Given an array A of N integers and a positive integer K. Only three operations can be performed on this array:
- Replace an integer with the negative value of the integer,
- Add index number (1-based indexing) of the element to the element itself and
- 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 : Yes
Explanation
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 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 implementation of this approach
C++
/* 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); // Subtracting 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 Functionbool 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 Codeint main(){ int a[] = { 1, 2, 3, 4 }; int n = 4, k = 5; (wrapper(n, k, a) ? (cout << "Yes") : (cout << "No")); return 0;} |
Java
/* Java Program to find if Array can have a sumof K by applying three types of possible operations on it */class GFG { static int MAX = 100; // Check if it is possible to achieve a sum with // three operation allowed. static int check(int i, int sum, int n, int k, int a[], int dp[][]) { // If sum is negative. if (sum <= 0) { return 0; } // If going out of bound. if (i >= n) { // If sum is achieved. if (sum == k) { return 1; } return 0; } // 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); // Subtracting 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 static int wrapper(int n, int k, int a[]) { int sum = 0; for (int i = 0; i < n; i++) { sum += a[i]; } int[][] dp = new int[MAX][MAX]; for (int i = 0; i < MAX; i++) { for (int j = 0; j < MAX; j++) { dp[i][j] = -1; } } return check(0, sum, n, k, a, dp); } // Driver Code public static void main(String[] args) { int a[] = {1, 2, 3, 4}; int n = 4, k = 5; if (wrapper(n, k, a) == 1) { System.out.println("Yes"); } else { System.out.println("No"); } }}// This code is contributed by Princi Singh |
C#
// C# Program to find if Array can have a sum// of K by applying three types of possible using System;class GFG{ static int MAX = 100; // Check if it is possible to achieve a sum with // three operation allowed. static int check(int i, int sum, int n, int k, int []a, int [,]dp) { // If sum is negative. if (sum <= 0) { return 0; } // If going out of bound. if (i >= n) { // If sum is achieved. if (sum == k) { return 1; } return 0; } // 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); // Subtracting 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 static int wrapper(int n, int k, int []a) { int sum = 0; for (int i = 0; i < n; i++) { sum += a[i]; } int[,] dp = new int[MAX,MAX]; for (int i = 0; i < MAX; i++) { for (int j = 0; j < MAX; j++) { dp[i, j] = -1; } } return check(0, sum, n, k, a, dp); } // Driver Code static public void Main () { int []a = {1, 2, 3, 4}; int n = 4, k = 5; if (wrapper(n, k, a) == 1) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } }}// This code is contributed by ajit_0023 |
Python3
# Python program to find if Array can have sum# of K by applying three types of possible# operations on itMAX = 100# Check if it is possible to achieve a sum with # three operation alloweddef check(i, add, n, k, a, dp): # if sum id negative. if add <= 0: return False # If going out of bound. if i >= n: if add == k: return True return False # If the current state is not evaluated yet. if dp[i][add] != -1: return dp[i][add] # Replacing element with negative value of # the element. dp[i][add] = (check(i+1, add-2*a[i], n, k, a, dp) or check(i+1, add, n, k, a, dp)) # Subtracting index number from the element. dp[i][add] = (check(i+1, add - (i+1), n, k, a, dp) or dp[i][add]) # Adding index number to the element. dp[i][add] = (check(i+1, add+i+1, n, k, a, dp) or dp[i][add]) return dp[i][add]# Wrapper Functiondef wrapper(n, k, a): add = 0 for i in range(n): add += a[i] dp = [-1]*MAX for i in range(MAX): dp[i] = [-1]*MAX return check(0, add, n, k, a, dp)# Driver Codeif __name__ == "__main__": a = [1,2,3,4] n = 4 k = 5 print("Yes") if wrapper(n, k, a) else print("No")# This code is contributed by# sanjeev2552 |
Javascript
<script>/* Javascript Program to find if Array can have a sum of K by applying three types of possible operations on it */var MAX = 100;// Check if it is possible // to achieve a sum with// three operation allowed.function check(i, sum, n, k, a, dp){ // 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); // Subtracting 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 Functionfunction wrapper(n, k, a){ var sum = 0; for (var i = 0; i < n; i++) sum += a[i]; var dp = Array.from(Array(MAX), ()=> Array(MAX).fill(-1)); return check(0, sum, n, k, a, dp);}// Driver Codevar a = [ 1, 2, 3, 4 ];var n = 4, k = 5;(wrapper(n, k, a) ? (document.write( "Yes")) : (document.write( "No")));</script> |
Output
Yes
Complexity Analysis:
- Time Complexity: O(N*MAX), where N is the number of elements in array and MAX is the defined constant .
- Auxiliary Space: O(MAX*MAX) , where MAX is constant define 100.
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