I have an array that looks like this:
arr = [[1,2,3],
[4,5,6],
[7,8,9]]
I have initialized an empty array and I want put the diagonals of the arr inside the new array. So i've tried this:
arr = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
new_arr = [];
tail = arr.length - 1;
for (let i = 0; i < arr.length; i++) {
for (let j = 0; j < arr[i].length; j++) {
if (j == i) {
new_arr.push(arr[i][j]);
}
if (j == tail) {
new_arr.push(arr[i][j]);
tail--;
}
}
}
console.log(new_arr)The logic seems to work but I can't seem to get the structure right. What I want is to nest two arrays inside the new array like this:
[[1,5,9],[3,5,7]]
But what I get is one array with the right values unordered. How to get the expected output? Any help is appreciated. Thanks!
var arr = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
];
var diagonal1 = [];
var diagonal2 = [];
for (var i = 0; i < arr.length; i++) {
diagonal1.push(arr[i][i]);
diagonal2.push(arr[i][arr.length - i - 1]);
}
var new_arr = [diagonal1, diagonal2];
console.log(new_arr)The following solution would work for you if the width and height of the number matrix is always equal.
const arr = [[1,2,3],
[4,5,6],
[7,8,9]];
const result = [[],[]];
arr.map((row,index) => {
result[0].push(row[0+index]);
result[1].push(row[row.length - index - 1]);
});
console.log(result); // [[1,5,9],[3,5,7]]
You need to have 2 separate temporary arrays. And you don't need nested loops. You can optimize the code like this with a single loop if you understand the math.
arr = [[1,2,3],
[4,5,6],
[7,8,9]];
function findDiagonals(arr) {
const diagonal_1 = [];
const diagonal_2 = [];
for( let i = 0; i < arr.length; i++ ) {
diagonal_1.push(arr[i][i]);
diagonal_2.push(arr[i][arr.length - (i+1)]);
}
return [diagonal_1, diagonal_2];
}
console.log(findDiagonals(arr));