Given here are n squares which touch each other externally, and are lined up in a row. The distance between the centers of the first and last square is given. The squares have equal side length. The task is to find the side of each square.
Examples:
Input: d = 42, n = 4 Output: The side of each square is 14 Input: d = 36, n = 5 Output: The side of each square is 9
Approach:
Suppose there are n squares each having side of length a.
Let, the distance between the first and last squares = d
From the figure, it is clear,
a/2 + a/2 + (n-2)*a = d
a + na – 2a = d
na – a = d
so, a = d/(n-1) 
C++
// C++ program to find side of the squares// which are lined in a row and distance between the// centers of first and last squares is given#include <bits/stdc++.h>using namespace std;void radius(int n, int d){ cout << "The side of each square is " << d / (n - 1) << endl;}// Driver codeint main(){ int d = 42, n = 4; radius(n, d); return 0;} |
Java
// Java program to find side of the squares// which are lined in a row and distance between the// centers of first and last squares is givenimport java.io.*;class GFG{static void radius(int n, int d){ System.out.print( "The side of each square is " + d / (n - 1));}// Driver codepublic static void main (String[] args) { int d = 42, n = 4; radius(n, d);}}// This code is contributed by vt_m. |
Python3
# Python program to find side of the squares# which are lined in a row and distance between the# centers of first and last squares is givendef radius(n, d): print("The side of each square is ", d / (n - 1));d = 42; n = 4;radius(n, d);# This code contributed by PrinciRaj1992 |
C#
// C# program to find side of the squares// which are lined in a row and distance between the// centers of first and last squares is givenusing System;class GFG{static void radius(int n, int d){ Console.Write( "The side of each square is " + d / (n - 1));}// Driver codepublic static void Main () { int d = 42, n = 4; radius(n, d);}}// This code is contributed by anuj_67.. |
Javascript
<script>// javascript program to find side of the squares// which are lined in a row and distance between the// centers of first and last squares is givenfunction radius(n , d){ document.write( "The side of each square is " + d / (n - 1));}// Driver codevar d = 42, n = 4;radius(n, d);// This code is contributed by 29AjayKumar </script> |
Output:
The side of each square is 14
Time Complexity: O(1)
Auxiliary Space: O(1)
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