Problem
Given a multi-dimensional array arr
and a depth n
, return a flattened version of that array.
A multi-dimensional array is a recursive data structure that contains integers or other multi-dimensional arrays.
A flattened array is a version of that array with some or all of the sub-arrays removed and replaced with the actual elements in that sub-array. This flattening operation should only be done if the current depth of nesting is less than n
. The depth of the elements in the first array are considered to be 0
.
Please solve it without the built-in Array.flat
method.
Example 1:
Input
arr = [1, 2, 3, [4, 5, 6], [7, 8, [9, 10, 11], 12], [13, 14, 15]]
n = 0
Output
[1, 2, 3, [4, 5, 6], [7, 8, [9, 10, 11], 12], [13, 14, 15]]
Explanation
Passing a depth of n=0 will always result in the original array. This is because the smallest possible depth of a subarray (0) is not less than n=0. Thus, no subarray should be flattened.
Example 2:
Input
arr = [1, 2, 3, [4, 5, 6], [7, 8, [9, 10, 11], 12], [13, 14, 15]]
n = 1
Output
[1, 2, 3, 4, 5, 6, 7, 8, [9, 10, 11], 12, 13, 14, 15]
Explanation
The subarrays starting with 4, 7, and 13 are all flattened. This is because their depth of 0 is less than 1. However [9, 10, 11] remains unflattened because its depth is 1.
Example 3:
Input
arr = [[1, 2, 3], [4, 5, 6], [7, 8, [9, 10, 11], 12], [13, 14, 15]]
n = 2
Output
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
Explanation
The maximum depth of any subarray is 1. Thus, all of them are flattened.
Constraints:
0 <= count of numbers in arr <= 105
0 <= count of subarrays in arr <= 105
maxDepth <= 1000
-1000 <= each number <= 1000
0 <= n <= 1000
Solution (Javascript)
/**
* @param {any[]} arr
* @param {number} depth
* @return {any[]}
*/
var flat = function (arr, n) {
if (n == 0) {
return arr;
}
let ans = [];
for (let i = 0; i < arr.length; i++) {
if (n > 0 && Array.isArray(arr[i])) {
ans.push(...flat(arr[i], n - 1));
} else {
ans.push(arr[i]);
}
}
return ans;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).