Find number of Positional Elements

Given a matrix of integers, task is to find out number of positional elements. A positional element is one which is either minimum or maximum in a row or in a column.

Examples:

Input : a = {{1, 3, 4}, {5, 2, 9}, {8, 7, 6}}
Output : 7
There are total 7 elements min elements are 1, 2, 6 and 4. And max elements are 9, 8 and 7.

Input : a = {{1, 1}, {1, 1}, {1, 1}}
Output : 4

Source :Goldman Sachs Interview Set



Idea is to store the maximum and minimum of every row and column and then check for the required condition.

Below is the implementation of above approach.

C++

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// CPP program to find positional elements in
// a matrix.
#include <bits/stdc++.h>
using namespace std;
  
const int MAX = 100;
  
int countPositional(int a[][MAX], int m, int n)
{
    // rwomax[i] is going to store maximum of
    // i-th row and other arrays have similar 
    // meaning
    int rowmax[m], rowmin[m];
    int colmax[n], colmin[n];
  
    // Find rminn, rmaxx, cminn and cmaxx
    // of every row and column.
    // k will help maintaining the column value, 
    // otherwise it will only work for square matrices.
    int k = 0;
    for (int i = 0; i < m; i++) {
        int rminn = INT_MAX;
        int rmaxx = INT_MIN;
        int cminn = INT_MAX;
        int cmaxx = INT_MIN;        
        for (int j = 0; j < n; j++) {
            if (a[i][j] > rmaxx)
                rmaxx = a[i][j];
            if (a[i][j] < rminn) 
                rminn = a[i][j];
            if (a[j][i] > cmaxx) 
                cmaxx = a[j][i];
            if (a[j][i] < cminn) 
                cminn = a[j][i];
        }
        rowmax[i] = rmaxx;
        rowmin[i] = rminn;
  
        colmax[k] = cmaxx;
        colmin[k] = cminn;
        k++;
    }
  
    // check for optimal element
    int coun = 0;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if ((a[i][j] == rowmax[i]) || 
                (a[i][j] == rowmin[i]) || 
                (a[i][j] == colmax[j]) || 
                (a[i][j] == colmin[j]))
                coun++;
        }
    }
    return coun;
}
  
int main()
{
    int a[][MAX] = { { 1, 3, 4 }, { 5, 2, 9 }, { 8, 7, 6 } };
    int m = 3, n = 3;
    cout << countPositional(a, m, n);
    return 0;
}

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Java

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// Java program to find positional elements in 
// a matrix. 
class GfG { 
  
static int MAX = 100
  
static int countPositional(int a[][], int m, int n) 
    // rwomax[i] is going to store maximum of 
    // i-th row and other arrays have similar 
    // meaning 
    int rowmax[] = new int[m];
    int rowmin[] = new int[m]; 
    int colmax[] = new int[n];
    int colmin[] = new int[n]; 
    int k = 0;
    // Find rminn, rmaxx, cminn and cmaxx 
    // of every row and column. 
    for (int i = 0; i < m; i++) { 
        int rminn = Integer.MAX_VALUE; 
        int rmaxx = Integer.MIN_VALUE; 
        int cminn = Integer.MAX_VALUE; 
        int cmaxx = Integer.MIN_VALUE;         
        for (int j = 0; j < n; j++) { 
            if (a[i][j] > rmaxx) 
                rmaxx = a[i][j]; 
            if (a[i][j] < rminn) 
                rminn = a[i][j]; 
            if (a[j][i] > cmaxx) 
                cmaxx = a[j][i]; 
            if (a[j][i] < cminn) 
                cminn = a[j][i]; 
        
        rowmax[i] = rmaxx; 
        rowmin[i] = rminn; 
  
        colmax[k] = cmaxx; 
        colmin[k] = cminn; 
    
  
    // check for optimal element 
    int coun = 0
    for (int i = 0; i < m; i++) { 
        for (int j = 0; j < n; j++) { 
            if ((a[i][j] == rowmax[i]) || 
                (a[i][j] == rowmin[i]) || 
                (a[i][j] == colmax[j]) || 
                (a[i][j] == colmin[j])) 
                coun++; 
        
    
    return coun; 
  
public static void main(String[] args) 
    int a[][] = new int[][]{{ 1, 3, 4 }, { 5, 2, 9 }, { 8, 7, 6 }}; 
    int m = 3, n = 3
    System.out.println(countPositional(a, m, n)); 
}

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C#

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// C# program to find positional elements in 
using System;
  
class GFG
{
      
static int countPositional(int [,]a, int m, int n) 
    // rwomax[i] is going to store maximum of 
    // i-th row and other arrays have similar 
    // meaning 
    int []rowmax = new int[m];
    int []rowmin = new int[m]; 
    int []colmax = new int[n];
    int []colmin = new int[n]; 
    int k = 0;
    // Find rminn, rmaxx, cminn and cmaxx 
    // of every row and column. 
    for (int i = 0; i < m; i++)
    
        int rminn = int.MaxValue; 
        int rmaxx = int.MinValue; 
        int cminn = int.MaxValue; 
        int cmaxx = int.MinValue;    
        for (int j = 0; j < n; j++)
        
            if (a[i,j] > rmaxx) 
                rmaxx = a[i,j]; 
            if (a[i,j] < rminn) 
                rminn = a[i,j]; 
            if (a[j,i] > cmaxx) 
                cmaxx = a[j,i]; 
            if (a[j,i] < cminn) 
                cminn = a[j,i]; 
        
        rowmax[i] = rmaxx; 
        rowmin[i] = rminn; 
  
        colmax[k] = cmaxx; 
        colmin[k] = cminn; 
    
  
    // check for optimal element 
    int coun = 0; 
    for (int i = 0; i < m; i++)
    
        for (int j = 0; j < n; j++)
        
            if ((a[i,j] == rowmax[i]) || 
                (a[i,j] == rowmin[i]) || 
                (a[i,j] == colmax[j]) || 
                (a[i,j] == colmin[j])) 
                coun++; 
        
    
    return coun; 
  
// Driver Code 
static public void Main ()
{
    int [,]a = new int[,]{{ 1, 3, 4 }, { 5, 2, 9 }, { 8, 7, 6 }}; 
    int m = 3, n = 3; 
    Console.WriteLine(countPositional(a, m, n)); 
}
  
// This code is contributed by Tushil.

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Output:

7


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