What is the time complexity of Find Beautiful Indices in the Given Array I?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Find Beautiful Indices in the Given Array I?

This problem uses Two Pointers, String, Binary Search, Rolling Hash, String Matching, Hash Function. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Find Beautiful Indices in the Given Array I?

Insight 1: The problem can be solved by first identifying all possible starting indices for string `a` and string `b` within the given string `s`. This can be done through simple string matching. Insight 2: After identifying all potential starting indices, the condition `|j - i| <= k` becomes the central element for determining beautiful indices. Essentially, we need to check for each `a` index if there is a `b` index within the range `[i-k, i+k]`.

Which companies ask LeetCode 3006. Find Beautiful Indices in the Given Array I in interviews?

This problem is frequently asked at Palantir Technologies, Samsara.

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Find Beautiful Indices in the Given Array I - LeetCode 3006 Solution

Q: How to solve Find Beautiful Indices in the Given Array I?

The solution first finds all starting indices of string `a` and string `b` within string `s`. Then, for each index `i` where `s[i:i + len(a)] == a`, it checks if there exists an index `j` where `s[j:j + len(b)] == b` and `|j - i| <= k`. The algorithm iterates through the `a` indices and uses a pointer to optimize the search for a valid `b` index within the specified range.

Q: Which companies ask LeetCode 3006. Find Beautiful Indices in the Given Array I in interviews?

This problem is frequently asked at Palantir Technologies, Samsara.

Find Beautiful Indices in the Given Array I - Complete Solution Guide

Find Beautiful Indices in the Given Array I is LeetCode problem 3006, 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 string s , a string a , a string b , and an integer k . An index i is beautiful if: 0 <= i <= s.length - a.length s[i..(i + a.length - 1)] == a There exists an index j such that: 0 <= j <= s.length - b.length s[j..(j + b.length - 1)] == b |j - i| <= k Return the array that contains beautiful indices in sorted order from smallest to largest . Example 1: Input: s = "isawsquirrelnearmysquirrelhouseohmy", a = "my", b = "squirrel", k = 15 Output: [16,33] Explanation: There a

Detailed Explanation

The problem asks us to find "beautiful indices" within a string `s`. An index `i` is considered beautiful if the substring of `s` starting at `i` and of length `a` is equal to the string `a`, AND there exists another index `j` such that the substring of `s` starting at `j` and of length `b` is equal to the string `b`, AND the absolute difference between `i` and `j` is less than or equal to `k`. The goal is to return a sorted list of all such beautiful indices.

Solution Approach

The solution first finds all starting indices of string `a` and string `b` within string `s`. Then, for each index `i` where `s[i:i + len(a)] == a`, it checks if there exists an index `j` where `s[j:j + len(b)] == b` and `|j - i| <= k`. The algorithm iterates through the `a` indices and uses a pointer to optimize the search for a valid `b` index within the specified range.

Step-by-Step Algorithm

Step 1: Find all starting indices of string `a` in `s` and store them in a list called `a_indices`.
Step 2: Find all starting indices of string `b` in `s` and store them in a list called `b_indices`.
Step 3: Initialize an empty list called `result` to store the beautiful indices.
Step 4: Initialize a pointer `ptr_b` to 0, which will be used to iterate through the `b_indices` list.
Step 5: Iterate through each index `i` in the `a_indices` list.
Step 6: While `ptr_b` is within the bounds of `b_indices` and `b_indices[ptr_b]` is less than `i - k`, increment `ptr_b`. This effectively skips all `b` indices that are too far to the left of the current `a` index.
Step 7: If `ptr_b` is still within the bounds of `b_indices` and `b_indices[ptr_b]` is less than or equal to `i + k`, it means we have found a valid `b` index within the required range. Add `i` to the `result` list.
Step 8: After iterating through all indices in `a_indices`, return the `result` list.

Key Insights

Insight 1: The problem can be solved by first identifying all possible starting indices for string `a` and string `b` within the given string `s`. This can be done through simple string matching.
Insight 2: After identifying all potential starting indices, the condition `|j - i| <= k` becomes the central element for determining beautiful indices. Essentially, we need to check for each `a` index if there is a `b` index within the range `[i-k, i+k]`.
Insight 3: Using two pointers to iterate through the indices of `a` and `b` can help to reduce the number of comparisons, optimizing the time complexity slightly.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Two Pointers, String, Binary Search, Rolling Hash, String Matching, Hash Function.

Companies

Asked at: Palantir Technologies, Samsara.

Find Beautiful Indices in the Given Array I | LeetCode 3006

Medium

Two Pointers String Binary Search Rolling Hash String Matching Hash Function

Time: O(n^2)Space: O(n)

Solution Code

Available in:

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

class Solution {
    public List<Integer> beautifulIndices(String s, String a, String b, int k) {
        int n = s.length();
        int len_a = a.length();
        int len_b = b.length();

        List<Integer> a_indices = new ArrayList<>();
        for (int i = 0; i <= n - len_a; i++) {
            if (s.substring(i, i + len_a).equals(a)) {
                a_indices.add(i);
            }
        }

        List<Integer> b_indices = new ArrayList<>();
        for (int i = 0; i <= n - len_b; i++) {
            if (s.substring(i, i + len_b).equals(b)) {
                b_indices.add(i);
            }
        }
        
        List<Integer> result = new ArrayList<>();
        int ptr_b = 0;
        for (int i : a_indices) {
            while (ptr_b < b_indices.size() && b_indices.get(ptr_b) < i - k) {
                ptr_b += 1;
            }
            
            if (ptr_b < b_indices.size() && b_indices.get(ptr_b) <= i + k) {
                result.add(i);
            }
        }
        
        return result;
    }
}

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Problem Info

Description

Problem 3006: Find Beautiful Indices in the Given Array I - Detailed explanation and solution approaches.

Topics

Two Pointers String Binary Search Rolling Hash String Matching Hash Function

Companies

Palantir TechnologiesSamsara

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:3006

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

8 steps

1
Step 1: Find all starting indices of string `a` in `s` and store them in a list called `a_indices`.
2
Step 2: Find all starting indices of string `b` in `s` and store them in a list called `b_indices`.
3
Step 3: Initialize an empty list called `result` to store the beautiful indices.
4
Step 4: Initialize a pointer `ptr_b` to 0, which will be used to iterate through the `b_indices` list.
5
Step 5: Iterate through each index `i` in the `a_indices` list.
6
Step 6: While `ptr_b` is within the bounds of `b_indices` and `b_indices[ptr_b]` is less than `i - k`, increment `ptr_b`. This effectively skips all `b` indices that are too far to the left of the current `a` index.
7
Step 7: If `ptr_b` is still within the bounds of `b_indices` and `b_indices[ptr_b]` is less than or equal to `i + k`, it means we have found a valid `b` index within the required range. Add `i` to the `result` list.
8
Step 8: After iterating through all indices in `a_indices`, return the `result` list.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the nested loops and the string comparison operations. The outer loop iterates through `a_indices`, which in the worst case can have `n` elements. The inner while loop, while nested, only increments `ptr_b` at most `n` times overall. Inside these loops are substring operations, which have a cost of O(len(a)) or O(len(b)). Therefore, the overall time complexity is O(n*(len(a) + len(b))). The space complexity is determined by the size of `a_indices` and `b_indices`, which can each grow to a maximum size of `n`.

📖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

Binary Search

Instead of using a two-pointer approach to find a `b` index within the range `[i-k, i+k]`, binary search could be used on the `b_indices` list for each `a` index `i`. This would involve searching for the first `b` index greater than or equal to `i-k` and then checking if it's within the range `[i-k, i+k]`.

Time: O(m log n) where m is the number of indices in 'a_indices' and n is the number of indices in 'b_indices'.Space: O(n) same as the original approach

Trade-offs: The binary search approach could be faster than the two-pointer approach if `a_indices` is significantly smaller than `b_indices`, as the logarithmic time complexity of binary search may outweigh the linear cost of the two-pointer traversal.

✨

Key Insights

1Insight 1: The problem can be solved by first identifying all possible starting indices for string `a` and string `b` within the given string `s`. This can be done through simple string matching.
2Insight 2: After identifying all potential starting indices, the condition `|j - i| <= k` becomes the central element for determining beautiful indices. Essentially, we need to check for each `a` index if there is a `b` index within the range `[i-k, i+k]`.
3Insight 3: Using two pointers to iterate through the indices of `a` and `b` can help to reduce the number of comparisons, optimizing the time complexity slightly.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: `a` or `b` is an empty string. The given problem constraints specify that a.length, b.length >=1, so no edge case needs to be considered.
Edge case 2: `k` is very large (close to the length of `s`). The algorithm correctly handles this as the inner while loop will ensure `ptr_b` iterates to the maximum possible index within the valid range (i.e., it won't go out of bounds).
Edge case 3: No beautiful indices exist. The algorithm will correctly return an empty list in this scenario.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly implementing the condition `|j - i| <= k`. Some might implement it as `j - i <= k` which would only search to the right and not the left.

✗Common mistake 2: Not considering the edge case where no beautiful indices exist, and potentially returning incorrect results or errors.

✗Common mistake 3: Forgetting to sort the `result` array before returning it (though the problem states it should be in sorted order, and the algorithm generates them in sorted order already).

✅

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: Finding patterns in DNA sequences. In bioinformatics, algorithms are used to identify specific sequences within large DNA strands. The given problem mimics the scenario to find specific sequence 'a' at location i within 'k' distance of sequence 'b'

🏦Real-world scenario 2: Log analysis. Identifying events in log files that occur within a specific time window of each other. For example, finding error messages that occur within 5 minutes of a user login.

💡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

The two-pointer technique is powerful for sorted arrays. The key insight is recognizing when elements can be processed from both ends simultaneously.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

The most common mistake is not handling the case when pointers meet. Always consider the termination condition.

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: October 5, 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

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

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

Frequently Asked Questions

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

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 Beautiful Indices in the Given Array I"?

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

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