Check if it is possible to perform the given Grid Division

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 conditions
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
int 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 approach
import 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 conditions
def 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 code
if __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 approach
using 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 conditions
function 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 code
var 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)