Follow the hints! - Part 2

If you follow the hints on the problem, it really becomes super straightforward. Calculate the max chain of good indices towards the left and towards the right in two different O(N)-time loops. Iterate a third time to create the return list. Rest is history. Code is down below, cheers, ACC.

Find All Good Indices - LeetCode

2420. Find All Good Indices

Medium

You are given a 0-indexed integer array nums of size n and a positive integer k.

We call an index i in the range k <= i < n - k good if the following conditions are satisfied:

• The k elements that are just before the index i are in non-increasing order.
• The k elements that are just after the index i are in non-decreasing order.

Return an array of all good indices sorted in increasing order.

 

Example 1:

Input: nums = [2,1,1,1,3,4,1], k = 2
Output: [2,3]
Explanation: There are two good indices in the array:
- Index 2. The subarray [2,1] is in non-increasing order, and the subarray [1,3] is in non-decreasing order.
- Index 3. The subarray [1,1] is in non-increasing order, and the subarray [3,4] is in non-decreasing order.
Note that the index 4 is not good because [4,1] is not non-decreasing.

Example 2:

Input: nums = [2,1,1,2], k = 2
Output: []
Explanation: There are no good indices in this array.

 

Constraints:

• n == nums.length
• 3 <= n <= 105
• 1 <= nums[i] <= 106
• 1 <= k <= n / 2

Accepted

10,729

Submissions

31,856

public IList GoodIndices(int[] nums, int k)
{
    int[] left = new int[nums.Length];
    left[0] = 0;
    left[1] = 1;

    for (int i = 2; i < nums.Length; i++)
        left[i] = (nums[i - 1] <= nums[i - 2]) ? left[i - 1] + 1 : 1;

    int[] right = new int[nums.Length];
    right[right.Length - 1] = 0;
    right[right.Length - 2] = 1;

    for (int i = nums.Length - 3; i >= 0; i--)
        right[i] = (nums[i + 1] <= nums[i + 2]) ? right[i + 1] + 1 : 1;

    List retVal = new List();
    for (int i = 0; i < nums.Length; i++)
        if (left[i] >= k && right[i] >= k) retVal.Add(i);

    return retVal;
}