Minimum Swaps to Group All 1's Together II

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ArraysSliding Windows

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] using 1 swap.
[0,1,1,1,0,0,0] using 1 swap.
[1,1,0,0,0,0,1] using 2 swaps (using the circular property of the array).
There is no way to group all 1's together with 0 swaps.
Thus, the minimum number of swaps required is 1.

Example 2:

Input: nums = [0,1,1,1,0,0,1,1,0]
Output: 2
Explanation: Here are a few of the ways to group all the 1's together:
[1,1,1,0,0,0,0,1,1] using 2 swaps (using the circular property of the array).
[1,1,1,1,1,0,0,0,0] using 2 swaps.
There is no way to group all 1's together with 0 or 1 swaps.
Thus, the minimum number of swaps required is 2.

Example 3:

Input: nums = [1,1,0,0,1]
Output: 0
Explanation: All the 1's are already grouped together due to the circular property of the array.
Thus, the minimum number of swaps required is 0.

Constraints:

1 <= nums.length <= 10⁵
nums[i] is either 0 or 1.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum size of the input array `nums`?
Can the input array `nums` be empty or contain null values?
If the input array `nums` contains no 1's, what should the function return?
Is it possible for the input array to contain values other than 0 and 1?
By 'adjacent elements', do you mean only elements at indices `i` and `i+1` can be swapped, even considering the circular nature?

Brute Force Solution

Approach

The brute force approach tries every possible arrangement of the elements to find the minimum swaps. Imagine physically moving elements around in all possible ways and counting the swaps each time. We select the arrangement with the fewest swaps.

Here's how the algorithm would work step-by-step:

First, count how many 1s we have in total. This number is fixed and will be our target.
Imagine a window that is just big enough to contain the target number of 1s.
Now, slide this window across all possible positions in the sequence.
For each position of the window, count how many 1s are already inside the window.
Calculate the difference between the total number of 1s and the number of 1s inside the window for the current window position. This tells us the minimum swaps needed for that position.
Do this for every possible window position.
Finally, find the smallest number of swaps across all the window positions we tried. This is our answer.

Code Implementation

def min_swaps_group_ones_brute_force(numbers):
    number_of_ones = sum(numbers)

    list_length = len(numbers)
    min_swaps = float('inf')

    # Iterate through all possible window positions.

    for start_index in range(list_length):
        current_swaps = 0
        ones_in_window = 0

        # Count 1s in the current window.
        for index in range(number_of_ones):
            current_index = (start_index + index) % list_length
            if numbers[current_index] == 1:
                ones_in_window += 1

        # Calculate swaps for this window. 
        current_swaps = number_of_ones - ones_in_window

        #Update minimum swaps.
        min_swaps = min(min_swaps, current_swaps)

    return min_swaps

Big(O) Analysis

Time Complexity

O(n) – First, we count the total number of 1s, which takes O(n) time. Then, we iterate through the array using a sliding window of a fixed size (the total count of 1s). Inside the sliding window, for each window position, we count the number of 1s, which takes O(1) time since the window size is fixed. Finally, we find the minimum swaps, which takes O(n) to compare swaps across all window positions. Therefore, the overall time complexity is O(n).

Space Complexity

O(1) – The provided solution calculates the minimum swaps by sliding a window across the input sequence. It primarily uses integer variables to store counts of 1s and track the minimum swaps found so far. No auxiliary data structures like arrays or hash maps are used. Therefore, the space required remains constant regardless of the input size N, resulting in O(1) space complexity.

Optimal Solution

Approach

The problem asks us to find the fewest moves to group all the ones together. Instead of moving ones around, imagine a fixed-size window sliding across the data. The optimal approach focuses on identifying the best window location.

Here's how the algorithm would work step-by-step:

First, count how many ones there are in total. This tells us the size of the group of ones we want to create.
Imagine a window that is exactly the size of the group of ones you identified.
Slide this window across the data, one position at a time.
At each position, count how many ones are currently inside the window.
The best window position is the one with the most ones already inside it. This means we'll need to move the fewest numbers to fill it with ones.
The fewest moves needed will be the window size, minus the maximum number of ones you found inside any window position.

Code Implementation

def min_swaps(data):
    total_ones = sum(data)

    # Handle edge case where all elements are 0 or 1
    if total_ones == 0 or total_ones == len(data):
        return 0

    window_size = total_ones
    max_ones_in_window = 0
    current_ones_in_window = 0

    # Initialize the first window
    for i in range(window_size):
        current_ones_in_window += data[i]

    max_ones_in_window = current_ones_in_window

    # Slide the window and update the max number of ones
    for i in range(window_size, len(data)): 
        current_ones_in_window += data[i] - data[i - window_size]
        max_ones_in_window = max(max_ones_in_window, current_ones_in_window)

    # The result is the window size minus the max ones in any window
    return window_size - max_ones_in_window

Big(O) Analysis

Time Complexity

O(n) – The algorithm first counts the total number of ones, which takes O(n) time, where n is the size of the input array. Then, it slides a window of fixed size (number of ones) across the array. In each position, it counts the number of ones within the window. Since each element is visited a constant number of times (once by the initial count and at most once by the sliding window), the overall time complexity is dominated by the single pass through the array. Therefore, the time complexity is O(n).

Space Complexity

O(1) – The algorithm calculates the total number of ones and then iterates through the input using a sliding window. It only stores a few integer variables: the total number of ones, the current number of ones in the window, and the maximum number of ones seen in any window. These variables consume a constant amount of extra space, irrespective of the input size N, where N is the number of elements in the input array. Therefore, the space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 since there are no 1's to group.
Array with all 0s	Return 0 since no swaps are needed; there are no 1's to group.
Array with all 1s	Return 0 since no swaps are needed; all 1's are already grouped.
Array with only one 1	Return 0 since no swaps are needed; a single 1 is considered grouped.
Array with only one 0	Return 0 swaps are needed since we only need to consider swapping around the number of ones and there will always be enough free space.
Array size is very large, causing potential integer overflow when calculating counts or window sizes	Use long data type for counts and window sizes to prevent integer overflow.
Circular array requires considering the wrap-around condition for the sliding window	Extend the array virtually by concatenating it with itself to handle the circular nature.
The number of 1s is greater than half the size of the array	The sliding window size calculation and subsequent minimum swaps calculation should correctly handle scenarios with a high density of 1s.

Null or empty input array

How to Handle:

Return 0 since there are no 1's to group.

Array with all 0s

How to Handle:

Return 0 since no swaps are needed; there are no 1's to group.

Array with all 1s

How to Handle:

Return 0 since no swaps are needed; all 1's are already grouped.

Array with only one 1

How to Handle:

Return 0 since no swaps are needed; a single 1 is considered grouped.

Array with only one 0

How to Handle:

Return 0 swaps are needed since we only need to consider swapping around the number of ones and there will always be enough free space.

Array size is very large, causing potential integer overflow when calculating counts or window sizes

How to Handle:

Use long data type for counts and window sizes to prevent integer overflow.

Circular array requires considering the wrap-around condition for the sliding window

How to Handle:

Extend the array virtually by concatenating it with itself to handle the circular nature.

The number of 1s is greater than half the size of the array

How to Handle:

The sliding window size calculation and subsequent minimum swaps calculation should correctly handle scenarios with a high density of 1s.