          for y in range(n + 1)] # Function callprint(a[0] + maxValue(a, n, 1, m, k, dp)) # This code is contributed by Shivam Singh
// C# Program to implement// the above approachusing System;class GFG{   // Function to find the maximum sum possible  // by given moves from the array  public static int maxValue(int []a, int n, int pos,                             int moves, int left,                             int [,]dp)  {    // Checking for boundary    if (moves == 0 || (pos > n - 1 || pos < 0))      return 0;     // If previously computed subproblem occurs    if (dp[pos, left] != -1)      return dp[pos, left];     int value = 0;     // If element can be moved left    if (left > 0 && pos >= 1)       // Calculate maximum possible sum      // by moving left from current index      value = Math.Max(      value, a[pos] + maxValue(a, n, pos - 1,                               moves - 1,                               left - 1, dp));     // If element can be moved right    if (pos <= n - 1)       // Calculate maximum possible sum      // by moving right from current index      // and update the maximum sum      value = Math.Max(      value, a[pos] + maxValue(a, n, pos + 1,                                 moves - 1,                               left, dp));     // Store the maximum sum    return dp[pos, left] = value;  }   // Driver Code  public static void Main(String []args)  {    int n = 5;    int []a = { 1, 5, 4, 3, 2 };    int k = 1;    int m = 4;     int [,]dp = new int[n + 1, k + 1];    for(int i = 0; i <= n; i++)      for(int j =0; j <= k; j++)        dp[i, j] = -1;     Console.WriteLine(     (a[0] + maxValue(a, n, 1, m, k, dp)));  }} // This code is contributed by Rajput-Ji
<script>// JavaScript program for the// above approach     // Function to find the maximum sum possible    // by given moves from the array    function maxValue(a, n, pos, moves, left,                               dp)    {        // Checking for boundary        if (moves == 0 || (pos > n - 1 || pos < 0))            return 0;          // If previously computed subproblem occurs        if (dp[pos][left] != -1)            return dp[pos][left];          let value = 0;          // If element can be moved left        if (left > 0 && pos >= 1)              // Calculate maximum possible sum            // by moving left from current index            value = Math.max(                value, a[pos] + maxValue(a, n, pos - 1,                                         moves - 1, left - 1, dp));          // If element can be moved right        if (pos <= n - 1)              // Calculate maximum possible sum            // by moving right from current index            // and update the maximum sum            value = Math.max(                value, a[pos]                           + maxValue(a, n, pos + 1,                                      moves - 1, left, dp));          // Store the maximum sum        return dp[pos][left] = value;    } // Driver Code          let n = 5;        let a = [ 1, 5, 4, 3, 2 ];        let k = 1;        let m = 4;          let dp = new Array(n + 1);        // Loop to create 2D array using 1D array        for (var i = 0; i < dp.length; i++) {            dp[i] = new Array(2);        }                 for (var i = 0; i < dp.length; i++) {            for (var j = 0; j < dp.length; j++) {            dp[i][j] = -1;        }        }          document.write(            (a[0] + maxValue(a, n, 1, m, k, dp))); </script>
19
 
Time Complexity: O(N * K) Auxiliary Space: O(N * K) 
SHIVAMSINGH67
Rajput-Ji
sapnasingh4991
susmitakundugoaldanga
array-traversal-question
Arrays
Dynamic Programming
Mathematical
Recursion
Arrays
Dynamic Programming
Mathematical
Recursion
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