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Formatted question description: https://leetcode.ca/all/2420.html

# 2420. Find All Good Indices

• Difficulty: Medium.
• Related Topics: .
• Similar Questions: Find Good Days to Rob the Bank, Abbreviating the Product of a Range.

## Problem

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 <= 10^5

• 1 <= nums[i] <= 10^6

• 1 <= k <= n / 2

## Solution (Java, C++, Python)

• 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) {
}
}
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
}


Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).