Given an array arr[] of positive integers where each element of the array represents the length of the rectangular blocks. The task is to find the largest length of the square which can be formed using the rectangular blocks.
Examples:
Input: arr[] = {3, 2, 1, 5, 2, 4}
Output: 3
Explanation:
Using rectangular block of length 3, 5 and 4, square of side length 3 can be constructed as shown below:
Input: arr[] = {1, 2, 3}
Output: 2
Approach:
- Sort the given array in decreasing order.
- Initialise maximum sidelength(say maxLength) as 0.
- Traverse the array arr[] and if arr[i] > maxLength then increment the maxLength and check this condition for next iteration.
- If the above condition doesn’t satisfy then break the loop and print the maxLength.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find maximum side // length of square void maxSide(int a[], int n) { int sideLength = 0; // Sort array in asc order sort(a, a + n); // Traverse array in desc order for (int i = n - 1; i >= 0; i--) { if (a[i] > sideLength) { sideLength++; } else { break; } } cout << sideLength << endl; } // Driver Code int main() { int N = 6; // Given array arr[] int arr[] = { 3, 2, 1, 5, 2, 4 }; // Function Call maxSide(arr, N); return 0; } |
Java
// Java implementation of the above approach import java.util.Arrays; class GFG{ // Function to find maximum side // length of square static void maxSide(int a[], int n) { int sideLength = 0; // Sort array in asc order Arrays.sort(a); // Traverse array in desc order for(int i = n - 1; i >= 0; i--) { if (a[i] > sideLength) { sideLength++; } else { break; } } System.out.println(sideLength); } // Driver code public static void main (String[] args) { int N = 6; // Given array arr[] int arr[] = new int[]{ 3, 2, 1, 5, 2, 4 }; // Function Call maxSide(arr, N); } } // This code is contributed by Pratima Pandey |
C#
// C# implementation of the above approach using System; class GFG{ // Function to find maximum side // length of square static void maxSide(int []a, int n) { int sideLength = 0; // Sort array in asc order Array.Sort(a); // Traverse array in desc order for(int i = n - 1; i >= 0; i--) { if (a[i] > sideLength) { sideLength++; } else { break; } } Console.Write(sideLength); } // Driver code public static void Main() { int N = 6; // Given array arr[] int []arr = new int[]{ 3, 2, 1, 5, 2, 4 }; // Function Call maxSide(arr, N); } } // This code is contributed by Code_Mech |
3
Time Complexity: O(N*log N)
Auxiliary Space: O(1)
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