# 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 : 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 ` `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)); ` `    ``} ` `} `

## 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. `

Output:

```7
```

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