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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2401_2500.s2420_find_all_good_indices;

// #Medium #Array #Dynamic_Programming #Prefix_Sum
// #2022_11_18_Time_13_ms_(82.76%)_Space_111.2_MB_(78.83%)

import java.util.ArrayList;
import java.util.List;

/**
 * 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`
**/
public class Solution {
    public List goodIndices(int[] nums, int k) {
        int amount = 1;
        List result = new ArrayList<>();
        if (k == 1) {
            for (int i = 1; i < nums.length - 1; i++) {
                result.add(i);
            }
            return result;
        }
        for (int left = 1, right = k + 2; right < nums.length; left++, right++) {
            if (nums[left - 1] >= nums[left] && nums[right - 1] <= nums[right]) {
                amount++;
                if (amount >= k) {
                    result.add(left + 1);
                }
            } else {
                amount = 1;
            }
        }
        return result;
    }
}