Given an integer area, the task is to find the length and breadth of a rectangle with the given area such that the difference between the length and the breadth is the minimum possible.
Examples:
Input: area = 99
Output: l = 11, b = 9
All possible rectangles (l, b) are (99, 1), (33, 3) and (11, 9)
And the one with the minimum |l – b| is (11, 9)Input: area = 25
Output: l = 5, b = 5
Approach: The task is to find two integers l and b such that l * b = area and |l – b| are as minimal as possible. Factorization can be used to solve the problem, but doing just simple factorization from 1 to N will take a long time to get the required output for larger values of N.
To overcome this, just iterate up to. Considering 
< l ? N, then for all values of l, b will always be < .
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to print the length (l)// and breadth (b) of the rectangle// having area = N and |l - b| as// minimum as possiblevoid find_rectangle(int area){ int l, b; int M = sqrt(area), ans; for (int i = M; i >= 1; i--) { // i is a factor if (area % i == 0) { // l >= sqrt(area) >= i l = (area / i); // so here l is +ve always b = i; break; } } // Here l and b are length and // breadth of the rectangle cout << "l = " << l << ", b = " << b << endl;}// Driver codeint main(){ int area = 99; find_rectangle(area); return 0;} |
Java
// Java implementation of the approachclass GFG { // Function to print the length (l) // and breadth (b) of the rectangle // having area = N and |l - b| as // minimum as possible static void find_rectangle(int area) { int l = 0, b = 0; int M = (int)Math.sqrt(area), ans; for (int i = M; i >= 1; i--) { // i is a factor if (area % i == 0) { // l >= sqrt(area) >= i l = (area / i); // so here l is +ve always b = i; break; } } // Here l and b are length and // breadth of the rectangle System.out.println("l = " + l + ", b = " + b); } // Driver code public static void main(String[] args) { int area = 99; find_rectangle(area); }}// This code is contributed by Ita_c. |
Python3
# Python3 implementation of the approachimport math as mt# Function to print the length (l)# and breadth (b) of the rectangle # having area = N and |l - b| as # minimum as possibledef find_rectangle(area): l, b = 0, 0 M = mt.ceil(mt.sqrt(area)) ans = 0 for i in range(M, 0, -1): # i is a factor if (area % i == 0): # l >= sqrt(area) >= i l = (area // i) # so here l is + ve always b = i break # Here l and b are length and # breadth of the rectangle print("l =", l, ", b =", b)# Driver codearea = 99find_rectangle(area)# This code is contributed by # Mohit kumar 29 |
C#
// C# implementation of the approachusing System;class GFG { // Function to print the length (l) // and breadth (b) of the rectangle // having area = N and |l - b| as // minimum as possible static void find_rectangle(int area) { int l = 0, b = 0; int M = (int)Math.Sqrt(area); for (int i = M; i >= 1; i--) { // i is a factor if (area % i == 0) { // l >= sqrt(area) >= i l = (area / i); // so here l is +ve always b = i; break; } } // Here l and b are length and // breadth of the rectangle Console.WriteLine("l = " + l + ", b = " + b); } // Driver code public static void Main() { int area = 99; find_rectangle(area); }}// This code is contributed by Mukul Singh. |
PHP
<?php// Php implementation of the approach // Function to print the length (l) // and breadth (b) of the rectangle // having area = N and |l - b| as // minimum as possible function find_rectangle($area) { $M = floor(sqrt($area)); for ($i = $M; $i >= 1; $i--) { // i is a factor if ($area % $i == 0) { // l >= sqrt(area) >= i $l = floor($area / $i); // so here l is +ve always $b = $i ; break; } } // Here l and b are length and // breadth of the rectangle echo "l = ", $l, ", b = ", $b, "\n"; } // Driver code $area = 99; find_rectangle($area); // This code is contributed by Ryuga?> |
Javascript
<script>// Javascript implementation of the approach // Function to print the length (l) // and breadth (b) of the rectangle // having area = N and |l - b| as // minimum as possible function find_rectangle(area) { let l = 0, b = 0; let M = Math.floor(Math.sqrt(area)), ans; for (let i = M; i >= 1; i--) { // i is a factor if (area % i == 0) { // l >= sqrt(area) >= i l = Math.floor(area / i); // so here l is +ve always b = i; break; } } // Here l and b are length and // breadth of the rectangle document.write("l = " + l + ", b = " + b); }// Driver code let area = 99; find_rectangle(area); </script> |
Output:
l = 11, b = 9
Time Complexity: O(
)
Auxiliary Space: O(1)
Below is a simple implementation.
CPP
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to print the length (l)// and breadth (b) of the rectangle// having area = N and |l - b| as// minimum as possiblevoid find_rectangle(int area){ for (int i = ceil(sqrt(area)); i <= area; i++) { if (area / i * i == area) { printf("%d %d", i, area / i); return; } }}// Driver codeint main(){ int area = 99; find_rectangle(area); return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG { // Function to print the length (l) // and breadth (b) of the rectangle // having area = N and |l - b| as // minimum as possible static void find_rectangle(int area) { for(int i = (int)Math.ceil(Math.sqrt(area)); i <= area; i++) { if(area / i * i == area) { System.out.println(i + " " + (int)(area / i)); return; } } } // Driver code public static void main (String[] args) { int area = 99; find_rectangle(area); }}// This code is contributed by rag2127 |
Python3
# Python3 implementation of the approach import math# Function to print the length (l) # and breadth (b) of the rectangle # having area = N and |l - b| as # minimum as possible def find_rectangle(area): for i in range(int(math.ceil(math.sqrt(area))) , area + 1): if((int(area / i) * i) == area): print(i, int(area / i)) return# Driver code area = 99find_rectangle(area)# This code is contributed by avanitrachhadiya2155 |
C#
// C# implementation of the approachusing System;class GFG{ // Function to print the length (l) // and breadth (b) of the rectangle // having area = N and |l - b| as // minimum as possible static void find_rectangle(int area) { for(int i = (int)Math.Ceiling(Math.Sqrt(area)); i <= area; i++) { if(area / i * i == area) { Console.WriteLine(i + " " + (int)(area / i)); return; } } } // Driver code static void Main() { int area = 99; find_rectangle(area); }}// This code is contributed by divyeshrabadiya07. |
Javascript
<script>// Javascript implementation of the approach// Function to print the length (l)// and breadth (b) of the rectangle// having area = N and |l - b| as// minimum as possiblefunction find_rectangle(are){ for(let i = Math.floor(Math.ceil(Math.sqrt(area))); i <= area; i++) { if(Math.floor(area / i) * i == area) { document.write(i + " " + Math.floor(area / i)); return; } }} // Driver codelet area = 99;find_rectangle(area); // This code is contributed by unknown2108</script> |
Output:
11 9
Time Complexity: O(log(area)), due to inbuild function sqrt()
Auxiliary Space: O(1)
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