What is the time complexity of Apply Operations to Make Two Strings Equal?

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

What data structures are used to solve Apply Operations to Make Two Strings Equal?

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

What are the key insights for solving Apply Operations to Make Two Strings Equal?

Insight 1: First, identify the indices where `s1` and `s2` differ. Only these differing indices need to be addressed. Insight 2: If the number of differing indices is odd, it's impossible to make the strings equal because each operation flips an even number of bits. So, return -1 in that case.

Which companies ask LeetCode 2896. Apply Operations to Make Two Strings Equal in interviews?

This problem is frequently asked at Zeta.

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Apply Operations to Make Two Strings Equal - LeetCode 2896 Solution

Q: How to solve Apply Operations to Make Two Strings Equal?

The provided solution uses dynamic programming to determine the minimum cost. It first identifies the indices where `s1` and `s2` differ and stores them in a list called `diffs`. If the number of differing indices is odd, it returns -1. Then, it initializes a DP table `dp` where `dp[i]` represents the minimum cost to fix the first `i` pairs of differing indices (meaning the first `2*i` differing bits). The algorithm iterates from `i = 1` to `num_pairs` (where `num_pairs` is half the number of differing indices). For each `i`, it considers two options: pairing the `(2*i - 2)`th and `(2*i - 1)`th differing indices directly (cost `min(x, diffs[2*i - 1] - diffs[2*i - 2])`) or pairing the `(2*i - 1)`th index with the `(2*i - 3)`th index instead. The latter option involves a chain of adjacent flips, whose cost is calculated and considered as well. The minimum of these two costs is stored in `dp[i]`. Finally, `dp[num_pairs]` contains the minimum cost to make the strings equal.

Apply Operations to Make Two Strings Equal - Complete Solution Guide

Apply Operations to Make Two Strings Equal is LeetCode problem 2896, 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 0-indexed binary strings s1 and s2 , both of length n , and a positive integer x . You can perform any of the following operations on the string s1 any number of times: Choose two indices i and j , and flip both s1[i] and s1[j] . The cost of this operation is x . Choose an index i such that i < n - 1 and flip both s1[i] and s1[i + 1] . The cost of this operation is 1 . Return the minimum cost needed to make the strings s1 and s2 equal, or return -1 if it is impossible. Note tha

Detailed Explanation

The problem asks us to find the minimum cost to make two binary strings, `s1` and `s2`, equal by applying two types of operations on `s1`. Operation 1: flip two arbitrary bits at cost `x`. Operation 2: flip two adjacent bits at cost 1. If it's impossible to make the strings equal, return -1. The strings `s1` and `s2` have the same length `n`, and `1 <= n, x <= 500`. We need to determine the sequence of operations with the lowest total cost to transform `s1` into `s2`.

Solution Approach

The provided solution uses dynamic programming to determine the minimum cost. It first identifies the indices where `s1` and `s2` differ and stores them in a list called `diffs`. If the number of differing indices is odd, it returns -1. Then, it initializes a DP table `dp` where `dp[i]` represents the minimum cost to fix the first `i` pairs of differing indices (meaning the first `2*i` differing bits). The algorithm iterates from `i = 1` to `num_pairs` (where `num_pairs` is half the number of differing indices). For each `i`, it considers two options: pairing the `(2*i - 2)`th and `(2*i - 1)`th differing indices directly (cost `min(x, diffs[2*i - 1] - diffs[2*i - 2])`) or pairing the `(2*i - 1)`th index with the `(2*i - 3)`th index instead. The latter option involves a chain of adjacent flips, whose cost is calculated and considered as well. The minimum of these two costs is stored in `dp[i]`. Finally, `dp[num_pairs]` contains the minimum cost to make the strings equal.

Step-by-Step Algorithm

Step 1: Identify differing indices: Iterate through `s1` and `s2` and store the indices where the characters differ in the `diffs` list.
Step 2: Check for impossibility: If the number of differing indices (`m`) is odd, return -1.
Step 3: Initialize DP table: Create a DP table `dp` of size `num_pairs + 1`, where `num_pairs = m // 2`. Initialize all elements to 0.
Step 4: Iterate and calculate costs: Iterate from `i = 1` to `num_pairs`. In each iteration, calculate the cost of two options:
Step 5: Option 1: Pair the current two differences (d_prev, d_curr) together. The cost is either using one type 1 op (cost x) or a chain of type 2 ops (cost d_curr - d_prev). cost1 = dp[i-1] + min(x, d_curr - d_prev)
Step 6: Option 2: Consider the last four differences. Instead of pairing (2i-4, 2i-3) and (2i-2, 2i-1) separately, we can form pairs like (2i-3, 2i-2). The cost for this operation is d_{2i-1} - d_{2i-3}. cost2 = dp[i-2] + (d_curr - d_pre_curr)
Step 7: Take the minimum: Set `dp[i] = min(cost1, cost2)`.
Step 8: Return the result: After the loop finishes, `dp[num_pairs]` contains the minimum cost to make the strings equal. Return this value.

Key Insights

Insight 1: First, identify the indices where `s1` and `s2` differ. Only these differing indices need to be addressed.
Insight 2: If the number of differing indices is odd, it's impossible to make the strings equal because each operation flips an even number of bits. So, return -1 in that case.
Insight 3: Dynamic programming is a suitable approach because we can break down the problem into smaller subproblems of minimizing the cost to fix the first i pairs of differing bits.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Dynamic Programming.

Companies

Asked at: Zeta.

Apply Operations to Make Two Strings Equal | LeetCode 2896

Medium

String Dynamic Programming

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

Solution Code

Available in:

import java.lang.Math;
import java.util.ArrayList;
import java.util.List;

class Solution {
    public int minOperations(String s1, String s2, int x) {
        List<Integer> diffs = new ArrayList<>();
        for (int i = 0; i < s1.length(); i++) {
            if (s1.charAt(i) != s2.charAt(i)) {
                diffs.add(i);
            }
        }

        int m = diffs.size();
        if (m % 2 != 0) {
            return -1;
        }

        if (m == 0) {
            return 0;
        }

        // dp[i] = min cost to fix the first i pairs of differences (i.e., first 2*i diffs)
        int num_pairs = m / 2;
        int[] dp = new int[num_pairs + 1];

        for (int i = 1; i <= num_pairs; i++) {
            int d_curr = diffs.get(2 * i - 1);
            int d_prev = diffs.get(2 * i - 2);

            // Option 1: Pair the current two differences (d_prev, d_curr) together.
            // The cost is either using one type 1 op (cost x) or a chain of type 2 ops (cost d_curr - d_prev).
            // We add this to the optimal cost for the first i-1 pairs.
            int cost1 = dp[i - 1] + Math.min(x, d_curr - d_prev);

            // Option 2: Consider the last four differences. Instead of pairing (2i-4, 2i-3) and (2i-2, 2i-1)
            // separately, we can form pairs like (2i-3, 2i-2).
            // Here, we model this by pairing d_{2i-1} with d_{2i-3} via a chain of adjacent flips.
            // The cost for this operation is d_{2i-1} - d_{2i-3}.
            // The remaining problem is to fix the first i-2 pairs.
            int cost2 = Integer.MAX_VALUE;
            if (i > 1) {
                int d_pre_curr = diffs.get(2 * i - 3);
                cost2 = dp[i - 2] + (d_curr - d_pre_curr);
            }

            dp[i] = Math.min(cost1, cost2);
        }

        return dp[num_pairs];
    }
}

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Problem Info

Description

Problem 2896: Apply Operations to Make Two Strings Equal - Detailed explanation and solution approaches.

Topics

String Dynamic Programming

Companies

Zeta

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2896

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: Identify differing indices: Iterate through `s1` and `s2` and store the indices where the characters differ in the `diffs` list.
2
Step 2: Check for impossibility: If the number of differing indices (`m`) is odd, return -1.
3
Step 3: Initialize DP table: Create a DP table `dp` of size `num_pairs + 1`, where `num_pairs = m // 2`. Initialize all elements to 0.
4
Step 4: Iterate and calculate costs: Iterate from `i = 1` to `num_pairs`. In each iteration, calculate the cost of two options:
5
Step 5: Option 1: Pair the current two differences (d_prev, d_curr) together. The cost is either using one type 1 op (cost x) or a chain of type 2 ops (cost d_curr - d_prev). cost1 = dp[i-1] + min(x, d_curr - d_prev)
6
Step 6: Option 2: Consider the last four differences. Instead of pairing (2i-4, 2i-3) and (2i-2, 2i-1) separately, we can form pairs like (2i-3, 2i-2). The cost for this operation is d_{2i-1} - d_{2i-3}. cost2 = dp[i-2] + (d_curr - d_pre_curr)
7
Step 7: Take the minimum: Set `dp[i] = min(cost1, cost2)`.
8
Step 8: Return the result: After the loop finishes, `dp[num_pairs]` contains the minimum cost to make the strings equal. Return this value.

💡

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 determined by the initial scan of the strings to populate the `diffs` vector (O(n)), and the dynamic programming loop, which iterates at most n/2 times (O(n)). The space complexity arises from storing the differing indices in the `diffs` vector (O(n)) and the `dp` table (O(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

Greedy Approach

A greedy approach might try to iteratively find adjacent differing bits and flip them, or find pairs of differing bits and flip them if the distance is greater than `x`. However, this approach is unlikely to yield the optimal solution because the choice to flip adjacent bits or pairs affects the overall cost and doesn't guarantee minimization.

Time: O(n^2) in worst case. Need to search for pairs of indices.Space: O(1)

Trade-offs: Greedy is easier to implement, but not guaranteed to be correct. DP finds the optimal solution but is a little harder to grasp.

✨

Key Insights

1Insight 1: First, identify the indices where `s1` and `s2` differ. Only these differing indices need to be addressed.
2Insight 2: If the number of differing indices is odd, it's impossible to make the strings equal because each operation flips an even number of bits. So, return -1 in that case.
3Insight 3: Dynamic programming is a suitable approach because we can break down the problem into smaller subproblems of minimizing the cost to fix the first i pairs of differing bits.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

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

Edge Cases

Edge case 1: Empty strings or identical strings. In this case, `m` is 0, and the solution correctly returns 0.
Edge case 2: The number of differing bits is odd. The code specifically checks for this case and returns -1, indicating that it's impossible to make the strings equal.
Edge case 3: x is very small. The solution correctly compares the cost of flipping pairs directly (cost `x`) versus flipping adjacent bits (cost `d_curr - d_prev`).
Edge case 4: The case when i <= 1. The code handles this edge case by setting cost2 to be infinity when i <= 1 to avoid accessing index out of bound of the diffs vector.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the number of differing bits is odd. This will lead to incorrect results or infinite loops.

✗Common mistake 2: Failing to consider both options (flipping pairs vs. flipping adjacent bits) at each step of the dynamic programming approach. Only considering one option may not lead to the minimum cost.

✗Common mistake 3: Integer overflow when calculating costs, especially when `x` is large or when `d_curr - d_prev` is large. Using appropriate data types and handling `Integer.MAX_VALUE` carefully is crucial.

✗Common mistake 4: Index out of bound when accessing the `diffs` vector. Ensure that the indices used to access the vector are within the valid range.

✅

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. When synchronizing two versions of a file or database, this algorithm can be used to determine the minimum number of operations (flips or changes) needed to make the two versions consistent.

🏦Real-world scenario 2: Error correction. In communication systems, this algorithm can be used to find the most efficient way to correct errors in a received message. The differing bits represent errors, and the operations represent ways to correct them with minimal cost.

🎮Real-world scenario 3: DNA Sequencing: Comparing DNA sequences and finding the minimum number of mutations (insertions, deletions, or substitutions) needed to transform one sequence into another. The parameter `x` could represent the cost of a specific type of mutation.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

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

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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: June 17, 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 14

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

@InterviewProJul 7

❓

Frequently Asked Questions

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.

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 "Apply Operations to Make Two Strings Equal"?

For this medium-level String 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.

🤖 Need help with this problem?