# 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.

Examples:

Input : N = 3, K = 2 A[] = { 1, 1, 1 } Output : YesExplanationReplace 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 first**i** 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; ` `} ` |

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**Output:**

Yes

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