Given two integers N and M which represent the size of a grid. Also given an integer array arr[] of size P which represent that the given grid is divided into P parts each consisting of arr[i] cells from the grid. The task is to check whether it is possible to divide the grid in the given manner or not.
Examples:
Input: arr[] = {6, 3, 2, 1}, N = 3, M = 4
Output: Yes
Input: arr[] = {4, 2, 2}, N = 3, M = 2
Output: No
Approach: In order for the division to be possible, the sum of the cells of all the parts must be equal to the total number of cells in the given grid i.e. the sum of all the array elements must be equal to N * M.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns true if it is possible to// divide the grid satisfying the given conditionsbool isPossible(int arr[], int p, int n, int m){ // To store the sum of all the // cells of the given parts int sum = 0; for (int i = 0; i < p; i++) sum += arr[i]; // If the sum is equal to the total number // of cells in the given grid if (sum == (n * m)) return true; return false;}// Driver codeint main(){ int n = 3, m = 4; int arr[] = { 6, 3, 2, 1 }; int p = sizeof(arr) / sizeof(arr[0]); if (isPossible(arr, p, n, m)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG {// Function that returns true if it is possible to // divide the grid satisfying the given conditions static boolean isPossible(int arr[], int p, int n, int m) { // To store the sum of all the // cells of the given parts int sum = 0; for (int i = 0; i < p; i++) sum += arr[i]; // If the sum is equal to the total number // of cells in the given grid if (sum == (n * m)) return true; return false; } // Driver code public static void main(String[] args){ int n = 3, m = 4; int arr[] = { 6, 3, 2, 1 }; int p = arr.length; if (isPossible(arr, p, n, m)) System.out.println("Yes"); else System.out.println("No");}} // This code is contributed by Princi Singh |
Python3
# Python3 implementation of the approach# Function that returns true if # it is possible to divide the grid # satisfying the given conditionsdef isPossible(arr, p, n, m): # To store the sum of all the # cells of the given parts sum = 0; for i in range(p): sum += arr[i]; # If the sum is equal to the total number # of cells in the given grid if (sum == (n * m)): return True; return False;# Driver codeif __name__ == '__main__': n = 3; m = 4; arr = [6, 3, 2, 1]; p = len(arr); if (isPossible(arr, p, n, m)): print("Yes"); else: print("No");# This code is contributed by Rajput-Ji |
C#
// C# implementation of the approachusing System; class GFG {// Function that returns true if it is possible to // divide the grid satisfying the given conditions static bool isPossible(int []arr, int p, int n, int m) { // To store the sum of all the // cells of the given parts int sum = 0; for (int i = 0; i < p; i++) sum += arr[i]; // If the sum is equal to the total number // of cells in the given grid if (sum == (n * m)) return true; return false; } // Driver code public static void Main(String[] args){ int n = 3, m = 4; int []arr = { 6, 3, 2, 1 }; int p = arr.Length; if (isPossible(arr, p, n, m)) Console.WriteLine("Yes"); else Console.WriteLine("No");}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the approach// Function that returns true if it is possible to// divide the grid satisfying the given conditionsfunction isPossible(arr, p, n, m){ // To store the sum of all the // cells of the given parts var sum = 0; for (var i = 0; i < p; i++) sum += arr[i]; // If the sum is equal to the total number // of cells in the given grid if (sum == (n * m)) return true; return false;}// Driver codevar n = 3, m = 4;var arr = [ 6, 3, 2, 1 ];var p = arr.length;if (isPossible(arr, p, n, m)) document.write( "Yes");else document.write( "No");</script> |
Output:
Yes
Time Complexity: O(p)
Auxiliary Space: O(1)
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