Given two integers L and B denoting the length and breadth of a rectangle respectively. The task is to calculate the sum of the area of all possible squares that comes into the rectangle.
Examples:
Input: L = 4, B = 3 Output: 54 Input: L = 2, B = 5 Output: 26
The idea is to observe the count of number of squares in a rectangle.
Now, the number of squares of side 1 will be 12 as there will be two cases one as squares of 1-unit sides along the horizontal(3) and second case as squares of 1-unit sides along the vertical(4). That gives us 3*4 = 12 squares.
When the side is 2 units, one case will be as squares of side of 2 units along only one place horizontally and second case as two places vertically. So the number of squares = 6
So we can deduce that,
Number of squares of size 1*1 will be L*B
Number of squares of size 2*2 will be (L-1)(B-1)
Therefore, the number of squares with size 
will be:
Number of square of size K = (L-K+1)*(B-K+1)
Therefore, area of total number of squares of size K will be:
Area of total number of square of size K = (L-K+1)*(B-K+1)*K*K
Below is the implementation of above idea:
C++
// CPP program to calculate the sum of area// of all possible squares that comes// inside the rectangle#include <bits/stdc++.h>using namespace std;// Function to calculate the sum of area// of all possible squares that comes// inside the rectangleint calculateAreaSum(int l, int b){ int size = 1; // Square with max size possible int maxSize = min(l,b); int totalArea = 0; for(int i=1; i <= maxSize; i++) { // calculate total square of a given size int totalSquares = (l-size+1)*(b-size+1); // calculate area of squares of a // particular size int area = totalSquares*size*size; // total area totalArea += area; // increment size size++; } return totalArea;}// Driver Codeint main(){ int l = 4, b = 3; cout<<calculateAreaSum(l,b); return 0;} |
Java
// Java program to calculate the // sum of area of all possible // squares that comes inside// the rectangleclass GFG {// Function to calculate the // sum of area of all possible // squares that comes inside// the rectanglestatic int calculateAreaSum(int l, int b){ int size = 1; // Square with max size possible int maxSize = Math.min(l, b); int totalArea = 0; for(int i = 1; i <= maxSize; i++) { // calculate total square // of a given size int totalSquares = (l - size + 1) * (b - size + 1); // calculate area of squares // of a particular size int area = totalSquares * size * size; // total area totalArea += area; // increment size size++; } return totalArea;}// Driver Codepublic static void main(String[] args) { int l = 4, b = 3; System.out.println(calculateAreaSum(l, b));}}// This code is contributed // by ChitraNayal |
Python 3
# Python 3 program to calculate# the sum of area of all possible # squares that comes inside # the rectangle# Function to calculate the # sum of area of all possible# squares that comes inside# the rectangledef calculateAreaSum(l, b): size = 1 # Square with max size possible maxSize = min(l, b) totalArea = 0 for i in range(1, maxSize + 1 ): # calculate total square # of a given size totalSquares = ((l - size + 1) * (b - size + 1)) # calculate area of squares # of a particular size area = (totalSquares * size * size) # total area totalArea += area # increment size size += 1 return totalArea# Driver Codeif __name__ == "__main__": l = 4 b = 3 print(calculateAreaSum(l,b))# This code is contributed # by ChitraNayal |
C#
// C# program to calculate the // sum of area of all possible // squares that comes inside// the rectangleusing System;class GFG {// Function to calculate the // sum of area of all possible // squares that comes inside // the rectanglestatic int calculateAreaSum(int l, int b){ int size = 1; // Square with max size possible int maxSize = Math.Min(l, b); int totalArea = 0; for(int i = 1; i <= maxSize; i++) { // calculate total square // of a given size int totalSquares = (l - size + 1) * (b - size + 1); // calculate area of squares // of a particular size int area = totalSquares * size * size; // total area totalArea += area; // increment size size++; } return totalArea;}// Driver Codepublic static void Main() { int l = 4, b = 3; Console.Write(calculateAreaSum(l,b));}}// This code is contributed// by ChitraNayal |
PHP
<?php // PHP program to calculate the // sum of area of all possible // squares that comes inside // the rectangle// Function to calculate the // sum of area of all possible // squares that comes inside // the rectanglefunction calculateAreaSum($l, $b){ $size = 1; // Square with max size possible $maxSize = min($l, $b); $totalArea = 0; for($i = 1; $i <= $maxSize; $i++) { // calculate total square // of a given size $totalSquares = ($l - $size + 1) * ($b - $size + 1); // calculate area of squares // of a particular size $area = $totalSquares * $size * $size; // total area $totalArea += $area; // increment size $size++; } return $totalArea;}// Driver Code$l = 4;$b = 3; echo calculateAreaSum($l,$b); // This code is contributed // by ChitraNayal?> |
Javascript
<script>// Javascript program to calculate the sum of area// of all possible squares that comes// inside the rectangle// Function to calculate the sum of area// of all possible squares that comes// inside the rectanglefunction calculateAreaSum(l, b){ var size = 1; // Square with max size possible var maxSize = Math.min(l,b); var totalArea = 0; for(var i=1; i <= maxSize; i++) { // calculate total square of a given size var totalSquares = (l-size+1)*(b-size+1); // calculate area of squares of a // particular size var area = totalSquares*size*size; // total area totalArea += area; // increment size size++; } return totalArea;}// Driver Codevar l = 4, b = 3;document.write( calculateAreaSum(l,b));// This code is contributed by noob2000.</script> |
Output
54
Complexity Analysis:
- Time complexity: O(min(l,b))
- Auxiliary complexity: O(1)
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