Leetcode #2420: Find All Good Indices
In this guide, we solve Leetcode #2420 Find All Good Indices in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Dynamic Programming, Prefix Sum
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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.
Python Solution
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]
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.