What is the time complexity of Minimum Adjacent Swaps for K Consecutive Ones?

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

What data structures are used to solve Minimum Adjacent Swaps for K Consecutive Ones?

This problem uses Array, Greedy, Sliding Window, Prefix Sum. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Adjacent Swaps for K Consecutive Ones?

Insight 1: The core idea is to identify the indices of all the 1s in the array. The problem then reduces to finding an optimal subarray of length 'k' within these indices. Insight 2: The number of swaps required to bring 'k' ones together can be calculated by figuring out how far each '1' needs to move from its ideal position in the consecutive sequence. This is minimized when the '1's are clustered around their median index.

Which companies ask LeetCode 1703. Minimum Adjacent Swaps for K Consecutive Ones in interviews?

This problem is frequently asked at Flipkart.

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Minimum Adjacent Swaps for K Consecutive Ones - LeetCode 1703 Solution

Minimum Adjacent Swaps for K Consecutive Ones - Complete Solution Guide

Minimum Adjacent Swaps for K Consecutive Ones is LeetCode problem 1703, a Hard 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 an integer array, nums , and an integer k . nums comprises of only 0 's and 1 's. In one move, you can choose two adjacent indices and swap their values. Return the minimum number of moves required so that nums has k consecutive 1 's . Example 1: Input: nums = [1,0,0,1,0,1], k = 2 Output: 1 Explanation: In 1 move, nums could be [1,0,0,0, 1 , 1 ] and have 2 consecutive 1's. Example 2: Input: nums = [1,0,0,0,0,0,1,1], k = 3 Output: 5 Explanation: In 5 moves, the leftmost 1 can be shi

Detailed Explanation

The problem asks us to find the minimum number of adjacent swaps needed to group 'k' consecutive 1's in a binary array (an array containing only 0s and 1s). We are allowed to swap adjacent elements. The goal is to rearrange the array so that we have a sequence of 'k' consecutive 1s using the fewest possible swaps.

Solution Approach

The solution identifies the indices of all ones in the input array. It then transforms the problem into finding the minimum cost (number of swaps) to arrange k consecutive ones. A sliding window of size 'k' is used to iterate through the indices of the ones. For each window, the median element is determined. The cost is computed as the sum of the absolute differences between each element in the window and the median. To efficiently compute the sum of absolute differences, prefix sums are used.

Step-by-Step Algorithm

Step 1: Find and store the indices of all the 1s in the input array 'nums'.
Step 2: Create a new array 'B' where B[i] = (index of the i-th 1) - i. This transformation allows us to reason about the number of zeros to the left of each 1.
Step 3: Compute the prefix sum of array 'B'. This allows for efficient calculation of the sum of elements within a window.
Step 4: Iterate through 'B' using a sliding window of size 'k'. For each window:
Step 5: Find the median element within the current window.
Step 6: Calculate the cost (number of swaps) required to group the 'k' ones around the median.
Step 7: Update the minimum cost found so far.
Step 8: Return the minimum cost.

Key Insights

Insight 1: The core idea is to identify the indices of all the 1s in the array. The problem then reduces to finding an optimal subarray of length 'k' within these indices.
Insight 2: The number of swaps required to bring 'k' ones together can be calculated by figuring out how far each '1' needs to move from its ideal position in the consecutive sequence. This is minimized when the '1's are clustered around their median index.
Insight 3: Using a sliding window technique and prefix sums enables efficient calculation of the number of swaps required for each possible subarray of 'k' consecutive 1s.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Greedy, Sliding Window, Prefix Sum.

Companies

Asked at: Flipkart.

Minimum Adjacent Swaps for K Consecutive Ones | LeetCode 1703

Hard

Array Greedy Sliding Window Prefix Sum

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

Solution Code

Available in:

class Solution {
    public int minMoves(int[] nums, int k) {
        List<Integer> onesIndices = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 1) {
                onesIndices.add(i);
            }
        }
        int m = onesIndices.size();

        // B[i] = ones_indices[i] - i represents the number of zeros before the i-th one.
        int[] B = new int[m];
        for (int i = 0; i < m; i++) {
            B[i] = onesIndices.get(i) - i;
        }

        // Use prefix sums for B to calculate window sums efficiently.
        long[] prefixB = new long[m + 1];
        for (int i = 0; i < m; i++) {
            prefixB[i + 1] = prefixB[i] + B[i];
        }

        long minCost = Long.MAX_VALUE;

        // The median of a k-sized window is at relative index 'radius_left'.
        int radiusLeft = (k - 1) / 2;
        // The number of elements to the right of the median.
        int radiusRight = k - 1 - radiusLeft;

        // Slide a window of size k over B.
        for (int i = 0; i <= m - k; i++) {
            // Window is B[i : i+k].
            // Median element is at absolute index i + radius_left.
            int medianIdx = i + radiusLeft;
            long medianVal = B[medianIdx];

            // Sum of elements to the right of the median in the window.
            long rightSum = prefixB[i + k] - prefixB[medianIdx + 1];

            // Sum of elements to the left of the median in the window.
            long leftSum = prefixB[medianIdx] - prefixB[i];

            // Cost = sum(|x - median|)
            //      = (sum_{right} (x - median)) + (sum_{left} (median - x))
            //      = (right_sum - radius_right * median) + (radius_left * median - left_sum)
            long cost = (rightSum - radiusRight * medianVal) + (radiusLeft * medianVal - leftSum);

            minCost = Math.min(minCost, cost);
        }

        return (int)minCost;
    }
}

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Problem Info

Description

Problem 1703: Minimum Adjacent Swaps for K Consecutive Ones - Detailed explanation and solution approaches.

Topics

Array Greedy Sliding Window Prefix Sum

Companies

Flipkart

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1703

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: Find and store the indices of all the 1s in the input array 'nums'.
2
Step 2: Create a new array 'B' where B[i] = (index of the i-th 1) - i. This transformation allows us to reason about the number of zeros to the left of each 1.
3
Step 3: Compute the prefix sum of array 'B'. This allows for efficient calculation of the sum of elements within a window.
4
Step 4: Iterate through 'B' using a sliding window of size 'k'. For each window:
5
Step 5: Find the median element within the current window.
6
Step 6: Calculate the cost (number of swaps) required to group the 'k' ones around the median.
7
Step 7: Update the minimum cost found so far.
8
Step 8: Return the minimum cost.

💡

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 dominant operations are the initial scan for ones, the creation of array B, and the sliding window iteration. All of these take at most linear time with respect to the size of the input array. The space used is proportional to the number of ones, which, in the worst case, can be the size of the original array.

📖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 combinations of 'k' consecutive positions and calculate the number of swaps required for each. Choose the combination with the minimum swaps.

Time: O(n^2 * m), where 'n' is the length of nums and 'm' is the number of ones. We iterate n^2 to find possible starting indexes, then for each, we need to compute the number of swaps.Space: O(1)

Trade-offs: This approach is simpler to understand but is inefficient for larger input sizes. It's better for very small inputs due to its simplicity. The sliding window approach is significantly more efficient for larger datasets.

✨

Key Insights

1Insight 1: The core idea is to identify the indices of all the 1s in the array. The problem then reduces to finding an optimal subarray of length 'k' within these indices.
2Insight 2: The number of swaps required to bring 'k' ones together can be calculated by figuring out how far each '1' needs to move from its ideal position in the consecutive sequence. This is minimized when the '1's are clustered around their median index.
3Insight 3: Using a sliding window technique and prefix sums enables efficient calculation of the number of swaps required for each possible subarray of 'k' consecutive 1s.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sliding Window

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

Edge Cases

Edge case 1: 'k' is equal to the number of ones in the array. In this case, the solution correctly calculates the minimum swaps required to group all ones consecutively.
Edge case 2: The input array already has 'k' consecutive ones. The solution will correctly return 0, as no swaps are needed.
Edge case 3: 'k' is 1. The algorithm will find the minimum number of swaps to move each '1' to be next to each other. This case is handled by the general sliding window approach.
Edge case 4: When all the elements of the input array are 0. The algorithm will correctly return INT_MAX after the loop because min_cost is never assigned any values. The prompt states 1 <= k <= sum(nums). This edge case is covered by the constraint.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the median. The median needs to be found correctly within each window. This implementation uses the middle element as median.

✗Common mistake 2: Not using prefix sums. Without prefix sums, calculating the sum of differences within each window becomes significantly more expensive, leading to a time complexity higher than O(n).

✗Common mistake 3: Not handling edge cases such as empty arrays or k larger than the number of 1's present in the array. The prompt states 1 <= k <= sum(nums), this constraint handles the second case for us.

✅

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: Resource allocation. Optimizing the placement of resources (e.g., servers in a data center) to minimize network latency, where swaps represent moving resources between adjacent locations.

🏦Real-world scenario 2: Production line optimization. Rearranging steps in a production line to minimize the movement of products between stations, where the 1s represent critical steps that need to be performed consecutively.

🎮Real-world scenario 3: DNA sequencing. Minimizing the number of rearrangements to align specific DNA sequences.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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: August 23, 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

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

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 Adjacent Swaps for K Consecutive Ones"?

For this hard-level Array 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.

🤖 Need help with this problem?