Given a square matrix of size N*N using numbers 1 to N^2, the task is to find the maximum of minimums of each layer of the matrix.
The layers of the matrix are the boundary elements of the sub-matrix starting at (i, i) and ending at (N – i + 1, N – i + 1), where 1<= i<= ceil(N/2).
Examples:
Input: Below is the given matrix:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Output: 6
Explanation: The layers are {1, 2, 3, 4, 8, 12, 16, 15, 14, 13, 9, 5} with minimum 1 and {6, 7, 10, 11} with minimum 6. The maximum of 1 and 6 is 6.Input: Below is the given matrix:
1 2 3
4 30 5
1 2 3
Output: 30
Explanation: The layers are {1, 2, 3, 5, 3, 2, 1, 4, 1} with minimum 1 and {30} with minimum 30. The maximum of 1 and 30 is 30.
Approach: The idea is to observe carefully, for the ith layer, the elements of the top, left, right and bottom boundary are at indexes:
- top boundary is at indexes (i, j)
- left boundary is at indexes (j, i)
- right boundary is at indexes (j, n – i + 1)
- bottom boundary is at indexes (n – i + 1, j), where i <= j <= n – i + 1
Thus, find the minimums of each layer and store the maximum of these minimums.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include<bits/stdc++.h>using namespace std;// Function to return the minimum// of the boundary elementsint getBoundaryMin(int a[][4], int n, int index){ int min1 = INT_MAX; for(int i = index; i < n - index; i++) { // Top boundary min1 = min(min1, a[index][i]); // Left boundary min1 = min(min1, a[i][index]); // Right boundary min1 = min(min1, a[i][n - index - 1]); // Bottom boundary min1 = min(min1, a[n - index - 1][i]); } return min1;}// Function to find the maximum of// minimums of all layersvoid MaximumOfMinimum(int a[][4], int n){ // Calculate the layers int layers = n / 2 + n % 2; int max1 = INT_MIN; // Iterate for all the layers for(int i = 0; i < layers; i++) { // Find maximum max1 = max(max1, getBoundaryMin(a, n, i)); } // Print the answer cout << (max1);}// Driver Codeint main(){ // Initialize the matrix int a[][4] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; int n = sizeof(a) / sizeof(a[0]); // Function call MaximumOfMinimum(a, n);}// This code is contributed by chitranayal |
Java
// Java Program for the above approachclass GFG { // Function to return the minimum // of the boundary elements static int getBoundaryMin(int a[][], int n, int index) { int min = Integer.MAX_VALUE; for (int i = index; i < n - index; i++) { // Top boundary min = Math.min( min, a[index][i]); // Left boundary min = Math.min( min, a[i][index]); // Right boundary min = Math.min( min, a[i][n - index - 1]); // Bottom boundary min = Math.min( min, a[n - index - 1][i]); } return min; } // Function to find the maximum of // minimums of all layers static void MaximumOfMinimum( int a[][], int n) { // Calculate the layers int layers = n / 2 + n % 2; int max = Integer.MIN_VALUE; // Iterate for all the layers for (int i = 0; i < layers; i++) { // Find maximum max = Math.max( max, getBoundaryMin(a, n, i)); } // Print the answer System.out.print(max); } // Driver Code public static void main(String[] args) { // Initialize the matrix int a[][] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; int n = a.length; // Function call MaximumOfMinimum(a, n); }} |
Python3
# Python3 program for the above approachimport sys# Function to return the minimum# of the boundary elementsdef getBoundaryMin(a, n, index): min1 = sys.maxsize for i in range(index, n - index): # Top boundary min1 = min(min1, a[index][i]) # Left boundary min1 = min(min1, a[i][index]) # Right boundary min1 = min(min1, a[i][n - index - 1]) # Bottom boundary min1 = min(min1, a[n - index - 1][i]) return min1# Function to find the maximum of# minimums of all layersdef MaximumOfMinimum(a, n): # Calculate the layers layers = n // 2 + n % 2 max1 = -sys.maxsize - 1 # Iterate for all the layers for i in range(0, layers): # Find maximum max1 = max(max1, getBoundaryMin(a, n, i)) # Print the answer print(max1)# Driver Code # Initialize the matrixa = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ] , [ 13, 14, 15, 16 ] ] n = len(a)# Function callMaximumOfMinimum(a, n)# This code is contributed by sanjoy_62 |
C#
// C# program for the above approachusing System;class GFG{// Function to return the minimum// of the boundary elementsstatic int getBoundaryMin(int[,]a, int n, int index){ int min = int.MaxValue; for(int i = index; i < n - index; i++) { // Top boundary min = Math.Min(min, a[index, i]); // Left boundary min = Math.Min(min, a[i, index]); // Right boundary min = Math.Min(min, a[i, n - index - 1]); // Bottom boundary min = Math.Min(min, a[n - index - 1, i]); } return min;}// Function to find the maximum of// minimums of all layersstatic void MaximumOfMinimum(int[,]a, int n){ // Calculate the layers int layers = n / 2 + n % 2; int max = int.MinValue; // Iterate for all the layers for(int i = 0; i < layers; i++) { // Find maximum max = Math.Max(max, getBoundaryMin(a, n, i)); } // Print the answer Console.Write(max);}// Driver Codepublic static void Main(String[] args){ // Initialize the matrix int[,]a = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; int n = a.GetLength(0); // Function call MaximumOfMinimum(a, n);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program for the above approach// Function to return the minimum// of the boundary elementsfunction getBoundaryMin(a, n, index){ min1 = 1000000000; for(var i = index; i < n - index; i++) { // Top boundary min1 = Math.min(min1, a[index][i]); // Left boundary min1 = Math.min(min1, a[i][index]); // Right boundary min1 = Math.min(min1, a[i][n - index - 1]); // Bottom boundary min1 = Math.min(min1, a[n - index - 1][i]); } return min1;} // Function to find the maximum of// minimums of all layersfunction MaximumOfMinimum(a, n){ // Calculate the layers var layers = parseInt(n / 2) + n % 2; var max1 = -1000000000; // Iterate for all the layers for(var i = 0; i < layers; i++) { // Find maximum max1 = Math.max(max1, getBoundaryMin(a, n, i)); } // Print the answer document.write(max1);}// Driver Code// Initialize the matrixvar a = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ] ];var n = a.length;// Function callMaximumOfMinimum(a, n);// This code is contributed by itsok</script> |
Output:
6
Time Complexity: O(N2)
Auxiliary Space: O(1)
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