LeetCode 2420: Find All Good Indices
LeetCode 2420 Solution Explanation
Explanation
To solve this problem, we iterate through the array and check each index satisfying the given conditions. We can use two pointers to traverse the k elements before and after the current index, ensuring they are in non-increasing and non-decreasing order respectively. If both conditions are met, we add the index to the result array.
- Start by initializing an empty list to store the good indices.
- Iterate through the array from index
kton-k. - For each index
i, check the k elements before and afterito ensure they are in the required order. - If both conditions are satisfied, add
ito the result list. - Finally, return the list of good indices sorted in increasing order.
The time complexity of this solution is O(n) as we iterate through the entire array once. The space complexity is O(1) as we are not using any additional space proportional to the input size.
LeetCode 2420 Solutions in Java, C++, Python
class Solution {
public List<Integer> findGoodIndices(int[] nums, int k) {
List<Integer> goodIndices = new ArrayList<>();
for (int i = k; i < nums.length - k; i++) {
if (checkGoodIndex(nums, i, k)) {
goodIndices.add(i);
}
}
return goodIndices;
}
private boolean checkGoodIndex(int[] nums, int idx, int k) {
for (int j = 1; j <= k; j++) {
if (nums[idx - j] < nums[idx - j + 1] || nums[idx + j] < nums[idx + j - 1]) {
return false;
}
}
return true;
}
}Interactive Code Editor for LeetCode 2420
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