Check if array sum can be made K by three operations on it

Given an array A of N integers and a positive integer K. Only three operations can be performed on this array:
1) Replace an integer with the negative value of the integer,
2) Add index number (1-based indexing) of the element to the element itself and
3) Subtract index number of the element from the element itself.
The task is to check if the given array can be transformed, using any of above three allowed operations performed only once on each element. such that the sum of the array becomes K.


Input : N = 3, K = 2
        A[] = { 1, 1, 1 }
Output : Yes
Replace index 0 element with -1. It will sum of array equal to k = 2.

Input : N = 4, K = 5
        A[] = { 1, 2, 3, 4 }
Output : Yes

Pre Requisites Dynamic Programming

Approach The idea is to use dynamic programming to solve the problem.
Declare a 2D Boolean array, dp[][], where dp[i][j] states if there is any way to obtain the sum of the array equal to j using some operations on the firsti elements of the array.
dp[i][j] will be true if the sum is possible else it will be False.
Also, it is possible that the intermediate sum of the array is negative, in that case do not perform any operation and ignore it thus letting the sum be always positive because k is always positive.

For calculating dp[i][j] we need the values of all states that can make a sum j if we apply an operation on a[i] and add it to the sum.

Below is the C++ implementation of this approach





/* C++ Program to find if Array can have a sum
   of K by applying three types of possible 
   operations on it */
#include <bits/stdc++.h>
using namespace std;
#define MAX 100
// Check if it is possible to achieve a sum with
// three operation allowed.
int check(int i, int sum, int n, int k, int a[],
                               int dp[MAX][MAX])
    // If sum is negative.
    if (sum <= 0)
        return false;
    // If going out of bound.
    if (i >= n) {
        // If sum is achieved.
        if (sum == k)
            return true;
        return false;
    // If the current state is not evaluated yet.
    if (dp[i][sum] != -1)
        return dp[i][sum];
    // Replacing element with negative value of
    // the element.
    dp[i][sum] = check(i + 1, sum - 2 * a[i], n, 
          k, a, dp) || check(i + 1, sum, n, k, a, dp);
    // Substracting index number from the element.
    dp[i][sum] = check(i + 1, sum - (i + 1), n, 
         k, a, dp) || dp[i][sum];
    // Adding index number to the element.
    dp[i][sum] = check(i + 1, sum + i + 1, n,
                      k, a, dp) || dp[i][sum];
    return dp[i][sum];
// Wrapper Function
bool wrapper(int n, int k, int a[])
    int sum = 0;
    for (int i = 0; i < n; i++) 
        sum += a[i];    
    int dp[MAX][MAX];
    memset(dp, -1, sizeof(dp));
    return check(0, sum, n, k, a, dp);
// Driver Code
int main()
    int a[] = { 1, 2, 3, 4 };
    int n = 4, k = 5;
    (wrapper(n, k, a) ? (cout << "Yes") : (cout << "No"));
    return 0;




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.