Javascript Program for Maximum and Minimum in a square matrix.

Given a square matrix of order n*n, find the maximum and minimum from the matrix given. 

Examples:

Input : arr[][] = {5, 4, 9,
                   2, 0, 6,
                   3, 1, 8};
Output : Maximum = 9, Minimum = 0

Input : arr[][] = {-5, 3, 
                   2, 4};
Output : Maximum = 4, Minimum = -5

Naive Method : 
We find maximum and minimum of matrix separately using linear search. Number of comparison needed is n² for finding minimum and n² for finding the maximum element. The total comparison is equal to 2n².

Javascript

// JavaScript Program for finding maximum
// and minimum in a matrix

function maxMin(arr, n){
    let min = +2147483647;
    let max = -2147483648;

    // loop to find the maximum element
    for(let i = 0; i<n; i++){
        for(let j = 0; j<n; j++){
            if(max < arr[i][j]) max = arr[i][j];
        }
    }

    // for finding the minimum element
    for(let i = 0; i<n; i++){
        for(let j = 0; j<n; j++){
            if(min > arr[i][j]) min = arr[i][j];
        }
    }

    console.log("Maximum = " + max + ", Minimum = " + min);
}

// driver code for above approach
let arr = [ [ 5, 9, 11 ], [ 25, 0, 14 ], [ 21, 6, 4 ] ];
maxMin(arr, 3);
// THIS CODE IS CONTRIBUTED BY KIRTI AGARWAL(KIRTIAGARWAL23121999)

Output

Maximum = 25, Minimum = 0

Time Complexity: O(N^2)

Auxiliary Space: O(1)

Pair Comparison (Efficient method): 
Select two elements from the matrix one from the start of a row of the matrix another from the end of the same row of the matrix, compare them and next compare smaller of them to the minimum of the matrix and larger of them to the maximum of the matrix. We can see that for two elements we need 3 compare so for traversing whole of the matrix we need total of 3/2 n² comparisons.

Note : This is extended form of method 3 of Maximum Minimum of Array.

Javascript

<script>

// Javascript program for finding maximum
// and minimum in a matrix.
let MAX = 100;

// Finds maximum and minimum
// in arr[0..n-1][0..n-1]
// using pair wise comparisons
function maxMin(arr,n)
{
    let min = +2147483647;
    let max = -2147483648;

    // Traverses rows one by one
    for(let i = 0; i < n; i++)
    {
        for(let j = 0; j <= n / 2; j++)
        {

            // Compare elements from beginning
            // and end of current row
            if (arr[i][j] > arr[i][n - j - 1])
            {
                if (min > arr[i][n - j - 1])
                    min = arr[i][n - j - 1];
                if (max< arr[i][j])
                    max = arr[i][j];
            }
            else
            {
                if (min > arr[i][j])
                    min = arr[i][j];
                if (max < arr[i][n - j - 1])
                    max = arr[i][n - j - 1];
            }
        }
    }
    document.write("Maximum = " + max +
                   ", Minimum = " + min);
}

// Driver Code
let arr = [ [ 5, 9, 11 ],
            [ 25, 0, 14 ],
            [ 21, 6, 4 ] ];

maxMin(arr, 3);

// This code is contributed by sravan kumar

</script>

Output:

Maximum = 25, Minimum = 0

Time Complexity: O(N^2)

Auxiliary Space: O(1)

Please refer complete article on Maximum and Minimum in a square matrix. for more details!