What is the time complexity of Minimum Suffix Flips?

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

What data structures are used to solve Minimum Suffix Flips?

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

What are the key insights for solving Minimum Suffix Flips?

Insight 1: The problem can be solved by iterating through the target string and counting the number of times the current desired state (0 or 1) needs to change. Each change corresponds to a flip operation. Insight 2: The initial state of 's' being all zeros is crucial. This means we begin in a state of '0'.

Which companies ask LeetCode 1529. Minimum Suffix Flips in interviews?

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

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Minimum Suffix Flips - LeetCode 1529 Solution

Q: How to solve Minimum Suffix Flips?

The solution uses a greedy approach. It simulates the process of flipping the string by tracking the current state of the suffix. It iterates through the target string, and whenever the current character in the target string differs from the current state, a flip is performed, and the flip count is incremented. The current state is also flipped to reflect the effect of the flip operation.

Minimum Suffix Flips - Complete Solution Guide

Minimum Suffix Flips is LeetCode problem 1529, 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 0-indexed binary string target of length n . You have another binary string s of length n that is initially set to all zeros. You want to make s equal to target . In one operation, you can pick an index i where 0 <= i < n and flip all bits in the inclusive range [i, n - 1] . Flip means changing '0' to '1' and '1' to '0' . Return the minimum number of operations needed to make s equal to target . Example 1: Input: target = "10111" Output: 3 Explanation: Initially, s = "00000". Cho

Detailed Explanation

The problem requires finding the minimum number of 'flips' needed to transform an initial string 's' (consisting of all zeros) into a target binary string 'target'. A flip operation involves selecting an index 'i' and inverting all bits from that index to the end of the string. The goal is to determine the fewest such operations to achieve the target string.

Solution Approach

The solution uses a greedy approach. It simulates the process of flipping the string by tracking the current state of the suffix. It iterates through the target string, and whenever the current character in the target string differs from the current state, a flip is performed, and the flip count is incremented. The current state is also flipped to reflect the effect of the flip operation.

Step-by-Step Algorithm

Step 1: Initialize a 'flips' counter to 0, representing the number of flip operations.
Step 2: Initialize 'current_state' to '0', reflecting the initial state of the string 's' (all zeros).
Step 3: Iterate through each character in the 'target' string.
Step 4: In each iteration, compare the current character in the 'target' string with the 'current_state'.
Step 5: If the character is different from the 'current_state', increment the 'flips' counter by 1 (meaning a flip operation is needed) and flip the 'current_state' (from '0' to '1' or vice versa).
Step 6: After iterating through the entire 'target' string, return the final 'flips' count.

Key Insights

Insight 1: The problem can be solved by iterating through the target string and counting the number of times the current desired state (0 or 1) needs to change. Each change corresponds to a flip operation.
Insight 2: The initial state of 's' being all zeros is crucial. This means we begin in a state of '0'.
Insight 3: Consecutive identical characters in the target string after the initial '0' state are irrelevant, as they can be achieved with a single flip.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Greedy.

Companies

Asked at: IBM, J.P. Morgan.

Minimum Suffix Flips | LeetCode 1529

Medium

String Greedy

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

Solution Code

Available in:

class Solution {
    public int minFlips(String target) {
        int flips = 0;
        char current_state = '0';
        for (int i = 0; i < target.length(); i++) {
            char char_ = target.charAt(i);
            if (char_ != current_state) {
                flips += 1;
                current_state = (current_state == '0') ? '1' : '0';
            }
        }
        return flips;
    }
}

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Problem Info

Description

Problem 1529: Minimum Suffix Flips - Detailed explanation and solution approaches.

Topics

String Greedy

Companies

IBMJ.P. Morgan

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1529

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 a 'flips' counter to 0, representing the number of flip operations.
2
Step 2: Initialize 'current_state' to '0', reflecting the initial state of the string 's' (all zeros).
3
Step 3: Iterate through each character in the 'target' string.
4
Step 4: In each iteration, compare the current character in the 'target' string with the 'current_state'.
5
Step 5: If the character is different from the 'current_state', increment the 'flips' counter by 1 (meaning a flip operation is needed) and flip the 'current_state' (from '0' to '1' or vice versa).
6
Step 6: After iterating through the entire 'target' string, return the final 'flips' 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 time complexity is linear because we only loop through the input string once. The space complexity is constant because we only use a few variables, and the number of variables doesn't change as the input size grows.

📖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 Difference

An alternative approach could involve calculating a prefix sum-like array representing the number of flips needed up to each index. This approach would require slightly more space but could offer different insights into the problem.

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

Trade-offs: The main approach is more space-efficient. This alternative might be useful if you want to precompute results for various substrings of the target string.

✨

Key Insights

1Insight 1: The problem can be solved by iterating through the target string and counting the number of times the current desired state (0 or 1) needs to change. Each change corresponds to a flip operation.
2Insight 2: The initial state of 's' being all zeros is crucial. This means we begin in a state of '0'.
3Insight 3: Consecutive identical characters in the target string after the initial '0' state are irrelevant, as they can be achieved with a single flip.

📚

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: target = "00000": The algorithm correctly returns 0, as no flips are needed.
Edge case 2: target = "11111": The algorithm correctly returns 1, as only one flip at index 0 is needed.
Edge case 3: target = "10101": The algorithm correctly calculates the number of flips needed, accounting for alternating sequences.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the initial state correctly. The initial state is all zeros, which is crucial to the logic of the algorithm.

✗Common mistake 2: Overlooking the consecutive identical characters. Only changes in the desired state require flip operations, so consecutive same chars don't require more flips.

✅

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. The problem is analogous to finding the minimum number of operations to transform one sequence of bits into another, which is a fundamental concept in data compression algorithms.

🏦Real-world scenario 2: Signal processing. The 'flip' operation can be viewed as inverting a signal within a specific time window. Minimizing the number of such inversions is relevant in optimizing signal transmission or processing.

💡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: August 29, 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

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

@TechLeadJul 13

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

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 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.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

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 Suffix Flips"?

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

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