2420

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

Companies: Goldman Sachs

Related Topics:
Array, Dynamic Programming, Prefix Sum

Similar Questions:

• Find Good Days to Rob the Bank (Medium)
• Abbreviating the Product of a Range (Hard)
• Count the Number of K-Big Indices (Hard)

Solution 1. Fixed-length Sliding Window

// OJ: https://leetcode.com/problems/find-all-good-indices
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(K)
class Solution {
public:
    vector<int> goodIndices(vector<int>& A, int k) {
        vector<int> ans;
        deque<int> left, right;
        for (int i = 0, N = A.size(); i < N; ++i) {
            if (i - k - 1 >= 0) {
                while (left.size() && A[left.back()] < A[i - k - 1]) left.pop_back();
                left.push_back(i - k - 1);
            }
            if (left.size() && left.front() < i - 2 * k) left.pop_front();
            while (right.size() && A[right.back()] > A[i]) right.pop_back();
            right.push_back(i);
            if (right.front() == i - k) right.pop_front();
            if (left.size() == k && right.size() == k) ans.push_back(i - k);
        }
        return ans;
    }
};

Solution 2. DP

// OJ: https://leetcode.com/problems/find-all-good-indices
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<int> goodIndices(vector<int>& A, int k) {
        int N = A.size();
        vector<int> ans, left(N, 1), right(N, 1);
        for (int i = 1; i < N; ++i) {
            if (A[i] <= A[i - 1]) left[i] = left[i - 1] + 1;
        }
        for (int i = N - 2; i >= 0; --i) {
            if (A[i] <= A[i + 1]) right[i] = right[i + 1] + 1;
        }
        for (int i = k; i < N - k; ++i) {
            if (left[i - 1] >= k && right[i + 1] >= k) ans.push_back(i);
        }
        return ans;
    }
};