How do you solve Minimum Swaps to Group All 1's Together II on LeetCode?

Minimum Swaps to Group All 1's Together II (LeetCode #2134) 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: Understanding the circular nature of the array is crucial for solving the problem efficiently.

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

The optimal solution for Minimum Swaps to Group All 1's Together II runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only traverse the array a constant number of times, and the space complexity is O(n) due to the extended array.

What is the brute force solution for Minimum Swaps to Group All 1's Together II?

The brute force approach for Minimum Swaps to Group All 1's Together II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subarray of size equal to the total number of 1's in the array. For each subarray, we count how many swaps are needed to make all elements 1's. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Swaps to Group All 1's Together II?

Minimum Swaps to Group All 1's Together II uses the following concepts: Array, Sliding Window. The recommended patterns to study are: Sliding Window, Array.

What are the interview tips for Minimum Swaps to Group All 1's Together II?

Always clarify the problem constraints and edge cases with the interviewer. Discuss your thought process out loud; it helps the interviewer understand your approach. Practice explaining your code as you write it; this simulates the interview environment.

What are common mistakes when solving Minimum Swaps to Group All 1's Together II?

Forgetting to handle the circular nature of the array, leading to incorrect subarray evaluations. Not counting the total number of 1's correctly, which can skew the swap calculations.

#2134

Minimum Swaps to Group All 1's Together II

Medium

ArraySliding WindowSliding WindowArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution uses a sliding window approach to efficiently calculate the minimum swaps needed. By keeping track of the number of 1's in a window of size equal to the total number of 1's, we can determine how many swaps are needed in constant time as we slide the window.

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Algorithm

5 steps

1Step 1: Count the total number of 1's in the array.
2Step 2: Create a sliding window of size equal to the total number of 1's.
3Step 3: Count the number of 1's in the initial window and calculate the swaps needed.
4Step 4: Slide the window across the array, updating the count of 1's and the swaps needed.
5Step 5: Keep track of the minimum swaps found.

solution.py15 lines

1# Full working Python code
2
3def minSwaps(nums):
4    total_ones = sum(nums)
5    n = len(nums)
6    nums = nums * 2  # To handle circular nature
7    current_ones = sum(nums[:total_ones])
8    min_swaps = total_ones - current_ones
9
10    for i in range(total_ones, n):
11        current_ones += nums[i] - nums[i - total_ones]
12        swaps = total_ones - current_ones
13        min_swaps = min(min_swaps, swaps)
14
15    return min_swaps

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array a constant number of times, and the space complexity is O(n) due to the extended array.

1Understanding the circular nature of the array is crucial for solving the problem efficiently.
2Using a sliding window approach allows us to minimize the number of swaps needed in linear time.

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