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Find all matrix elements which are minimum in their row and maximum in their column

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  • Difficulty Level : Easy
  • Last Updated : 28 Dec, 2022
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Given a matrix mat[][] of size M * N, the task is to find all matrix elements which are minimum in their respective row and maximum in their respective column. If no such element is present, print -1.

Examples:

Input: mat[][] = {{1, 10, 4}, {9, 3, 8}, {15, 16, 17}}
Output: 15
Explanation:
15 is the only element which is maximum in its column {1, 9, 15} and minimum in its row {15, 16, 17}.

Input: m[][] = {{10, 41}, {3, 5}, {16, 2}}
Output: -1

Approach: Follow the steps below to solve the problem:

  1. Create an unordered_set and store the minimum element of each row of the matrix.
  2. Traverse the matrix and find the maximum element of each column. For every column, check if the maximum obtained is already present in the unordered_set or not.
  3. If found to be true, print that number. If no such matrix element is found, print -1.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Functionto find all the matrix elements
// which are minimum in its row and maximum
// in its column
vector<int> minmaxNumbers(vector<vector<int> >& matrix,
                          vector<int>& res)
{
 
    // Initialize unordered set
    unordered_set<int> set;
 
    // Traverse the matrix
    for (int i = 0; i < matrix.size(); i++) {
        int minr = INT_MAX;
        for (int j = 0; j < matrix[i].size(); j++) {
 
            // Update the minimum
            // element of current row
            minr = min(minr, matrix[i][j]);
        }
 
        // Insert the minimum
        // element of the row
        set.insert(minr);
    }
 
    for (int j = 0; j < matrix[0].size(); j++) {
        int maxc = INT_MIN;
 
        for (int i = 0; i < matrix.size(); i++) {
 
            // Update the maximum
            // element of current column
            maxc = max(maxc, matrix[i][j]);
        }
 
        // Checking if it is already present
        // in the unordered_set or not
        if (set.find(maxc) != set.end()) {
            res.push_back(maxc);
        }
    }
 
    return res;
}
 
// Driver Code
int main()
{
    vector<vector<int> > mat
        = { { 1, 10, 4 },
            { 9, 3, 8 },
            { 15, 16, 17 } };
 
    vector<int> ans;
 
    // Function call
    minmaxNumbers(mat, ans);
 
    // If no such matrix
    // element is found
    if (ans.size() == 0)
        cout << "-1" << endl;
 
    for (int i = 0; i < ans.size(); i++)
        cout << ans[i] << endl;
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Functionto find all the matrix elements
// which are minimum in its row and maximum
// in its column
public static Vector<Integer>minmaxNumbers(int[][] matrix,
              Vector<Integer> res)        
{
     
    // Initialize unordered set
    Set<Integer> set = new HashSet<Integer>();
   
    // Traverse the matrix
    for(int i = 0; i < matrix.length; i++)
    {
        int minr = Integer.MAX_VALUE;
        for(int j = 0; j < matrix[i].length; j++)
        {
             
            // Update the minimum
            // element of current row
            minr = Math.min(minr, matrix[i][j]);
        }
   
        // Insert the minimum
        // element of the row
        set.add(minr);
    }
   
    for(int j = 0; j < matrix[0].length; j++)
    {
        int maxc = Integer.MIN_VALUE;
        for(int i = 0; i < matrix.length; i++)
        {
             
            // Update the maximum
            // element of current column
            maxc = Math.max(maxc, matrix[i][j]);
        }
   
        // Checking if it is already present
        // in the unordered_set or not
        if (set.contains(maxc))
        {
            res.add(maxc);
        }
    }
    return res;
}
 
// Driver code
public static void main(String[] args)
{
    int[][] mat = { { 1, 10, 4 }, 
                    { 9, 3, 8 }, 
                    { 15, 16, 17 } };
 
    Vector<Integer> ans = new Vector<Integer>();
   
    // Function call
    ans = minmaxNumbers(mat, ans);
   
    // If no such matrix
    // element is found
    if (ans.size() == 0)
        System.out.println("-1");
   
    for(int i = 0; i < ans.size(); i++)
        System.out.println(ans.get(i));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3




# Python3 program for the above approach
import sys
 
# Functionto find all the matrix elements
# which are minimum in its row and maximum
# in its column
def minmaxNumbers(matrix, res):
     
    # Initialize unordered set
    s = set()
 
    # Traverse the matrix
    for i in range(0, len(matrix), 1):
        minr = sys.maxsize
        for j in range(0, len(matrix[i]), 1):
             
            # Update the minimum
            # element of current row
            minr = min(minr, matrix[i][j])
 
        # Insert the minimum
        # element of the row
        s.add(minr)
 
    for j in range(0, len(matrix[0]), 1):
        maxc = -sys.maxsize - 1
 
        for i in range(0, len(matrix), 1):
             
            # Update the maximum
            # element of current column
            maxc = max(maxc, matrix[i][j])
 
        # Checking if it is already present
        # in the unordered_set or not
        if (maxc in s):
            res.append(maxc)
 
    return res
 
# Driver Code
if __name__ == '__main__':
     
    mat = [ [ 1, 10, 4 ],
            [ 9, 3, 8 ],
            [ 15, 16, 17 ] ]
 
    ans = []
 
    # Function call
    minmaxNumbers(mat, ans)
 
    # If no such matrix
    # element is found
    if (len(ans) == 0):
        print("-1")
 
    for i in range(len(ans)):
        print(ans[i])
         
# This code is contributed by SURENDRA_GANGWAR

C#




// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
     
// Functionto find all
// the matrix elements
// which are minimum
// in its row and
//  maximum in its column
public static List<int> minmaxNumbers(int[,] matrix,
                                      List<int> res)        
{    
  // Initialize unordered set
  HashSet<int> set = new HashSet<int>();
 
  // Traverse the matrix
  for(int i = 0; i < matrix.GetLength(0); i++)
  {
    int minr = int.MaxValue;
    for(int j = 0; j < matrix.GetLength(1); j++)
    {
      // Update the minimum
      // element of current row
      minr = Math.Min(minr, matrix[i, j]);
    }
 
    // Insert the minimum
    // element of the row
    set.Add(minr);
  }
 
  for(int j = 0; j < matrix.GetLength(0); j++)
  {
    int maxc = int.MinValue;
    for(int i = 0; i < matrix.GetLength(1); i++)
    {
      // Update the maximum
      // element of current column
      maxc = Math.Max(maxc, matrix[i, j]);
    }
 
    // Checking if it is already present
    // in the unordered_set or not
    if (set.Contains(maxc))
    {
      res.Add(maxc);
    }
  }
  return res;
}
 
// Driver code
public static void Main(String[] args)
{
  int[,] mat = {{1, 10, 4}, 
                {9, 3, 8}, 
                {15, 16, 17}};
 
  List<int> ans = new List<int>();
 
  // Function call
  ans = minmaxNumbers(mat, ans);
 
  // If no such matrix
  // element is found
  if (ans.Count == 0)
    Console.WriteLine("-1");
 
  for(int i = 0; i < ans.Count; i++)
    Console.WriteLine(ans[i]);
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// Javascript program for the above approach
 
// Functionto find all the matrix elements
// which are minimum in its row and maximum
// in its column
function minmaxNumbers(matrix, res)
{
 
    // Initialize unordered set
    var set = new Set();
 
    // Traverse the matrix
    for (var i = 0; i < matrix.length; i++) {
        var minr = 1000000000;
        for (var j = 0; j < matrix[i].length; j++) {
 
            // Update the minimum
            // element of current row
            minr = Math.min(minr, matrix[i][j]);
        }
 
        // Insert the minimum
        // element of the row
        set.add(minr);
    }
 
    for (var j = 0; j < matrix[0].length; j++) {
        var maxc = -1000000000;
 
        for (var i = 0; i < matrix.length; i++) {
 
            // Update the maximum
            // element of current column
            maxc = Math.max(maxc, matrix[i][j]);
        }
 
        // Checking if it is already present
        // in the unordered_set or not
        if (set.has(maxc)) {
            res.push(maxc);
        }
    }
 
    return res;
}
 
// Driver Code
var mat
    = [ [ 1, 10, 4 ],
        [ 9, 3, 8 ],
        [ 15, 16, 17 ] ];
var ans = [];
 
// Function call
minmaxNumbers(mat, ans);
 
// If no such matrix
// element is found
if (ans.length == 0)
    document.write( "-1");
for (var i = 0; i < ans.length; i++)
    document.write( ans[i] + "<br>");
 
// This code is contributed by rrrtnx.
</script>

Output: 

15

 

Time Complexity: O(M * N) 
Auxiliary Space: O(M + N)


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