What is the time complexity of Can Make Palindrome from Substring?

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

What data structures are used to solve Can Make Palindrome from Substring?

This problem uses Array, Hash Table, String, Bit Manipulation, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Can Make Palindrome from Substring?

Insight 1: A string can be rearranged into a palindrome if and only if the number of characters appearing an odd number of times is at most 1. Since we can replace characters, we can tolerate more characters appearing an odd number of times, as long as we can 'fix' them with the allowed replacements. Insight 2: The number of characters appearing an odd number of times in a substring can be efficiently calculated using bit manipulation with prefix XOR. Each bit in an integer can represent whether a character ('a' to 'z') appears an odd or even number of times. XORing the prefix masks allows us to compute the counts for each substring.

Which companies ask LeetCode 1177. Can Make Palindrome from Substring in interviews?

This problem is frequently asked at Akuna Capital.

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Can Make Palindrome from Substring - LeetCode 1177 Solution

Can Make Palindrome from Substring - Complete Solution Guide

Can Make Palindrome from Substring is LeetCode problem 1177, 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 string s and array queries where queries[i] = [left i , right i , k i ] . We may rearrange the substring s[left i ...right i ] for each query and then choose up to k i of them to replace with any lowercase English letter. If the substring is possible to be a palindrome string after the operations above, the result of the query is true . Otherwise, the result is false . Return a boolean array answer where answer[i] is the result of the i th query queries[i] . Note that each letter

Detailed Explanation

The problem asks us to determine, for each query in a given list of queries, whether a substring of a string `s` can be rearranged into a palindrome after replacing at most `k` characters with any lowercase English letters. Each query consists of a left index `left_i`, a right index `right_i`, and an integer `k_i`. We need to return a boolean array where each element indicates whether the substring `s[left_i...right_i]` can be made a palindrome with at most `k_i` replacements.

Solution Approach

The solution utilizes the prefix XOR technique. It computes a prefix mask array where each element `prefix_masks[i]` represents the XOR of all characters from `s[0]` to `s[i-1]`. For each query, the XOR of `prefix_masks[r+1]` and `prefix_masks[l]` gives a mask representing the substring `s[l...r]`. The number of set bits (bits equal to 1) in this substring mask corresponds to the number of characters that appear an odd number of times in the substring. If half of this number is less than or equal to `k`, then the substring can be made a palindrome with the given number of replacements.

Step-by-Step Algorithm

Step 1: Create a prefix mask array `prefix_masks` of size `n + 1`, initialized to 0, where `n` is the length of the string `s`.
Step 2: Iterate through the string `s` from `i = 0` to `n - 1`. For each character `s[i]`, update `prefix_masks[i + 1]` by XORing it with `prefix_masks[i]` and the bit corresponding to the character `s[i]` (calculated as `1 << (ord(s[i]) - ord('a'))`).
Step 3: Create a boolean array `result` to store the results of the queries.
Step 4: Iterate through the `queries` array. For each query `[l, r, k]`, calculate the substring mask by XORing `prefix_masks[r + 1]` and `prefix_masks[l]`.
Step 5: Count the number of set bits (1s) in the `substring_mask`. This represents the number of characters that appear an odd number of times in the substring.
Step 6: Check if `odd_count / 2 <= k`. If it is, then the substring can be made a palindrome with `k` replacements, so append `true` to the `result` array. Otherwise, append `false`.
Step 7: Return the `result` array.

Key Insights

Insight 1: A string can be rearranged into a palindrome if and only if the number of characters appearing an odd number of times is at most 1. Since we can replace characters, we can tolerate more characters appearing an odd number of times, as long as we can 'fix' them with the allowed replacements.
Insight 2: The number of characters appearing an odd number of times in a substring can be efficiently calculated using bit manipulation with prefix XOR. Each bit in an integer can represent whether a character ('a' to 'z') appears an odd or even number of times. XORing the prefix masks allows us to compute the counts for each substring.
Insight 3: The number of required replacements to make a string palindrome is half the number of characters that occur an odd number of times (rounded up if needed, but we can handle by integer division since we need only to make sure if the replacements required <= k).

Complexity Analysis

Time Complexity: O(m + n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, String, Bit Manipulation, Prefix Sum.

Companies

Asked at: Akuna Capital.

Can Make Palindrome from Substring | LeetCode 1177

Medium

Array Hash Table String Bit Manipulation Prefix Sum

Time: O(m + n)Space: O(n)

Solution Code

Available in:

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

class Solution {
    public List<Boolean> canMakePaliQueries(String s, int[][] queries) {
        int n = s.length();
        int[] prefixMasks = new int[n + 1];

        for (int i = 0; i < n; i++) {
            prefixMasks[i + 1] = prefixMasks[i] ^ (1 << (s.charAt(i) - 'a'));
        }

        List<Boolean> result = new ArrayList<>();
        for (int[] query : queries) {
            int l = query[0];
            int r = query[1];
            int k = query[2];

            int substringMask = prefixMasks[r + 1] ^ prefixMasks[l];
            int oddCount = Integer.bitCount(substringMask);
            result.add(oddCount / 2 <= k);
        }

        return result;
    }
}

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Problem Info

Description

Problem 1177: Can Make Palindrome from Substring - Detailed explanation and solution approaches.

Topics

Array Hash Table String Bit Manipulation Prefix Sum

Companies

Akuna Capital

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1177

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

7 steps

1
Step 1: Create a prefix mask array `prefix_masks` of size `n + 1`, initialized to 0, where `n` is the length of the string `s`.
2
Step 2: Iterate through the string `s` from `i = 0` to `n - 1`. For each character `s[i]`, update `prefix_masks[i + 1]` by XORing it with `prefix_masks[i]` and the bit corresponding to the character `s[i]` (calculated as `1 << (ord(s[i]) - ord('a'))`).
3
Step 3: Create a boolean array `result` to store the results of the queries.
4
Step 4: Iterate through the `queries` array. For each query `[l, r, k]`, calculate the substring mask by XORing `prefix_masks[r + 1]` and `prefix_masks[l]`.
5
Step 5: Count the number of set bits (1s) in the `substring_mask`. This represents the number of characters that appear an odd number of times in the substring.
6
Step 6: Check if `odd_count / 2 <= k`. If it is, then the substring can be made a palindrome with `k` replacements, so append `true` to the `result` array. Otherwise, append `false`.
7
Step 7: Return the `result` array.

💡

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(m + n)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because we iterate through the string and the queries once. The space complexity is linear because we store a prefix mask array of size n+1.

📖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

Character Counting with HashMap

Instead of using bit manipulation, we can use a hash map to count the occurrences of each character in the substring. Then, count the number of characters with odd occurrences and check if `odd_count / 2 <= k`.

Time: O(m * len), where 'len' is the average length of the substring for each query, since we need to construct the HashMap for each substring. In worst case len is n (length of s), this would be O(m*n).Space: O(len), where 'len' is the average length of the substring for each query. In worst case len is n (length of s), this would be O(n).

Trade-offs: The hash map approach is more intuitive but has a higher time complexity than the bit manipulation approach, especially if the substring lengths are significant. It's only worthwhile if memory is extremely constrained and the code needs to be understandable.

✨

Key Insights

1Insight 1: A string can be rearranged into a palindrome if and only if the number of characters appearing an odd number of times is at most 1. Since we can replace characters, we can tolerate more characters appearing an odd number of times, as long as we can 'fix' them with the allowed replacements.
2Insight 2: The number of characters appearing an odd number of times in a substring can be efficiently calculated using bit manipulation with prefix XOR. Each bit in an integer can represent whether a character ('a' to 'z') appears an odd or even number of times. XORing the prefix masks allows us to compute the counts for each substring.
3Insight 3: The number of required replacements to make a string palindrome is half the number of characters that occur an odd number of times (rounded up if needed, but we can handle by integer division since we need only to make sure if the replacements required <= k).

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

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

Edge Cases

Edge case 1: Empty substring (l == r). The substring consists of only one character which is always a palindrome. The odd count will be 1 and 1//2 <= k is true if k >=0, which the constraints ensure is always the case.
Edge case 2: k = 0. If k is 0, we can only make the substring a palindrome if the number of characters appearing an odd number of times is at most 1. The algorithm correctly handles this case since it checks if `odd_count / 2 <= k`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not understanding the palindrome condition. Forgetting that a string is a palindrome if the number of characters occurring an odd number of times is 0 or 1. Also failing to realize we can fix up to `k` character mismatches by replacing them. Thus the condition of required replacements being less than or equal to k must be checked.

✗Common mistake 2: Incorrectly calculating the substring mask. Not using prefix XOR correctly or using the wrong indices for XORing the prefix masks, leading to incorrect counts of odd occurrences. Off-by-one errors in calculating the range of the substring.

✗Common mistake 3: Forgetting to integer divide the `odd_count` by 2 before comparing with k. Without this division, the condition `odd_count <= k` is wrong. We need to make `odd_count/2` replacements. Remember if a character occurs odd number of times, we can use two of them to make a pair.

✅

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: Text editors with palindrome detection. This algorithm can be used to quickly check if a selected text region can be transformed into a palindrome with a given number of changes, improving the user experience.

🏦Real-world scenario 2: DNA sequence analysis. In bioinformatics, similar techniques can be applied to analyze DNA sequences and identify regions that are close to being palindromic with a certain number of mutations, which can be relevant for understanding gene regulation or identifying potential drug targets.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

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

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: September 6, 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

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

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Can Make Palindrome from Substring"?

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 hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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