How do you solve Minimum Adjacent Swaps for K Consecutive Ones on LeetCode?

Minimum Adjacent Swaps for K Consecutive Ones (LeetCode #1703) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Identifying positions of 1's helps in calculating swaps efficiently.

What is the brute force solution for Minimum Adjacent Swaps for K Consecutive Ones?

The brute force approach for Minimum Adjacent Swaps for K Consecutive Ones is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible combination of k consecutive 1's and calculating how many swaps are needed to group them together. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Adjacent Swaps for K Consecutive Ones?

Minimum Adjacent Swaps for K Consecutive Ones uses the following concepts: Array, Greedy, Sliding Window, Prefix Sum. The recommended patterns to study are: Sliding Window, Greedy.

What are the interview tips for Minimum Adjacent Swaps for K Consecutive Ones?

Always clarify the problem and constraints before jumping into coding. Discuss your thought process and potential optimizations with the interviewer. Practice dry-running your code with examples to ensure correctness.

What are common mistakes when solving Minimum Adjacent Swaps for K Consecutive Ones?

Not considering edge cases where k is larger than the number of 1's. Failing to optimize the calculation of swaps leading to timeouts on larger inputs.

#1703

Minimum Adjacent Swaps for K Consecutive Ones

Hard

ArrayGreedySliding WindowPrefix SumSliding WindowGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a sliding window approach to efficiently calculate the number of swaps needed to group k consecutive 1's. This method reduces the time complexity significantly by avoiding redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Identify the positions of all 1's in the array.
2Step 2: Use a sliding window of size k to calculate the number of swaps needed to group the 1's.
3Step 3: For each window, calculate the cost of moving the 1's to the center of the window and update the minimum swaps.

solution.py15 lines

1# Full working Python code
2
3def min_swaps(nums, k):
4    ones = [i for i in range(len(nums)) if nums[i] == 1]
5    total_ones = len(ones)
6    if total_ones < k:
7        return 0
8    min_swaps = 0
9    for i in range(k):
10        min_swaps += ones[i] - (ones[0] + i)
11    result = min_swaps
12    for i in range(1, total_ones - k + 1):
13        min_swaps += (ones[i + k - 1] - ones[i - 1]) - k
14        result = min(result, min_swaps)
15    return result

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array a few times to gather positions of 1's and calculate swaps, making it efficient for larger inputs.

1Identifying positions of 1's helps in calculating swaps efficiently.
2Using a sliding window reduces the need for redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.