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2420. Find All Good Indices

Description

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

Solutions

Solution 1: Recursion

We define two arrays decr and incr, which represent the longest non-increasing and non-decreasing subarray lengths from left to right and from right to left, respectively.

We traverse the array, updating the decr and incr arrays.

Then we sequentially traverse the index $i$ (where $k\le i \lt n - k$), if $decr[i] \geq k$ and $incr[i] \geq k$, then $i$ is a good index.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array.

  • class Solution {
    public List<Integer> goodIndices(int[] nums, int k) {
        int n = nums.length;
        int[] decr = new int[n];
        int[] incr = new int[n];
        Arrays.fill(decr, 1);
        Arrays.fill(incr, 1);
        for (int i = 2; i < n - 1; ++i) {
            if (nums[i - 1] <= nums[i - 2]) {
                decr[i] = decr[i - 1] + 1;
            }
        }
        for (int i = n - 3; i >= 0; --i) {
            if (nums[i + 1] <= nums[i + 2]) {
                incr[i] = incr[i + 1] + 1;
            }
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = k; i < n - k; ++i) {
            if (decr[i] >= k && incr[i] >= k) {
                ans.add(i);
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    vector<int> goodIndices(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> decr(n, 1);
        vector<int> incr(n, 1);
        for (int i = 2; i < n; ++i) {
            if (nums[i - 1] <= nums[i - 2]) {
                decr[i] = decr[i - 1] + 1;
            }
        }
        for (int i = n - 3; ~i; --i) {
            if (nums[i + 1] <= nums[i + 2]) {
                incr[i] = incr[i + 1] + 1;
            }
        }
        vector<int> ans;
        for (int i = k; i < n - k; ++i) {
            if (decr[i] >= k && incr[i] >= k) {
                ans.push_back(i);
            }
        }
        return ans;
    }
};

  • class Solution:
    def goodIndices(self, nums: List[int], k: int) -> List[int]:
        n = len(nums)
        decr = [1] * (n + 1)
        incr = [1] * (n + 1)
        for i in range(2, n - 1):
            if nums[i - 1] <= nums[i - 2]:
                decr[i] = decr[i - 1] + 1
        for i in range(n - 3, -1, -1):
            if nums[i + 1] <= nums[i + 2]:
                incr[i] = incr[i + 1] + 1
        return [i for i in range(k, n - k) if decr[i] >= k and incr[i] >= k]


  • func goodIndices(nums []int, k int) []int {
	n := len(nums)
	decr := make([]int, n)
	incr := make([]int, n)
	for i := range decr {
		decr[i] = 1
		incr[i] = 1
	}
	for i := 2; i < n; i++ {
		if nums[i-1] <= nums[i-2] {
			decr[i] = decr[i-1] + 1
		}
	}
	for i := n - 3; i >= 0; i-- {
		if nums[i+1] <= nums[i+2] {
			incr[i] = incr[i+1] + 1
		}
	}
	ans := []int{}
	for i := k; i < n-k; i++ {
		if decr[i] >= k && incr[i] >= k {
			ans = append(ans, i)
		}
	}
	return ans
}

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