Problem Description

You are given a binary circular array nums (containing only 0s and 1s). In a circular array, the first and last elements are considered adjacent to each other.

A swap operation means taking two distinct positions in the array and exchanging their values.

Your task is to find the minimum number of swaps needed to group all the 1s together at any location in the array.

For example, if you have nums = [1,0,1,0,1], you could group all 1s together to form [0,0,1,1,1] or [1,1,1,0,0] (remember it's circular, so the 1s at the beginning and end are adjacent).

The key insight is that since the array is circular, you need to find the best consecutive segment of length k (where k is the total number of 1s) that already contains the most 1s. The minimum swaps needed would be k minus the number of 1s already in that optimal segment.

The solution uses a sliding window approach to check all possible windows of size k in the circular array, keeping track of which window contains the maximum number of 1s. The window slides through positions [0, k-1], [1, k], ..., and wraps around using modulo arithmetic to handle the circular nature of the array.

Intuition

Since we want to group all 1s together, let's think about what the final result looks like. If we have k ones in total, then in the final configuration, these k ones will occupy k consecutive positions in the circular array.

The key realization is that we don't need to determine exactly how to perform the swaps. Instead, we can think about it differently: if we choose any segment of length k in the circular array to be our "target zone" where all 1s should be grouped, how many 1s are already in that zone?

If a segment already contains m ones, then there are k - m ones outside this segment that need to be swapped in. Each of these outside 1s needs exactly one swap with a 0 inside the segment. Therefore, the number of swaps needed is k - m.

To minimize the number of swaps, we want to maximize m - find the segment of length k that already contains the most 1s. This transforms our problem into finding the maximum sum of a sliding window of size k in a circular binary array.

The circular nature of the array means we need to consider windows that wrap around. For instance, in array [0,1,0,1,1] with window size 3, we need to check windows like [1,1,0] (positions 3,4,0). We handle this by extending our iteration to n + k positions and using modulo arithmetic to wrap around: nums[i % n] gives us the circular access pattern we need.

This sliding window approach efficiently checks all possible target zones and finds the one requiring the minimum number of swaps.

Learn more about Sliding Window patterns.

Solution Approach

First, we count the total number of 1s in the array using k = nums.count(1). This tells us the size of the window we need to consider - all 1s must eventually be grouped into a consecutive segment of length k.

Next, we initialize our sliding window by calculating the sum of the first k elements: mx = cnt = sum(nums[:k]). Here, cnt represents the current count of 1s in our window, and mx tracks the maximum count we've seen so far.

Now we slide the window through the circular array. The loop runs from position k to n + k - 1, which ensures we check all possible windows in the circular array:

for i in range(k, n + k):
    cnt += nums[i % n]
    cnt -= nums[(i - k + n) % n]
    mx = max(mx, cnt)

For each iteration:

• nums[i % n] adds the new element entering the window (using modulo to handle circular indexing)
• nums[(i - k + n) % n] removes the element leaving the window (the one that's k positions behind)
• We update mx to keep track of the maximum number of 1s found in any window

The expression (i - k + n) % n ensures proper circular indexing for the element being removed. Adding n before taking modulo handles cases where i - k might be negative.

Finally, we return k - mx. Since mx represents the maximum number of 1s already positioned correctly in some window of size k, we need k - mx swaps to bring the remaining 1s into that window.

Time complexity: O(n) where n is the length of the array, as we slide through the array once. Space complexity: O(1) as we only use a constant amount of extra space.

Example Walkthrough

Let's walk through the solution with nums = [0,1,0,1,1,0,0].

Step 1: Count total 1s

• Count all 1s in the array: k = 3
• This means we need to group all three 1s into a consecutive segment of length 3

Step 2: Initialize the first window

• First window of size 3: [0,1,0] (positions 0-2)
• Initial count: cnt = 0 + 1 + 0 = 1
• Maximum so far: mx = 1

Step 3: Slide the window through the circular array

Window at position [1,2,3]: [1,0,1]

• Add nums[3] = 1: cnt = 1 + 1 = 2
• Remove nums[0] = 0: cnt = 2 - 0 = 2
• Update mx = max(1, 2) = 2

Window at position [2,3,4]: [0,1,1]

• Add nums[4] = 1: cnt = 2 + 1 = 3
• Remove nums[1] = 1: cnt = 3 - 1 = 2
• Update mx = max(2, 2) = 2

Window at position [3,4,5]: [1,1,0]

• Add nums[5] = 0: cnt = 2 + 0 = 2
• Remove nums[2] = 0: cnt = 2 - 0 = 2
• Update mx = max(2, 2) = 2

Window at position [4,5,6]: [1,0,0]

• Add nums[6] = 0: cnt = 2 + 0 = 2
• Remove nums[3] = 1: cnt = 2 - 1 = 1
• Update mx = max(2, 1) = 2

Window at position [5,6,0] (wrapping around): [0,0,0]

• Add nums[7%7] = nums[0] = 0: cnt = 1 + 0 = 1
• Remove nums[4] = 1: cnt = 1 - 1 = 0
• Update mx = max(2, 0) = 2

Window at position [6,0,1] (wrapping around): [0,0,1]

• Add nums[8%7] = nums[1] = 1: cnt = 0 + 1 = 1
• Remove nums[5] = 0: cnt = 1 - 0 = 1
• Update mx = max(2, 1) = 2

Step 4: Calculate minimum swaps

• Maximum 1s found in any window of size 3: mx = 2
• Minimum swaps needed: k - mx = 3 - 2 = 1

The best window found was [1,0,1] (or [0,1,1]) which already contains 2 ones. We only need 1 swap to bring the remaining 1 into this window. For example, we could swap the 0 at position 2 with the 1 at position 4 to get all 1s grouped together.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the array nums.

The algorithm performs the following operations:

• nums.count(1) traverses the entire array once: O(n)
• sum(nums[:k]) sums at most n elements: O(k) where k ≤ n
• The for loop iterates exactly k times: O(k) where k ≤ n
• Inside each iteration, we perform constant time operations (array access, addition, subtraction, comparison): O(1)

Total time complexity: O(n) + O(k) + O(k) × O(1) = O(n) since k ≤ n.

Space Complexity: O(1)

The algorithm uses only a fixed number of variables:

• k: stores the count of 1s
• mx: tracks the maximum count
• cnt: maintains the current window sum
• n: stores the array length
• i: loop counter

No additional data structures are created that scale with the input size, resulting in constant space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Forgetting the Circular Nature of the Array

Pitfall: A common mistake is treating the array as linear and only checking windows from index 0 to n-k, missing potential optimal windows that wrap around the array.

Example: For nums = [1,1,0,0,1], if you only check windows [0:3], [1:4], [2:5], you'd miss the optimal circular window [3:5] + [0:1] which contains all three 1s.

Solution: Always iterate through n complete windows by extending the loop to range(k, n + k) and using modulo arithmetic for indexing.

2. Incorrect Modulo Calculation for Window Start

Pitfall: When calculating the element leaving the window, using (i - k) % n instead of (i - k + n) % n can produce negative indices in Python, leading to incorrect array access.

Example: When i = 2, k = 3, n = 5:

• Wrong: (2 - 3) % 5 = -1 (accesses the last element incorrectly)
• Correct: (2 - 3 + 5) % 5 = 4 (correctly wraps to index 4)

Solution: Always add n before taking modulo to ensure positive indices: (i - k + n) % n.

3. Off-by-One Error in Initial Window

Pitfall: Incorrectly initializing the first window sum, such as using sum(nums[:k-1]) or sum(nums[:k+1]).

Example: If k = 3 and nums = [1,0,1,0,1]:

• Wrong: sum(nums[:2]) only sums 2 elements
• Correct: sum(nums[:3]) correctly sums the first 3 elements

Solution: Use sum(nums[:k]) for the initial window, remembering that Python slicing is exclusive of the end index.

4. Edge Case: All 1s or All 0s

Pitfall: Not handling arrays that contain all 1s or all 0s, which could lead to unnecessary calculations or incorrect results.

Example:

• nums = [1,1,1,1]: Should return 0 immediately
• nums = [0,0,0,0]: Should return 0 immediately

Solution: Add early return conditions:

if total_ones == 0 or total_ones == array_length:
    return 0

5. Using Wrong Variable Names

Pitfall: In the provided code, variable names could be confusing. Using generic names like i, k, mx, cnt makes the code harder to debug and understand.

Solution: Use descriptive variable names:

total_ones = nums.count(1)
current_ones_in_window = sum(nums[:total_ones])
max_ones_in_window = current_ones_in_window