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Find number of Positional Elements
  • Difficulty Level : Easy
  • Last Updated : 07 May, 2021

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 : 6

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++




// 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 and rmaxx for every row
    for (int i = 0; i < m; i++) {
        int rminn = INT_MAX;
        int rmaxx = 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];
        }
        rowmax[i] = rmaxx;
        rowmin[i] = rminn;
    }
 
    // Find cminn and cmaxx for every column
    for (int j = 0; j < n; j++) {
        int cminn = INT_MAX;
        int cmaxx = INT_MIN;
        for (int i = 0; i < m; i++) {
            if (a[i][j] > cmaxx)
                cmaxx = a[i][j];
            if (a[i][j] < cminn)
                cminn = a[i][j];
        }
 
        colmax[j] = cmaxx;
        colmin[j] = cminn;
    }
 
    // Check for optimal element
    int count = 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])) {
                count++;
            }
        }
    }
 
    return count;
}
 
// Driver code
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;
}

Java




// 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];
 
        // Find rminn and rmaxx for every row
        for (int i = 0; i < m; i++) {
            int rminn = Integer.MAX_VALUE;
            int rmaxx = 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];
            }
            rowmax[i] = rmaxx;
            rowmin[i] = rminn;
        }
 
        // Find cminn and cmaxx for every column
        for (int j = 0; j < n; j++) {
            int cminn = Integer.MAX_VALUE;
            int cmaxx = Integer.MIN_VALUE;
            for (int i = 0; i < m; i++) {
                if (a[i][j] > cmaxx)
                    cmaxx = a[i][j];
                if (a[i][j] < cminn)
                    cminn = a[i][j];
            }
 
            colmax[j] = cmaxx;
            colmin[j] = cminn;
        }
 
        // Check for optimal element
        int count = 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])) {
                    count++;
                }
            }
        }
 
        return count;
    }
 
    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));
    }
}

Python3




# Python3 program to find positional elements in a matrix.
import sys
 
MAX = 100
 
def countPositional(a, m, n):
 
    # rwomax[i] is going to store maximum of
    # i-th row and other arrays have similar
    # meaning
    rowmax = [0] * m
    rowmin = [0] * m
    colmax = [0] * n
    colmin = [0] * n
 
    # Find rminn and rmaxx for every row
    for i in range(m) :
        rminn = sys.maxsize
        rmaxx = -sys.maxsize
        for j in range(n) :
            if (a[i][j] > rmaxx) :
                rmaxx = a[i][j]
            if (a[i][j] < rminn) :
                rminn = a[i][j]
     
        rowmax[i] = rmaxx
        rowmin[i] = rminn
 
    # Find cminn and cmaxx for every column
    for j in range(n) :
        cminn = sys.maxsize
        cmaxx = -sys.maxsize
        for i in range(m) :
            if (a[i][j] > cmaxx) :
                cmaxx = a[i][j]
            if (a[i][j] < cminn) :
                cminn = a[i][j]
 
        colmax[j] = cmaxx
        colmin[j] = cminn
 
    # Check for optimal element
    count = 0
    for i in range(m) :
        for j in range(n) :
            if ((a[i][j] == rowmax[i]) or (a[i][j] == rowmin[i])
                or (a[i][j] == colmax[j])
                or (a[i][j] == colmin[j])) :
                count += 1
 
    return count
 
# Driver code
a = [ [ 1, 3, 4 ], [ 5, 2, 9 ], [ 8, 7, 6 ] ]
m, n = 3, 3
print(countPositional(a, m, n))
 
# This code is contributed by divyeshrabadiya07

C#




// 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];
 
        // Find rminn and rmaxx for every row
        for (int i = 0; i < m; i++) {
            int rminn = int.MaxValue;
            int rmaxx = 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];
            }
            rowmax[i] = rmaxx;
            rowmin[i] = rminn;
        }
 
        // Find cminn and cmaxx for every column
        for (int j = 0; j < n; j++) {
            int cminn = int.MaxValue;
            int cmaxx = int.MinValue;
            for (int i = 0; i < m; i++) {
                if (a[i, j] > cmaxx)
                    cmaxx = a[i, j];
                if (a[i, j] < cminn)
                    cminn = a[i, j];
            }
 
            colmax[j] = cmaxx;
            colmin[j] = cminn;
        }
 
        // Check for optimal element
        int count = 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])) {
                    count++;
                }
            }
        }
 
        return count;
    }
 
    // 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.

Javascript




<script>
 
// Javascript program to find positional
// elements in a matrix.
let MAX = 100;
 
function countPositional(a, m, n)
{
     
    // rwomax[i] is going to store maximum of
    // i-th row and other arrays have similar
    // meaning
    let rowmax = new Array(m);
    let rowmin = new Array(m);
    let colmax = new Array(n);
    let colmin = new Array(n);
 
    // Find rminn and rmaxx for every row
    for(let i = 0; i < m; i++)
    {
        let rminn = Number.MAX_VALUE;
        let rmaxx = Number.MIN_VALUE;
        for(let j = 0; j < n; j++)
        {
            if (a[i][j] > rmaxx)
                rmaxx = a[i][j];
            if (a[i][j] < rminn)
                rminn = a[i][j];
        }
        rowmax[i] = rmaxx;
        rowmin[i] = rminn;
    }
 
    // Find cminn and cmaxx for every column
    for(let j = 0; j < n; j++)
    {
        let cminn = Number.MAX_VALUE;
        let cmaxx = Number.MIN_VALUE;
        for(let i = 0; i < m; i++)
        {
            if (a[i][j] > cmaxx)
                cmaxx = a[i][j];
            if (a[i][j] < cminn)
                cminn = a[i][j];
        }
 
        colmax[j] = cmaxx;
        colmin[j] = cminn;
    }
 
    // Check for optimal element
    let count = 0;
    for(let i = 0; i < m; i++)
    {
        for(let 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]))
            {
                count++;
            }
        }
    }
 
    return count;
}
 
// Driver code
let a = [ [ 1, 3, 4 ],
          [ 5, 2, 9 ],
          [ 8, 7, 6 ] ];
let m = 3, n = 3;
document.write(countPositional(a, m, n));
 
// This code is contributed by suresh07
 
</script>
Output: 
7

 

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