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Area of plot remaining at the end

Last Updated : 12 Oct, 2022
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Given two integers N and M that represent the length and breadth of a rectangular plot and another integer K that represent the number of persons. Each person will divide the plot into two parts such that it will be the largest possible square from the plot and will leave the second part for others, it continues until the plot is over or every person gets the plot. Now, The task is to determine the area of the plot left in the end.
Examples: 
 

Input: N = 5, M = 3, K = 2 
Output:
1st person divides the 5×3 plot into 2 parts i.e 3×3 and 2×3 
and will get the plot having dimension 3×3. The other person divides the 2×3 plot again into 
two parts i.e 2×2 and 1×2 and will get the plot having dimension 2×2. Now, the remaining 
part of plot is having dimension 1×2 and area as 2 units.
Input: N = 4, M = 8, K = 4 
Output:
 

 

Approach: 
 

  1. If Length is greater than breadth then subtract breadth from length.
  2. If breadth is greater than length then subtract length from breadth.
  3. Repeat the above steps for all the persons while there area of the remaining plot is greater than 0.
  4. Print the area of the remaining plot in the end i.e. length * breadth.

Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the area
// of the remaining plot
int remainingArea(int N, int M, int K)
{
 
    // Continue while plot has positive area
    // and there are persons left
    while (K-- && N && M) {
 
        // If length > breadth
        // then subtract breadth from length
        if (N > M)
            N = N - M;
 
        // Else subtract length from breadth
        else
            M = M - N;
    }
 
    if (N > 0 && M > 0)
        return N * M;
    else
        return 0;
}
 
// Driver code
int main()
{
    int N = 5, M = 3, K = 2;
 
    cout << remainingArea(N, M, K);
 
    return 0;
}


Java




// Java implementation of the approach
class GFG {
 
    // Function to return the area
    // of the remaining plot
    static int remainingArea(int N, int M, int K)
    {
 
        // Continue while plot has positive area
        // and there are persons left
        while (K-- > 0 && N > 0 && M > 0) {
 
            // If length > breadth
            // then subtract breadth from length
            if (N > M)
                N = N - M;
 
            // Else subtract length from breadth
            else
                M = M - N;
        }
 
        if (N > 0 && M > 0)
            return N * M;
        else
            return 0;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int N = 5, M = 3, K = 2;
 
        System.out.println(remainingArea(N, M, K));
    }
}
 
/* This code contributed by PrinciRaj1992 */


Python3




# Python3 implementation of the approach
 
# Function to return the area
# of the remaining plot
def remainingArea(N, M, K):
 
    # Continue while plot has positive area
    # and there are persons left
    while (K > 0 and N > 0 and M > 0):
 
        # If length > breadth
        # then subtract breadth from length
        if (N > M):
            N = N - M;
 
        # Else subtract length from breadth
        else:
            M = M - N;
        K = K - 1;
    if (N > 0 and M > 0):
        return N * M;
    else:
        return 0;
 
# Driver code
N = 5;
M = 3;
K = 2;
 
print(remainingArea(N, M, K));
 
# This code contributed by Rajput-Ji


C#




// C# implementation of the approach
using System;
 
class GFG {
 
    // Function to return the area
    // of the remaining plot
    static int remainingArea(int N, int M, int K)
    {
 
        // Continue while plot has positive area
        // and there are persons left
        while (K-- > 0 && N > 0 && M > 0) {
 
            // If length > breadth
            // then subtract breadth from length
            if (N > M)
                N = N - M;
 
            // Else subtract length from breadth
            else
                M = M - N;
        }
 
        if (N > 0 && M > 0)
            return N * M;
        else
            return 0;
    }
 
    // Driver code
    static public void Main()
    {
        int N = 5, M = 3, K = 2;
 
        Console.WriteLine(remainingArea(N, M, K));
    }
}
 
/* This code contributed by ajit */


PHP




<?php
// PHP implementation of the approach
 
// Function to return the area
// of the remaining plot
function remainingArea($N, $M, $K)
{
 
    // Continue while plot has positive area
    // and there are persons left
    while ($K-- && $N && $M)
    {
 
        // If length > breadth
        // then subtract breadth from length
        if ($N > $M)
            $N = $N - $M;
 
        // Else subtract length from breadth
        else
            $M = $M - $N;
    }
 
    if ($N > 0 && $M > 0)
        return $N * $M;
    else
        return 0;
}
 
// Driver code
$N = 5;
$M = 3;
$K = 2;
 
echo remainingArea($N, $M, $K);
 
// This code is contributed by AnkitRai01
 
?>


Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the area
// of the remaining plot
function remainingArea(N, M, K)
{
 
    // Continue while plot has positive area
    // and there are persons left
    while (K-- && N && M) {
 
        // If length > breadth
        // then subtract breadth from length
        if (N > M)
            N = N - M;
 
        // Else subtract length from breadth
        else
            M = M - N;
    }
 
    if (N > 0 && M > 0)
        return N * M;
    else
        return 0;
}
 
// Driver code
var N = 5, M = 3, K = 2;
document.write( remainingArea(N, M, K));
 
</script>


Output: 

2

 

Time Complexity: O(K)

Auxiliary Space: O(1)



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