Rotate Image

// Note: Try to solve this task in-place (with O(1) additional memory), since this is what you'll be asked to do during an interview.

// You are given an n x n 2D matrix that represents an image. Rotate the image by 90 degrees (clockwise).

// Example

// For

// a = [[1, 2, 3],
//      [4, 5, 6],
//      [7, 8, 9]]
// the output should be

// solution(a) =
//    [[7, 4, 1],
//     [8, 5, 2],
//     [9, 6, 3]]

//Input/Output

// [execution time limit] 4 seconds (js)

// [input] array.array.integer a

// Guaranteed constraints:
// 1 ≤ a.length ≤ 100,
// a[i].length = a.length,
// 1 ≤ a[i][j] ≤ 104.

// [output] array.array.integer

function solution(a) {
    // First we want to create a for loop
    // that travels through our matrix
    // with two pointers
    for(let i = 0; i < a.length; i++)
    {
        for(let j = i; j < a.length; j++)
        {
            // we want to first turn the first
            // row of the matrix into the first 
            // column of the matrix. We will swap 
            // a[i][j] (Row-Column) with a[j][i]
            // (Column-Row).  Using example 1, we 
            // will have a 2D array that looks like:
            // [[1,4,7]]
            // [[2,5,8]]
            // [[3,6,9]]
            let temp = a[i][j];
            a[i][j] = a[j][i];
            a[j][i] = temp;
        }
    }
    // now that we have turned the nth row into the 
    // nth column, the nth+1 row into the nth+2 column,
    // etc...we now need to start swapping the numbers
    // from the edges of the matrix with one another and
    // move inward from there, swapping as we go. 
    // [[7,4,1]]
    // [[8,5,2]]
    // [[9,6,3]]
    for(let i = 0; i < a.length; i++)
    {
        // remember that we are only going halfway
        // through the length of the array because
        // we want to work our way to the center
        for(let j = 0; j < (a.length / 2); j++)
        {
            // a[i][j] is our pointer on the start
            // and a[i][a.length - 1 - j] is the 
            // pointer that moves from the end of the 
            // array by starting at a.length - 1 (which
            // gets us at the endpoint) and minus j 
            // (which allows us to move backwards towards
            // the middle of the rows)
            let temp = a[i][j];
            a[i][j] = a[i][a.length - 1 - j];
            a[i][a.length - 1 - j] = temp;
        }
    }
    return a
}