What is the time complexity of Delete Operation for Two Strings?

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

What data structures are used to solve Delete Operation for Two Strings?

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

What are the key insights for solving Delete Operation for Two Strings?

Insight 1: The problem can be rephrased as finding the length of the Longest Common Subsequence (LCS) of the two strings. The number of deletions needed is then the sum of the lengths of the two strings minus twice the length of their LCS. Insight 2: Dynamic programming is a suitable technique to solve this problem because the optimal solution for larger substrings can be built from optimal solutions for smaller substrings.

Which companies ask LeetCode 583. Delete Operation for Two Strings in interviews?

This problem is frequently asked at ByteDance.

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Delete Operation for Two Strings - LeetCode 583 Solution

Q: How to solve Delete Operation for Two Strings?

The solution employs dynamic programming to find the length of the Longest Common Subsequence (LCS) implicitly. It uses a 1D DP array `dp` of size `n+1`, where `n` is the length of `word2`. `dp[j]` represents the minimum number of deletions needed to make `word1[:i]` and `word2[:j]` equal. The algorithm iterates through the characters of `word1` and `word2`. If `word1[i-1]` and `word2[j-1]` are equal, it means these characters can be part of the LCS, and no deletion is needed. Otherwise, we need to delete either `word1[i-1]` or `word2[j-1]`, and we choose the minimum deletions obtained by deleting either character.

Delete Operation for Two Strings - Complete Solution Guide

Delete Operation for Two Strings is LeetCode problem 583, 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

Given two strings word1 and word2 , return the minimum number of steps required to make word1 and word2 the same . In one step , you can delete exactly one character in either string. Example 1: Input: word1 = "sea", word2 = "eat" Output: 2 Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea". Example 2: Input: word1 = "leetcode", word2 = "etco" Output: 4 Constraints: 1 <= word1.length, word2.length <= 500 word1 and word2 consist of only lowercase English l

Detailed Explanation

The problem asks us to find the minimum number of deletion steps needed to make two given strings, `word1` and `word2`, identical. A 'step' consists of deleting a single character from either string. We are given two strings as input, and we need to return an integer representing the minimum number of deletions.

Solution Approach

The solution employs dynamic programming to find the length of the Longest Common Subsequence (LCS) implicitly. It uses a 1D DP array `dp` of size `n+1`, where `n` is the length of `word2`. `dp[j]` represents the minimum number of deletions needed to make `word1[:i]` and `word2[:j]` equal. The algorithm iterates through the characters of `word1` and `word2`. If `word1[i-1]` and `word2[j-1]` are equal, it means these characters can be part of the LCS, and no deletion is needed. Otherwise, we need to delete either `word1[i-1]` or `word2[j-1]`, and we choose the minimum deletions obtained by deleting either character.

Step-by-Step Algorithm

Step 1: Initialize a 1D DP array `dp` of size `n+1` with values from 0 to n, representing the minimum deletions to make `word2[:j]` equal to an empty string.
Step 2: Iterate through `word1` (outer loop from `i = 1` to `m`).
Step 3: In the outer loop, initialize `prev_row_j_minus_1` with `dp[0]` which will store `dp[i-1][j-1]` for the next row calculation and set `dp[0] = i` as it requires `i` deletions to match `word1[:i]` with an empty string.
Step 4: Iterate through `word2` (inner loop from `j = 1` to `n`).
Step 5: Inside the inner loop, store the current value of `dp[j]` in a temporary variable `temp` because it is needed for the calculation of the next jth value. This is equivalent to storing `dp[i-1][j]`
Step 6: If `word1[i-1]` equals `word2[j-1]`, it means the characters match. Then set `dp[j] = prev_row_j_minus_1`, which corresponds to `dp[i][j] = dp[i-1][j-1]` since no deletion is needed.
Step 7: If `word1[i-1]` is not equal to `word2[j-1]`, then set `dp[j] = 1 + min(dp[j], dp[j-1])`. This chooses the minimum between deleting `word1[i-1]` (corresponding to `dp[i-1][j]`) or deleting `word2[j-1]` (corresponding to `dp[i][j-1]`).
Step 8: Set `prev_row_j_minus_1 = temp` to keep track of the previous row's values for the next iteration.
Step 9: After iterating through both strings, `dp[n]` contains the minimum number of deletions to make the strings equal.

Key Insights

Insight 1: The problem can be rephrased as finding the length of the Longest Common Subsequence (LCS) of the two strings. The number of deletions needed is then the sum of the lengths of the two strings minus twice the length of their LCS.
Insight 2: Dynamic programming is a suitable technique to solve this problem because the optimal solution for larger substrings can be built from optimal solutions for smaller substrings.
Insight 3: Space optimization is possible in the DP solution. Since we only need the previous row to compute the current row, we can reduce the space complexity from O(m*n) to O(n), where m and n are the lengths of the input strings.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(n)

Topics

This problem involves: String, Dynamic Programming.

Companies

Asked at: ByteDance.

Delete Operation for Two Strings | LeetCode 583

Medium

String Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length();
        int n = word2.length();

        // This is a classic dynamic programming problem.
        // Let dp[i][j] be the minimum number of deletions to make
        // word1[:i] and word2[:j] the same.

        // The relationship is as follows:
        // 1. If word1[i-1] == word2[j-1], the last characters are the same.
        //    No deletion is needed for these characters.
        //    dp[i][j] = dp[i-1][j-1]
        // 2. If word1[i-1] != word2[j-1], we must delete one of them.
        //    - Delete word1[i-1]: 1 + dp[i-1][j]
        //    - Delete word2[j-1]: 1 + dp[i][j-1]
        //    We take the minimum: dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1])

        // Base cases:
        // dp[i][0] = i (to make word1[:i] and an empty string same, delete i chars)
        // dp[0][j] = j (to make an empty string and word2[:j] same, delete j chars)

        // We can optimize space from O(m*n) to O(n) by only keeping track of
        // the previous row's DP values.

        int[] dp = new int[n + 1];
        for (int j = 0; j <= n; j++) {
            dp[j] = j;
        }

        for (int i = 1; i <= m; i++) {
            int prev_row_j_minus_1 = dp[0];
            dp[0] = i;
            for (int j = 1; j <= n; j++) {
                int temp = dp[j];
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                    dp[j] = prev_row_j_minus_1;
                } else {
                    dp[j] = 1 + Math.min(dp[j], dp[j - 1]);
                }
                prev_row_j_minus_1 = temp;
            }
        }

        return dp[n];
    }
}

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Problem Info

Description

Problem 583: Delete Operation for Two Strings - Detailed explanation and solution approaches.

Topics

String Dynamic Programming

Companies

ByteDance

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:583

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

9 steps

1
Step 1: Initialize a 1D DP array `dp` of size `n+1` with values from 0 to n, representing the minimum deletions to make `word2[:j]` equal to an empty string.
2
Step 2: Iterate through `word1` (outer loop from `i = 1` to `m`).
3
Step 3: In the outer loop, initialize `prev_row_j_minus_1` with `dp[0]` which will store `dp[i-1][j-1]` for the next row calculation and set `dp[0] = i` as it requires `i` deletions to match `word1[:i]` with an empty string.
4
Step 4: Iterate through `word2` (inner loop from `j = 1` to `n`).
5
Step 5: Inside the inner loop, store the current value of `dp[j]` in a temporary variable `temp` because it is needed for the calculation of the next jth value. This is equivalent to storing `dp[i-1][j]`
6
Step 6: If `word1[i-1]` equals `word2[j-1]`, it means the characters match. Then set `dp[j] = prev_row_j_minus_1`, which corresponds to `dp[i][j] = dp[i-1][j-1]` since no deletion is needed.
7
Step 7: If `word1[i-1]` is not equal to `word2[j-1]`, then set `dp[j] = 1 + min(dp[j], dp[j-1])`. This chooses the minimum between deleting `word1[i-1]` (corresponding to `dp[i-1][j]`) or deleting `word2[j-1]` (corresponding to `dp[i][j-1]`).
8
Step 8: Set `prev_row_j_minus_1 = temp` to keep track of the previous row's values for the next iteration.
9
Step 9: After iterating through both strings, `dp[n]` contains the minimum number of deletions to make the strings equal.

💡

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 comes from the nested for loops that iterates over `m` and `n` respectively. The space complexity comes from the DP array of size `n+1`. We are able to reduce the space complexity from O(m*n) to O(n) by observing that we only need to store the results of the previous row.

📖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

2D DP Array Approach

Using a 2D DP array of size (m+1)x(n+1) where dp[i][j] represents the minimum deletion cost of word1[:i] and word2[:j]. This provides a more straightforward implementation without space optimization.

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

Trade-offs: This approach is easier to understand but has a higher space complexity compared to the 1D DP approach. If memory is not a constraint, this approach can be preferred for its simplicity.

Recursion with Memoization

A recursive solution can be formulated with memoization to avoid redundant calculations. This involves defining a recursive function that calculates the minimum deletion steps for two substrings. A memoization table is used to store the results of already computed subproblems.

Time: O(m*n)Space: O(m*n) due to the memoization table and recursion stack

Trade-offs: This approach provides a more intuitive understanding of the problem but can be slower than the iterative DP approach due to the overhead of function calls. It also might encounter stack overflow errors for large input sizes.

✨

Key Insights

1Insight 1: The problem can be rephrased as finding the length of the Longest Common Subsequence (LCS) of the two strings. The number of deletions needed is then the sum of the lengths of the two strings minus twice the length of their LCS.
2Insight 2: Dynamic programming is a suitable technique to solve this problem because the optimal solution for larger substrings can be built from optimal solutions for smaller substrings.
3Insight 3: Space optimization is possible in the DP solution. Since we only need the previous row to compute the current row, we can reduce the space complexity from O(m*n) to O(n), where m and n are the lengths of the input strings.

📚

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: One or both strings are empty. If `word1` is empty, the number of deletions needed is the length of `word2`. Similarly, if `word2` is empty, the number of deletions needed is the length of `word1`. The code correctly handles this case due to the initial values assigned to the dp array.
Edge case 2: Both strings are identical. In this case, the length of the LCS will be equal to the length of either string, and the number of deletions will be zero. The algorithm handles this correctly since `dp[j]` will be assigned to the `prev_row_j_minus_1` when `word1[i-1] == word2[j-1]` and the final value will converge to 0 if strings are equal.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting the base cases in the DP array initialization, like not initializing dp[i][0] = i and dp[0][j] = j. This results in incorrect calculations of the minimum deletions.

✗Common mistake 2: Incorrectly updating the DP array when the characters match or don't match. Confusing the indices while accessing word1 and word2. The correct formula to update dp[i][j] if the characters match is `dp[j] = prev_row_j_minus_1` and `dp[j] = 1 + min(dp[j], dp[j - 1])` otherwise.

✗Common mistake 3: Not optimizing the space complexity. Using the 2D DP approach when the problem constraints allow for O(n) space complexity.

✅

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: Spell checking. Finding the minimum number of edits (insertions, deletions, substitutions) required to transform a misspelled word into a correct word is a similar problem (edit distance).

🏦Real-world scenario 2: DNA sequencing. Finding the similarity between two DNA sequences can be done by finding their longest common subsequence. The number of deletions and insertions can then be used as a measure of dissimilarity.

💡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: July 14, 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

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

@CodeNinjaJul 14

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

@TechLeadJul 7

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

@DevExpert92Jun 30

❓

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.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

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 "Delete Operation for Two Strings"?

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?