Skip to content
Related Articles

Related Articles

Maximize minimum of array generated by maximums of same indexed elements of two rows of a given Matrix
  • Last Updated : 21 Apr, 2021

Given a matrix mat[][] of N rows and M columns, the task is to choose any two rows(i, j) (0 ≤ i, j ≤ N – 1) and construct a new array A[] of size M, where A[k] = max(mat[i][k], mat[j][k]) such that minimum of A[] is maximum possible.

Examples:

Input: mat[][] = {{5, 0, 3, 1, 2}, {1, 8, 9, 1, 3}, {1, 2, 3, 4, 5}, {9, 1, 0, 3, 7}, {2, 3, 0, 6, 3}, {6, 4, 1, 7, 0}}
Output: 3
Explanation:
Choose two rows i = 0 and j = 4. Then, A[] = {5, 3, 3, 6, 3} and min(A[]) is 3, which is maximum possible.

Input: mat[][] = {{2, 13, 41, 5}, {91, 11, 10, 13}, {12, 3, 28, 6}}
Output: 13
Explanation:
Choose two rows i = 0 and j = 1. Then, A[] = {91, 13, 41, 13} and min(A[]) is 13, which is maximum possible.

Approach: Follow the steps below to solve the problem:



  • Initialize a variable global_max with INT_MAX, which stores the maximum of a minimum of array constructed from any two rows of mat[][].
  • Iterate through every possible combination of rows i and j, and find the minimum of array constructed from them as row_min.
  • In each iteration, update global_max to the maximum of row_min and itself.
  • After the above steps, print the value of global_max as the result.

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 the maximum of
// minimum of array constructed from
// any two rows of the given matrix
int getMaximum(int N, int M,
               vector<vector<int> > mat)
{
    // Initialize global max as INT_MIN
    int global_max = INT_MIN;
 
    // Iterate through the rows
    for (int i = 0; i < N; i++) {
 
        // Iterate through remaining rows
        for (int j = i + 1; j < N; j++) {
 
            // Initialize row_min as INT_MAX
            int row_min = INT_MAX;
 
            // Iterate through the column
            // values of two rows
            for (int k = 0; k < M; k++) {
 
                // Find max of two elements
                int m = max(mat[i][k],
                            mat[j][k]);
 
                // Update the row_min
                row_min = min(row_min, m);
            }
 
            // Update the global_max
            global_max = max(global_max,
                             row_min);
        }
    }
 
    // Print the global max
    return global_max;
}
 
// Driver Code
int main()
{
 
    // Given matrix mat[][]
    vector<vector<int> > mat
        = { { 5, 0, 3, 1, 2 },
            { 1, 8, 9, 1, 3 },
            { 1, 2, 3, 4, 5 },
            { 9, 1, 0, 3, 7 },
            { 2, 3, 0, 6, 3 },
            { 6, 4, 1, 7, 0 } };
 
    // Given number of rows and columns
    int N = 6, M = 5;
 
    // Function Call
    cout << getMaximum(N, M, mat);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to find the maximum of
// minimum of array constructed from
// any two rows of the given matrix
static int getMaximum(int N, int M, int[][] mat)
{
     
    // Initialize global max as INT_MIN
    int global_max = Integer.MIN_VALUE;
     
    // Iterate through the rows
    for(int i = 0; i < N; i++)
    {
         
        // Iterate through remaining rows
        for(int j = i + 1; j < N; j++)
        {
             
            // Initialize row_min as INT_MAX
            int row_min = Integer.MAX_VALUE;
             
            // Iterate through the column
            // values of two rows
            for(int k = 0; k < M; k++)
            {
                 
                // Find max of two elements
                int m = Math.max(mat[i][k],
                                 mat[j][k]);
 
                // Update the row_min
                row_min = Math.min(row_min, m);
            }
 
            // Update the global_max
            global_max = Math.max(global_max,
                                  row_min);
        }
    }
 
    // Print the global max
    return global_max;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given matrix mat[][]
    int[][] mat = { { 5, 0, 3, 1, 2 },
                    { 1, 8, 9, 1, 3 },
                    { 1, 2, 3, 4, 5 },
                    { 9, 1, 0, 3, 7 },
                    { 2, 3, 0, 6, 3 },
                    { 6, 4, 1, 7, 0 } };
 
    // Given number of rows and columns
    int N = 6, M = 5;
 
    // Function Call
    System.out.println(getMaximum(N, M, mat));
}
}
 
// This code is contributed by akhilsaini

Python3




# Python3 program for the above approach
import sys
 
# Function to find the maximum of
# minimum of array constructed from
# any two rows of the given matrix
def getMaximum(N, M, mat):
     
    # Initialize global max as INT_MIN
    global_max = -1 * (sys.maxsize)
 
    # Iterate through the rows
    for i in range(0, N):
         
        # Iterate through remaining rows
        for j in range(i + 1, N):
 
            # Initialize row_min as INT_MAX
            row_min = sys.maxsize
 
            # Iterate through the column
            # values of two rows
            for k in range(0, M):
 
                # Find max of two elements
                m = max(mat[i][k], mat[j][k])
 
                # Update the row_min
                row_min = min(row_min, m)
 
            # Update the global_max
            global_max = max(global_max,
                             row_min)
 
    # Print the global max
    return global_max
 
# Driver Code
if __name__ == '__main__':
 
    # Given matrix mat[][]
    mat = [ [ 5, 0, 3, 1, 2 ],
            [ 1, 8, 9, 1, 3 ],
            [ 1, 2, 3, 4, 5 ],
            [ 9, 1, 0, 3, 7 ],
            [ 2, 3, 0, 6, 3 ],
            [ 6, 4, 1, 7, 0 ] ]
 
    # Given number of rows and columns
    N = 6
    M = 5
 
    # Function Call
    print(getMaximum(N, M, mat))
 
# This code is contributed by akhilsaini

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the maximum of
// minimum of array constructed from
// any two rows of the given matrix
static int getMaximum(int N, int M, int[,] mat)
{
     
    // Initialize global max as INT_MIN
    int global_max = int.MinValue;
     
    // Iterate through the rows
    for(int i = 0; i < N; i++)
    {
         
        // Iterate through remaining rows
        for(int j = i + 1; j < N; j++)
        {
             
            // Initialize row_min as INT_MAX
            int row_min = int.MaxValue;
 
            // Iterate through the column
            // values of two rows
            for(int k = 0; k < M; k++)
            {
                 
                // Find max of two elements
                int m = Math.Max(mat[i, k],
                                 mat[j, k]);
                                  
                // Update the row_min
                row_min = Math.Min(row_min, m);
            }
 
            // Update the global_max
            global_max = Math.Max(global_max,
                                  row_min);
        }
    }
 
    // Print the global max
    return global_max;
}
 
// Driver Code
public static void Main()
{
     
    // Given matrix mat[][]
    int[,] mat = { { 5, 0, 3, 1, 2 },
                   { 1, 8, 9, 1, 3 },
                   { 1, 2, 3, 4, 5 },
                   { 9, 1, 0, 3, 7 },
                   { 2, 3, 0, 6, 3 },
                   { 6, 4, 1, 7, 0 } };
 
    // Given number of rows and columns
    int N = 6, M = 5;
 
    // Function Call
    Console.WriteLine(getMaximum(N, M, mat));
}
}
 
// This code is contributed by akhilsaini

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to find the maximum of
// minimum of array constructed from
// any two rows of the given matrix
function getMaximum(N, M, mat)
{
     
    // Initialize global max as let_MIN
    let global_max = Number.MIN_VALUE;
      
    // Iterate through the rows
    for(let i = 0; i < N; i++)
    {
          
        // Iterate through remaining rows
        for(let j = i + 1; j < N; j++)
        {
              
            // Initialize row_min as let_MAX
            let row_min = Number.MAX_VALUE;
              
            // Iterate through the column
            // values of two rows
            for(let k = 0; k < M; k++)
            {
                  
                // Find max of two elements
                let m = Math.max(mat[i][k],
                                 mat[j][k]);
  
                // Update the row_min
                row_min = Math.min(row_min, m);
            }
  
            // Update the global_max
            global_max = Math.max(global_max,
                                  row_min);
        }
    }
  
    // Prlet the global max
    return global_max;
}
 
// Driver code
 
// Given matrix mat[][]
let mat = [ [ 5, 0, 3, 1, 2 ],
            [ 1, 8, 9, 1, 3 ],
            [ 1, 2, 3, 4, 5 ],
            [ 9, 1, 0, 3, 7 ],
            [ 2, 3, 0, 6, 3 ],
            [ 6, 4, 1, 7, 0 ] ];
 
// Given number of rows and columns
let N = 6, M = 5;
 
// Function Call
document.write(getMaximum(N, M, mat));
 
// This code is contributed by avijitmondal1998
 
</script>
Output: 
3

 

Time Complexity: O(M*N2)
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up
Recommended Articles
Page :