How do you solve Find All Good Indices on LeetCode?

Find All Good Indices (LeetCode #2420) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Precomputation can significantly reduce the time complexity.

What is the brute force solution for Find All Good Indices?

The brute force approach for Find All Good Indices is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each index to see if the conditions for being a good index are met. This involves checking the k elements before and after the index for the required order. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find All Good Indices?

Find All Good Indices uses the following concepts: Array, Dynamic Programming, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Prefix Sum.

What are the interview tips for Find All Good Indices?

Explain your thought process clearly while coding. Consider edge cases and discuss them with the interviewer. Ask clarifying questions if the problem statement is not clear.

What are common mistakes when solving Find All Good Indices?

Not considering edge cases where k is close to n. Overlooking the need for separate checks for left and right conditions.

#2420

Find All Good Indices

Medium

ArrayDynamic ProgrammingPrefix SumDynamic ProgrammingPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can optimize by precomputing the non-increasing and non-decreasing conditions for the elements before and after each index. This allows us to check the conditions in constant time for each index.

⚙️

Algorithm

4 steps

1Step 1: Create two arrays, left and right, to store the lengths of non-increasing and non-decreasing sequences.
2Step 2: Fill the left array by iterating from left to right, counting how many elements are non-increasing.
3Step 3: Fill the right array by iterating from right to left, counting how many elements are non-decreasing.
4Step 4: Iterate through the indices from k to n-k-1 and check if left[i] >= k and right[i] >= k.

solution.py21 lines

1# Full working Python code
2
3def findGoodIndices(nums, k):
4    n = len(nums)
5    left = [0] * n
6    right = [0] * n
7    good_indices = []
8
9    for i in range(1, n):
10        if nums[i] <= nums[i - 1]:
11            left[i] = left[i - 1] + 1
12
13    for i in range(n - 2, -1, -1):
14        if nums[i] <= nums[i + 1]:
15            right[i] = right[i + 1] + 1
16
17    for i in range(k, n - k):
18        if left[i] >= k and right[i] >= k:
19            good_indices.append(i)
20
21    return good_indices

ℹ

Complexity note: The time complexity is O(n) because we only make three passes through the array: one for the left array, one for the right array, and one to collect good indices.

1Precomputation can significantly reduce the time complexity.
2Understanding the problem constraints helps in designing efficient algorithms.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.