What is the time complexity of Minimum Swaps to Group All 1's Together II?

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

What data structures are used to solve Minimum Swaps to Group All 1's Together II?

This problem uses Array, Sliding Window. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Swaps to Group All 1's Together II?

Insight 1: The problem is best approached using a sliding window. We need to find the window of size equal to the total number of 1's which contains the maximum number of 1's. Insight 2: Since the array is circular, we need to use the modulo operator (%) to handle the wraparound when the window reaches the end of the array.

Which companies ask LeetCode 2134. Minimum Swaps to Group All 1's Together II in interviews?

This problem is frequently asked at IBM, josh technology.

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Minimum Swaps to Group All 1's Together II - LeetCode 2134 Solution

Minimum Swaps to Group All 1's Together II - Complete Solution Guide

Minimum Swaps to Group All 1's Together II is LeetCode problem 2134, 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

A swap is defined as taking two distinct positions in an array and swapping the values in them. A circular array is defined as an array where we consider the first element and the last element to be adjacent . Given a binary circular array nums , return the minimum number of swaps required to group all 1 's present in the array together at any location . Example 1: Input: nums = [0,1,0,1,1,0,0] Output: 1 Explanation: Here are a few of the ways to group all the 1's together: [0, 0 , 1 ,1,1,0,0] u

Detailed Explanation

The problem asks us to find the minimum number of swaps needed to group all 1's together in a binary circular array. A circular array means the last element is considered adjacent to the first element. We need to find the arrangement of 1's that requires the fewest swaps. A swap involves exchanging the positions of two distinct elements in the array.

Solution Approach

The solution employs a sliding window technique. First, we count the total number of 1's in the array. Then, we initialize a window of size equal to the total number of 1's. We slide this window across the array, calculating the number of 1's within the current window. We maintain a variable to store the maximum number of 1's found in any window so far. After iterating through all possible window positions, the minimum number of swaps is the difference between the total number of 1's and the maximum number of 1's found in a window.

Step-by-Step Algorithm

Step 1: Calculate the total number of 1's in the array.
Step 2: Handle the edge cases where the array contains either all 0's or all 1's. In these cases, no swaps are needed, so return 0.
Step 3: Initialize a window of size equal to the total number of 1's, starting from the beginning of the array. Calculate the number of 1's in this initial window.
Step 4: Initialize `max_ones` to the number of 1's in the initial window.
Step 5: Iterate through the array, sliding the window one position at a time. For each iteration, update the number of 1's in the current window by subtracting the element that leaves the window and adding the element that enters the window (using the modulo operator to handle the circular property).
Step 6: Update `max_ones` if the current window has more 1's than the previous maximum.
Step 7: After the loop finishes, calculate the minimum number of swaps as `total_ones - max_ones`. This represents the number of 0's in the window with the most 1's.
Step 8: Return the calculated minimum number of swaps.

Key Insights

Insight 1: The problem is best approached using a sliding window. We need to find the window of size equal to the total number of 1's which contains the maximum number of 1's.
Insight 2: Since the array is circular, we need to use the modulo operator (%) to handle the wraparound when the window reaches the end of the array.
Insight 3: The minimum number of swaps is equal to the number of 0's within the best window. This is equivalent to the window size (total number of 1's) minus the maximum number of 1's we can find in a window of that size.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Sliding Window.

Companies

Asked at: IBM, josh technology.

Minimum Swaps to Group All 1's Together II | LeetCode 2134

Medium

Array Sliding Window

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

Solution Code

Available in:

class Solution {
    public int minSwaps(int[] nums) {
        int n = nums.length;
        int total_ones = 0;
        for (int num : nums) {
            total_ones += num;
        }

        if (total_ones == 0 || total_ones == n) {
            return 0;
        }

        int current_ones = 0;
        for (int i = 0; i < total_ones; i++) {
            current_ones += nums[i];
        }
        int max_ones = current_ones;

        for (int i = 1; i < n; i++) {
            // Slide the window by one position
            // The element at index i-1 leaves the window
            // The element at index (i + total_ones - 1) % n enters the window
            current_ones = current_ones - nums[i - 1] + nums[(i + total_ones - 1) % n];
            if (current_ones > max_ones) {
                max_ones = current_ones;
            }
        }
        
        // The number of swaps needed is the number of 0s in the best window.
        // This is the window size (total_ones) minus the maximum number of 1s found.
        return total_ones - max_ones;
    }
}

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Problem Info

Description

Problem 2134: Minimum Swaps to Group All 1's Together II - Detailed explanation and solution approaches.

Topics

Array Sliding Window

Companies

IBMjosh technology

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2134

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: Calculate the total number of 1's in the array.
2
Step 2: Handle the edge cases where the array contains either all 0's or all 1's. In these cases, no swaps are needed, so return 0.
3
Step 3: Initialize a window of size equal to the total number of 1's, starting from the beginning of the array. Calculate the number of 1's in this initial window.
4
Step 4: Initialize `max_ones` to the number of 1's in the initial window.
5
Step 5: Iterate through the array, sliding the window one position at a time. For each iteration, update the number of 1's in the current window by subtracting the element that leaves the window and adding the element that enters the window (using the modulo operator to handle the circular property).
6
Step 6: Update `max_ones` if the current window has more 1's than the previous maximum.
7
Step 7: After the loop finishes, calculate the minimum number of swaps as `total_ones - max_ones`. This represents the number of 0's in the window with the most 1's.
8
Step 8: Return the calculated minimum number of swaps.

💡

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 the sliding window iteration which occurs once over the input array. The other operations like summing the number of ones or computing the initial window's sum take O(n) time, but are performed only once. Therefore the complexity is O(n). Constant extra space is used, leading to O(1) space complexity.

📖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

Brute Force

Try all possible starting positions for the group of 1's and calculate the number of swaps required for each position. Choose the position with the minimum swaps.

Time: O(n^2)Space: O(1)

Trade-offs: This approach is less efficient than the sliding window approach due to the nested loops. It is easy to understand but not suitable for large arrays.

✨

Key Insights

1Insight 1: The problem is best approached using a sliding window. We need to find the window of size equal to the total number of 1's which contains the maximum number of 1's.
2Insight 2: Since the array is circular, we need to use the modulo operator (%) to handle the wraparound when the window reaches the end of the array.
3Insight 3: The minimum number of swaps is equal to the number of 0's within the best window. This is equivalent to the window size (total number of 1's) minus the maximum number of 1's we can find in a window of that size.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Sliding Window

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

Edge Cases

Edge case 1: Array contains all 0's or all 1's. The solution correctly handles this case by returning 0 swaps.
Edge case 2: Array length is 1. The general algorithm works correctly for this case as well.
Edge case 3: Total number of ones is equal to the array size. The solution will correctly return 0 as all elements are already grouped together.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the circular nature of the array correctly. Forgetting to use the modulo operator (%) when calculating the index of the element that enters the window.

✗Common mistake 2: Incorrectly calculating the number of swaps. The number of swaps is the difference between the window size (number of ones) and the maximum number of ones found in the window, NOT the number of zeros in the array.

✅

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: Optimizing data storage on a circular buffer or ring buffer. Minimizing fragmentation by grouping similar data together.

🏦Real-world scenario 2: Resource allocation in a circular queue. Efficiently allocating contiguous blocks of resources in a circular manner to minimize overhead.

💡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

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

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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 2, 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

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I determine the window size in sliding window problems?

For fixed-size windows (like max sum of k elements), window size is given. For variable-size windows (like smallest subarray with sum >= target), expand right pointer to meet condition, then shrink left pointer to find optimal. Track the best result during shrinking.

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 "Minimum Swaps to Group All 1's Together II"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether sliding window techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?