What is the time complexity of Minimum Swaps to Make Strings Equal?

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

What data structures are used to solve Minimum Swaps to Make Strings Equal?

This problem uses Math, String, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Swaps to Make Strings Equal?

Insight 1: The problem is fundamentally about counting mismatches. Specifically, we need to count the number of 'x' in `s1` that correspond to 'y' in `s2` and vice-versa. These counts determine the swap operations needed. Insight 2: The sum of mismatched 'x' and 'y' must be even for the strings to be potentially equalizable. If the sum is odd, it's impossible to equalize the strings because each swap corrects two mismatches.

Which companies ask LeetCode 1247. Minimum Swaps to Make Strings Equal in interviews?

This problem is frequently asked at J.P. Morgan.

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Minimum Swaps to Make Strings Equal - LeetCode 1247 Solution

Minimum Swaps to Make Strings Equal - Complete Solution Guide

Minimum Swaps to Make Strings Equal is LeetCode problem 1247, 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 two strings s1 and s2 of equal length consisting of letters "x" and "y" only . Your task is to make these two strings equal to each other. You can swap any two characters that belong to different strings, which means: swap s1[i] and s2[j] . Return the minimum number of swaps required to make s1 and s2 equal, or return -1 if it is impossible to do so. Example 1: Input: s1 = "xx", s2 = "yy" Output: 1 Explanation: Swap s1[0] and s2[1], s1 = "yx", s2 = "yx". Example 2: Input: s1 = "xy"

Detailed Explanation

The problem requires finding the minimum number of swaps between characters in two strings, `s1` and `s2`, to make them equal. The strings consist only of 'x' and 'y' characters, and swaps can only occur between `s1` and `s2` (not within a single string). If it's impossible to make the strings equal, return -1. Strings have equal length.

Solution Approach

The solution counts the number of 'xy' mismatches (`xy_mismatches`) and 'yx' mismatches (`yx_mismatches`). If the total number of mismatches is odd, it's impossible to equalize the strings, and -1 is returned. Otherwise, it calculates the number of swaps needed. Pairs of 'xy' and 'yx' mismatches are handled directly (e.g., two 'xy' mismatches require one swap). Finally, if there is a single remaining 'xy' and a single remaining 'yx' (resulting in xy_mismatches % 2 == 1), it necessitates two additional swaps.

Step-by-Step Algorithm

Step 1: Initialize `xy_mismatches` and `yx_mismatches` to 0.
Step 2: Iterate through the strings `s1` and `s2` simultaneously using `zip` (or index-based iteration in other languages).
Step 3: In each iteration, compare the characters at the current index. If they are different, increment `xy_mismatches` if `s1[i]` is 'x', or increment `yx_mismatches` if `s1[i]` is 'y'.
Step 4: After the loop, check if the sum of `xy_mismatches` and `yx_mismatches` is odd. If it is, return -1.
Step 5: Calculate the minimum number of swaps as `(xy_mismatches // 2) + (yx_mismatches // 2)`. This accounts for pairs of 'xy' and 'yx' mismatches requiring one swap each.
Step 6: If `xy_mismatches % 2 == 1`, it means there's one remaining 'xy' mismatch and one remaining 'yx' mismatch, which requires two additional swaps. Add 2 to the swap count.
Step 7: Return the calculated swap count.

Key Insights

Insight 1: The problem is fundamentally about counting mismatches. Specifically, we need to count the number of 'x' in `s1` that correspond to 'y' in `s2` and vice-versa. These counts determine the swap operations needed.
Insight 2: The sum of mismatched 'x' and 'y' must be even for the strings to be potentially equalizable. If the sum is odd, it's impossible to equalize the strings because each swap corrects two mismatches.
Insight 3: Two 'xy' mismatches can be corrected with one swap. Two 'yx' mismatches can also be corrected with one swap. If we have a single 'xy' and a single 'yx' mismatch, it requires two swaps.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Math, String, Greedy.

Companies

Asked at: J.P. Morgan.

Minimum Swaps to Make Strings Equal | LeetCode 1247

Medium

Math String Greedy

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minimumSwap(String s1, String s2) {
        int xy_mismatches = 0;
        int yx_mismatches = 0;

        for (int i = 0; i < s1.length(); i++) {
            char char1 = s1.charAt(i);
            char char2 = s2.charAt(i);

            if (char1 != char2) {
                if (char1 == 'x') {
                    xy_mismatches++;
                } else {
                    yx_mismatches++;
                }
            }
        }

        if ((xy_mismatches + yx_mismatches) % 2 != 0) {
            return -1;
        }

        int swaps = (xy_mismatches / 2) + (yx_mismatches / 2);
        
        if (xy_mismatches % 2 == 1) {
            swaps += 2;
        }

        return swaps;
    }
}

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Problem Info

Description

Problem 1247: Minimum Swaps to Make Strings Equal - Detailed explanation and solution approaches.

Topics

Math String Greedy

Companies

J.P. Morgan

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1247

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: Initialize `xy_mismatches` and `yx_mismatches` to 0.
2
Step 2: Iterate through the strings `s1` and `s2` simultaneously using `zip` (or index-based iteration in other languages).
3
Step 3: In each iteration, compare the characters at the current index. If they are different, increment `xy_mismatches` if `s1[i]` is 'x', or increment `yx_mismatches` if `s1[i]` is 'y'.
4
Step 4: After the loop, check if the sum of `xy_mismatches` and `yx_mismatches` is odd. If it is, return -1.
5
Step 5: Calculate the minimum number of swaps as `(xy_mismatches // 2) + (yx_mismatches // 2)`. This accounts for pairs of 'xy' and 'yx' mismatches requiring one swap each.
6
Step 6: If `xy_mismatches % 2 == 1`, it means there's one remaining 'xy' mismatch and one remaining 'yx' mismatch, which requires two additional swaps. Add 2 to the swap count.
7
Step 7: Return the calculated swap count.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operation is iterating through the strings, which takes O(n) time. The space used by the variables `xy_mismatches`, `yx_mismatches`, and `swaps` does not depend on the input size, making the space complexity O(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

Using Counter (Python)

Could concatenate s1 and s2 into a single string and then use collections.Counter to count the occurrences of x at different parity indexes. The idea is to count x's at even and odd indexes and infer mismatched pairs from this count.

Time: O(n), same as the main solutionSpace: O(1), since the counter will only store counts for 'x' and 'y'

Trade-offs: This approach might be considered slightly less readable than the main solution as it obscures the core logic of comparing characters directly. It's still O(n), but the constant factor might be slightly larger.

✨

Key Insights

1Insight 1: The problem is fundamentally about counting mismatches. Specifically, we need to count the number of 'x' in `s1` that correspond to 'y' in `s2` and vice-versa. These counts determine the swap operations needed.
2Insight 2: The sum of mismatched 'x' and 'y' must be even for the strings to be potentially equalizable. If the sum is odd, it's impossible to equalize the strings because each swap corrects two mismatches.
3Insight 3: Two 'xy' mismatches can be corrected with one swap. Two 'yx' mismatches can also be corrected with one swap. If we have a single 'xy' and a single 'yx' mismatch, it requires two swaps.

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

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

Edge Cases

Edge case 1: Empty strings: The provided code implicitly handles empty strings correctly because the loop won't execute, and both mismatch counts will remain 0. This leads to 0 swaps.
Edge case 2: Strings are already equal: In this case, both `xy_mismatches` and `yx_mismatches` will be 0, and the algorithm will correctly return 0 swaps.
Edge case 3: Impossible to equalize (odd number of total mismatches): The code explicitly checks for this condition and returns -1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the number of swaps when there is a single remaining 'xy' and a single remaining 'yx'. Forgetting to add 2 in this case.

✗Common mistake 2: Failing to check if the sum of mismatches is even. This results in incorrect output or runtime errors.

✗Common mistake 3: Trying to optimize prematurely. The core algorithm is already O(n) and fairly straightforward; excessive optimization efforts might introduce bugs and reduce readability.

✅

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: Data synchronization: Ensuring two datasets are consistent by identifying and correcting discrepancies using a minimal number of operations (swaps).

🏦Real-world scenario 2: Error correction: In communication systems, similar techniques can be used to correct bit errors in transmitted data by identifying and flipping incorrect bits, with the goal of minimizing the number of flips.

💡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

Always test your solution with edge cases: empty input, single element, and boundary values.

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: November 15, 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 13

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

@CodeNinjaJul 6

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

@TechLeadJun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

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 "Minimum Swaps to Make Strings Equal"?

For this medium-level Math 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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