What is the time complexity of Get Equal Substrings Within Budget?

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

What data structures are used to solve Get Equal Substrings Within Budget?

This problem uses String, Binary Search, Sliding Window, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Get Equal Substrings Within Budget?

Insight 1: The problem can be solved efficiently using the sliding window technique because we are looking for a substring that satisfies a given condition (cost <= maxCost). Insight 2: The core idea is to maintain a 'window' over the string, calculate the cost within the window, and expand the window to the right. If the cost exceeds `maxCost`, shrink the window from the left until the cost is within the budget.

Which companies ask LeetCode 1208. Get Equal Substrings Within Budget in interviews?

This problem is frequently asked at IBM.

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Get Equal Substrings Within Budget - LeetCode 1208 Solution

Q: How to solve Get Equal Substrings Within Budget?

The provided code implements the sliding window technique. It maintains a window defined by `left` and `right` pointers. The `current_cost` variable keeps track of the cost of changing the substring `s[left:right+1]` to `t[left:right+1]`. The `right` pointer expands the window one character at a time, increasing the `current_cost`. If `current_cost` exceeds `maxCost`, the `left` pointer is incremented, shrinking the window from the left, and correspondingly reducing the `current_cost`, until the `current_cost` is within the budget. The `max_len` variable stores the maximum length encountered so far.

Q: Which companies ask LeetCode 1208. Get Equal Substrings Within Budget in interviews?

This problem is frequently asked at IBM.

Get Equal Substrings Within Budget - Complete Solution Guide

Get Equal Substrings Within Budget is LeetCode problem 1208, 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 s and t of the same length and an integer maxCost . You want to change s to t . Changing the i th character of s to i th character of t costs |s[i] - t[i]| (i.e., the absolute difference between the ASCII values of the characters). Return the maximum length of a substring of s that can be changed to be the same as the corresponding substring of t with a cost less than or equal to maxCost . If there is no substring from s that can be changed to its corresponding substrin

Detailed Explanation

The problem asks us to find the maximum length of a substring within a given string `s` that can be transformed into the corresponding substring of another string `t` with a cost less than or equal to `maxCost`. The cost of changing a character in `s` to the corresponding character in `t` is the absolute difference of their ASCII values. We need to return the length of the longest such substring. If no such substring exists, we return 0. The inputs are the two strings `s` and `t` of equal length, and an integer `maxCost`. The output is the maximum length of the substring that satisfies the cost constraint.

Solution Approach

The provided code implements the sliding window technique. It maintains a window defined by `left` and `right` pointers. The `current_cost` variable keeps track of the cost of changing the substring `s[left:right+1]` to `t[left:right+1]`. The `right` pointer expands the window one character at a time, increasing the `current_cost`. If `current_cost` exceeds `maxCost`, the `left` pointer is incremented, shrinking the window from the left, and correspondingly reducing the `current_cost`, until the `current_cost` is within the budget. The `max_len` variable stores the maximum length encountered so far.

Step-by-Step Algorithm

Step 1: Initialize `left = 0`, `current_cost = 0`, and `max_len = 0`. These variables represent the left pointer of the sliding window, the current cost of the window, and the maximum length found so far, respectively.
Step 2: Iterate through the string `s` using the `right` pointer from 0 to `len(s) - 1`.
Step 3: In each iteration, calculate the cost of changing the character at index `right` in `s` to the character at index `right` in `t` and add it to `current_cost`.
Step 4: While `current_cost` is greater than `maxCost`, subtract the cost of changing the character at index `left` in `s` to the character at index `left` in `t` from `current_cost`, and increment `left`. This shrinks the window from the left.
Step 5: Update `max_len` to the maximum of its current value and the length of the current window (`right - left + 1`).
Step 6: After iterating through the entire string `s`, return `max_len`.

Key Insights

Insight 1: The problem can be solved efficiently using the sliding window technique because we are looking for a substring that satisfies a given condition (cost <= maxCost).
Insight 2: The core idea is to maintain a 'window' over the string, calculate the cost within the window, and expand the window to the right. If the cost exceeds `maxCost`, shrink the window from the left until the cost is within the budget.
Insight 3: The problem constraints suggest an O(n) solution, hinting at the suitability of a sliding window approach. The window represents a potential substring.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Binary Search, Sliding Window, Prefix Sum.

Companies

Asked at: IBM.

Get Equal Substrings Within Budget | LeetCode 1208

Medium

String Binary Search Sliding Window Prefix Sum

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

Solution Code

Available in:

class Solution {
    public int equalSubstring(String s, String t, int maxCost) {
        int left = 0;
        int current_cost = 0;
        int max_len = 0;
        
        for (int right = 0; right < s.length(); right++) {
            current_cost += Math.abs(s.charAt(right) - t.charAt(right));
            
            while (current_cost > maxCost) {
                current_cost -= Math.abs(s.charAt(left) - t.charAt(left));
                left += 1;
            }
            
            max_len = Math.max(max_len, right - left + 1);
        }
        
        return max_len;
    }
}

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Problem Info

Description

Problem 1208: Get Equal Substrings Within Budget - Detailed explanation and solution approaches.

Topics

String Binary Search Sliding Window Prefix Sum

Companies

IBM

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1208

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

6 steps

1
Step 1: Initialize `left = 0`, `current_cost = 0`, and `max_len = 0`. These variables represent the left pointer of the sliding window, the current cost of the window, and the maximum length found so far, respectively.
2
Step 2: Iterate through the string `s` using the `right` pointer from 0 to `len(s) - 1`.
3
Step 3: In each iteration, calculate the cost of changing the character at index `right` in `s` to the character at index `right` in `t` and add it to `current_cost`.
4
Step 4: While `current_cost` is greater than `maxCost`, subtract the cost of changing the character at index `left` in `s` to the character at index `left` in `t` from `current_cost`, and increment `left`. This shrinks the window from the left.
5
Step 5: Update `max_len` to the maximum of its current value and the length of the current window (`right - left + 1`).
6
Step 6: After iterating through the entire string `s`, return `max_len`.

💡

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 sliding window technique allows us to examine each substring efficiently. The outer loop iterates through each character of the string `s` once, and the inner loop (while loop) shrinks the window from the left. The crucial point is that the `left` pointer never goes backward, ensuring that the number of iterations in the inner loop, in total, is at most n. Therefore, the time complexity is O(n). The space used is only for a few variables like left, right, current_cost, and max_len. The space usage does not depend on the input size, leading to a space complexity of 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

Prefix Sum

Calculate the prefix sum of the costs (absolute difference of characters). Then, for each possible substring length, check if the cost of that substring is within the budget using the prefix sum array. Use binary search to optimize finding the maximum substring length.

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

Trade-offs: The prefix sum approach with binary search can be useful if modifications to the cost calculation are frequently required, as the prefix sum array can be easily updated. However, the sliding window approach is generally more efficient for this specific problem due to its linear time complexity.

✨

Key Insights

1Insight 1: The problem can be solved efficiently using the sliding window technique because we are looking for a substring that satisfies a given condition (cost <= maxCost).
2Insight 2: The core idea is to maintain a 'window' over the string, calculate the cost within the window, and expand the window to the right. If the cost exceeds `maxCost`, shrink the window from the left until the cost is within the budget.
3Insight 3: The problem constraints suggest an O(n) solution, hinting at the suitability of a sliding window approach. The window represents a potential substring.

📚

Key Terms Explained

For Students

Binary Search

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

Sliding Window

A moving "window" over data to process subsets efficiently.

String

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

Edge Cases

Edge case 1: `maxCost` is 0. The code correctly handles this case by finding the longest substring where `s[i] == t[i]`.
Edge case 2: `maxCost` is a very large number. The code correctly handles this case by finding the longest substring where the sum of the absolute differences between the characters of `s` and `t` is less than or equal to `maxCost`.
Edge case 3: `s` and `t` are identical. The code will correctly identify the entire string as the maximum substring, since the cost would be 0, which is less than or equal to maxCost.
Edge case 4: Empty String (although constraints specify length > 0). It would be handled correctly as the loop wouldn't execute, returning 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the cost of changing characters. Forgetting to use the absolute difference between the ASCII values can lead to wrong results.

✗Common mistake 2: Failing to shrink the window when the `current_cost` exceeds `maxCost`. This will result in an incorrect `max_len`.

✗Common mistake 3: Not understanding the sliding window technique and trying to solve the problem with brute force, which would lead to a time complexity of O(n^2) or higher, exceeding the time limit.

✗Common mistake 4: Integer overflow if the cost calculation isn't careful and maxCost is large.

✅

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 compression. Finding similar substrings within large data sets for efficient data storage.

🏦Real-world scenario 2: DNA sequencing. Identifying regions of similarity between DNA sequences with a certain tolerance for mutations (cost).

🎮Real-world scenario 3: Spell checking. Finding the longest segment of the word that is correctly spelled, given a max edit distance (cost) from the dictionary.

📱Real-world scenario 4: Image processing. Comparing regions in two images for similarity, where 'cost' is a metric representing the differences in pixel values.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: October 7, 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 13

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

@TechLeadJul 6

❓

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 avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

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 "Get Equal Substrings Within Budget"?

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

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