What is the time complexity of Find All Good Indices?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Find All Good Indices?

This problem uses Array, Dynamic Programming, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Find All Good Indices?

Insight 1: The core idea is to precompute two arrays: one storing the lengths of non-increasing sequences ending at each index from the left, and another storing the lengths of non-decreasing sequences ending at each index from the right. This avoids repeatedly checking the sequences around each possible index. Insight 2: Dynamic Programming can be efficiently applied to build the non-increasing and non-decreasing arrays. This reduces redundant computations.

Which companies ask LeetCode 2420. Find All Good Indices in interviews?

This problem is frequently asked at ByteDance, Goldman Sachs.

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Find All Good Indices - LeetCode 2420 Solution

Q: How to solve Find All Good Indices?

The solution uses dynamic programming to efficiently determine if each index is "good". It calculates two arrays: `non_inc` and `non_dec`. `non_inc[i]` stores the length of the longest non-increasing sequence ending at index `i`. `non_dec[i]` stores the length of the longest non-decreasing sequence starting at index `i`. Then, the code iterates through all possible "good" indices (from `k` to `n - k - 1`) and checks if `non_inc[i - 1]` and `non_dec[i + 1]` are both greater than or equal to `k`. If they are, the index `i` is added to the result list.

Find All Good Indices - Complete Solution Guide

Find All Good Indices is LeetCode problem 2420, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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. 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

Detailed Explanation

The problem asks us to identify "good" indices in an array `nums`. An index `i` is considered "good" if it satisfies two conditions: 1) The `k` elements immediately before index `i` are in non-increasing order (each element is less than or equal to the preceding element), and 2) The `k` elements immediately after index `i` are in non-decreasing order (each element is greater than or equal to the preceding element). The index `i` must also be within the range `k <= i < n - k`, where `n` is the length of `nums`. The function should return a sorted list of all good indices.

Solution Approach

The solution uses dynamic programming to efficiently determine if each index is "good". It calculates two arrays: `non_inc` and `non_dec`. `non_inc[i]` stores the length of the longest non-increasing sequence ending at index `i`. `non_dec[i]` stores the length of the longest non-decreasing sequence starting at index `i`. Then, the code iterates through all possible "good" indices (from `k` to `n - k - 1`) and checks if `non_inc[i - 1]` and `non_dec[i + 1]` are both greater than or equal to `k`. If they are, the index `i` is added to the result list.

Step-by-Step Algorithm

Step 1: Initialize two arrays, `non_inc` and `non_dec`, of the same size as `nums`, with all elements set to 1. This represents the base case where each element by itself forms a sequence of length 1.
Step 2: Calculate the `non_inc` array from left to right. For each index `i`, if `nums[i] <= nums[i - 1]`, then the non-increasing sequence extends, and `non_inc[i]` becomes `non_inc[i - 1] + 1`. Otherwise, `non_inc[i]` remains 1, indicating a new sequence.
Step 3: Calculate the `non_dec` array from right to left. For each index `i` (starting from `n - 2`), if `nums[i] <= nums[i + 1]`, then the non-decreasing sequence extends, and `non_dec[i]` becomes `non_dec[i + 1] + 1`. Otherwise, `non_dec[i]` remains 1, indicating a new sequence.
Step 4: Iterate through the indices from `k` to `n - k - 1`. For each index `i`, check if `non_inc[i - 1] >= k` and `non_dec[i + 1] >= k`. If both conditions are true, add index `i` to the result list.
Step 5: Return the result list, which contains all the good indices.

Key Insights

Insight 1: The core idea is to precompute two arrays: one storing the lengths of non-increasing sequences ending at each index from the left, and another storing the lengths of non-decreasing sequences ending at each index from the right. This avoids repeatedly checking the sequences around each possible index.
Insight 2: Dynamic Programming can be efficiently applied to build the non-increasing and non-decreasing arrays. This reduces redundant computations.
Insight 3: Only indices `k <= i < n - k` need to be checked, as indices outside this range cannot satisfy the condition of having `k` elements before and after them.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Prefix Sum.

Companies

Asked at: ByteDance, Goldman Sachs.

Find All Good Indices | LeetCode 2420

Medium

Array Dynamic Programming Prefix Sum

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<Integer> goodIndices(int[] nums, int k) {
        int n = nums.length;

        int[] non_inc = new int[n];
        for (int i = 0; i < n; i++) {
            non_inc[i] = 1;
        }
        for (int i = 1; i < n; i++) {
            if (nums[i] <= nums[i - 1]) {
                non_inc[i] = non_inc[i - 1] + 1;
            }
        }

        int[] non_dec = new int[n];
        for (int i = 0; i < n; i++) {
            non_dec[i] = 1;
        }
        for (int i = n - 2; i >= 0; i--) {
            if (nums[i] <= nums[i + 1]) {
                non_dec[i] = non_dec[i + 1] + 1;
            }
        }

        List<Integer> result = new ArrayList<>();
        for (int i = k; i < n - k; i++) {
            if (non_inc[i - 1] >= k && non_dec[i + 1] >= k) {
                result.add(i);
            }
        }

        return result;
    }
}

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Problem Info

Description

Problem 2420: Find All Good Indices - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Prefix Sum

Companies

ByteDanceGoldman Sachs

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2420

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Initialize two arrays, `non_inc` and `non_dec`, of the same size as `nums`, with all elements set to 1. This represents the base case where each element by itself forms a sequence of length 1.
2
Step 2: Calculate the `non_inc` array from left to right. For each index `i`, if `nums[i] <= nums[i - 1]`, then the non-increasing sequence extends, and `non_inc[i]` becomes `non_inc[i - 1] + 1`. Otherwise, `non_inc[i]` remains 1, indicating a new sequence.
3
Step 3: Calculate the `non_dec` array from right to left. For each index `i` (starting from `n - 2`), if `nums[i] <= nums[i + 1]`, then the non-decreasing sequence extends, and `non_dec[i]` becomes `non_dec[i + 1] + 1`. Otherwise, `non_dec[i]` remains 1, indicating a new sequence.
4
Step 4: Iterate through the indices from `k` to `n - k - 1`. For each index `i`, check if `non_inc[i - 1] >= k` and `non_dec[i + 1] >= k`. If both conditions are true, add index `i` to the result list.
5
Step 5: Return the result list, which contains all the good indices.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because the dominant operations are the three loops that iterate through the array `nums`. The space complexity is O(n) due to the storage of `non_inc` and `non_dec` which scale linearly with input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

For each index `i` in the range `k <= i < n - k`, explicitly check if the `k` elements before and after `i` are in non-increasing and non-decreasing order, respectively.

Time: O(n*k), where n is the length of nums.Space: O(1). No extra space is used.

Trade-offs: This approach is simpler to understand but less efficient, especially when `k` is large. The main solution using dynamic programming is more efficient as it precomputes the non-increasing/decreasing sequences.

✨

Key Insights

1Insight 1: The core idea is to precompute two arrays: one storing the lengths of non-increasing sequences ending at each index from the left, and another storing the lengths of non-decreasing sequences ending at each index from the right. This avoids repeatedly checking the sequences around each possible index.
2Insight 2: Dynamic Programming can be efficiently applied to build the non-increasing and non-decreasing arrays. This reduces redundant computations.
3Insight 3: Only indices `k <= i < n - k` need to be checked, as indices outside this range cannot satisfy the condition of having `k` elements before and after them.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `nums` is empty or `k` is greater than `n / 2`. In such cases, the loops will not be executed, and an empty list will be returned which is the correct behavior since no indices will be good.
Edge case 2: `k` is 1. This scenario effectively means we need to check if adjacent elements on either side of an index are non-increasing/non-decreasing. The algorithm correctly handles this by initializing `non_inc` and `non_dec` to 1 and updating based on adjacent element comparison.
Edge case 3: `nums` contains only one element. In this case, because of the constraint `3 <= n`, this cannot occur. If `n` was allowed to be less than 3, the algorithm would correctly handle the case by returning an empty array because the condition `k <= i < n - k` would never be true, thus the loop iterating from k to n-k-1 would not execute.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the `non_inc` and `non_dec` arrays. It's crucial to ensure that the logic for extending the sequences is correct (i.e., using `<=` for both non-increasing and non-decreasing conditions after switching direction).

✗Common mistake 2: Failing to properly consider the range of possible "good" indices. The indices must satisfy `k <= i < n - k`.

✗Common mistake 3: Not initializing the `non_inc` and `non_dec` arrays with 1, causing incorrect sequence length calculations.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Analyzing stock prices to identify periods of consistent decline (non-increasing) followed by a period of consistent increase (non-decreasing). This can help identify potential turning points in the market.

🏦Real-world scenario 2: Processing sensor data to detect patterns of decreasing values followed by increasing values. This can be used in anomaly detection or predictive maintenance.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 20, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Find All Good Indices"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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