Day 8 of Studying LeetCode Solution until I Can Solve One on My Own: Problem221.Maximal Square(Medium/JavaScript)

Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]Output: 4

Input: matrix = [["0","1"],["1","0"]]Output: 1

Input: matrix = [["0"]]Output: 0

1 1 11 1 11 1 1

1 1 11 2 11 1 1

1 1 11 2 21 2 1

1 1 11 2 21 2 3

function maximalSquare(matrix) {    if(!matrix || !matrix[0]) return 0//To deal with edge cases where an empty matrix is given. If// 'matrix' is false/doesn't exist (note 1) return 0.    let cache = [...matrix],//create a copy (note 2) of given array 'matrix'        height = matrix.length,        width = matrix[0].length,//define height and width of the array. Height of the array is//the length of the array (note 3). And width of the array is the//length of the first element of the array (note 3 & 4). For//example, matrix array shown below has width and length of 2.//Because the length of the matrix array is 2 (there are two//arrays nested in the matrix array). The length of the first//array ["0","1"] is also 2, which makes up the width of the//matrix array.//       [["0","1"],//        ["1","0"]]        solution = Math.max(...matrix[0])//solution = length of the largest square.//set the initial value of the solution as the maximum (note 6)//value of first array element. This is for the edge case where//there is only one element in the matrix array. And because our//iteration below starts from index 1 (the second element) of both//row and column; if there is only one array in the matrix, the//solution would be the max value of array. For example, if we//have a matrix of [["0","1"]], the largest square that contains 1//will be 1*1=1.     for (let i = 0; i < matrix.length; i++) {         solution = Math.max(solution, matrix[i][0])     }//This is for the edge case where there are two elements in the//matrix array and each element is a single element array. For//example, [["0"],["1"]]. Because our iteration below starts from//index 1 (the second element) of both row and column; if both//elements are single element array, the solution would be the max//value between two elements. For example, if we have a matrix of//[["0"],["1"]], the max of array[0] is 0 and the max of array[1]//is one, that will give us the max between two arrays of 1.      for (let row = 1; row < height; row++) {         for (let col = 1; col < width; col++) {//start interating from second elment of second array (note 7)             if(matrix[row][col] === "1") {                 cache[row][col] = Math.min(cache[row-1][col],                     cache[row][col-1],cache[row-1][col-1])+1;//if "1" if found, then compare it with it's surrounding element//and save minimum (note 5) of these elements plus 1 and save it//in the new maxtrix "Cashe" (see explaination above for reason//behind this step.               solution = Math.max(cache[row][col], solution);//update max solution             }         }     }    return solution **2//the area of a square is the product of the length of each side//with itself (note 8)}