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Minimum Swaps to Group All 1's Together

Medium

Asked by:

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Topics:

ArraysSliding Windows

Given a binary array data, return the minimum number of swaps required to group all 1’s present in the array together in any place in the array.

Example 1:

Input: data = [1,0,1,0,1]
Output: 1
Explanation: There are 3 1’s in data. The best place to group them is in the center. The minimum swap needed is 1.

Example 2:

Input: data = [0,0,0,1,0]
Output: 0
Explanation: There is only one 1 in data.

Example 3:

Input: data = [1,0,1,0,1,0,0,1,1,0,1]
Output: 3

Constraints:

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

Try coding it in LeetCode

Minimum Swaps to Group All 1's Together

Medium

Asked by:

11 views

Topics:

ArraysSliding Windows

Given a binary array data, return the minimum number of swaps required to group all 1’s present in the array together in any place in the array.

Example 1:

Input: data = [1,0,1,0,1]
Output: 1
Explanation: There are 3 1’s in data. The best place to group them is in the center. The minimum swap needed is 1.

Example 2:

Input: data = [0,0,0,1,0]
Output: 0
Explanation: There is only one 1 in data.

Example 3:

Input: data = [1,0,1,0,1,0,0,1,1,0,1]
Output: 3

Constraints:

1 <= data.length <= 10⁵
data[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, and what is the expected performance in terms of time and space complexity?
What should I return if the input array is null or empty?
Does the array contain only 0s and 1s as stated, or could there be other integer values present?
If there are no 1s in the array, what should the return value be?
If multiple subarrays containing all the 1's with the same minimum swaps are possible, is any one of them acceptable?

Brute Force Solution

Approach

The brute force approach is like trying every possible arrangement of items to find the best one. For this problem, we will be looking at all possible sections within the list to find one that needs the least amount of changes. We're looking for a section that has as many 1's already, which means it would need less changes.

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

First, figure out how many 1's there are in the entire list.
Then, consider a section of the list that is exactly the same size as the number of 1's we found.
Move this section across the entire list, one position at a time.
At each position, count how many 1's are already in that section.
Calculate the number of swaps needed for each section by subtracting the number of 1's in the section from the total number of 1's in the list.
Keep track of the lowest number of swaps needed that we've seen so far.
After checking all the positions, the lowest number of swaps we kept track of will be the answer.

Code Implementation

def min_swaps_to_group_ones_brute_force(data):
    number_of_ones = sum(data)

    minimum_swaps_needed = float('inf')

    # Iterate through all possible subarrays of length number_of_ones
    for i in range(len(data) - number_of_ones + 1):
        current_window = data[i:i + number_of_ones]

        # Count the number of ones in the current window
        ones_in_current_window = sum(current_window)

        # Calculate swaps needed: total ones - ones in window
        swaps_needed = number_of_ones - ones_in_current_window

        # Update the minimum swaps seen so far
        minimum_swaps_needed = min(minimum_swaps_needed, swaps_needed)

    return minimum_swaps_needed

Big(O) Analysis

Time Complexity

O(n) – The algorithm first counts the total number of 1's in the input array of size n. Then, it iterates through the array using a sliding window of size equal to the count of 1's. Within each window, it counts the number of 1's. Therefore, the algorithm performs a single pass through the array and the cost is driven by the size of the input array n. This results in a time complexity of O(n).

Space Complexity

O(1) – The algorithm calculates the total number of 1's, counts 1's within a sliding window, and tracks the minimum number of swaps needed. These operations utilize a few integer variables to store counts and the minimum swaps value. The space required for these variables remains constant regardless of the input array size (N). Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The goal is to find the fewest number of adjustments needed to bring all the ones together in a sequence. We do this by focusing on the biggest possible group of ones that we could fit into a sliding window and calculating how many adjustments that would take.

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

First, count the total number of ones in the entire sequence. This tells us the size of the ideal group we want to create.
Imagine a window sliding across the sequence, with the window's length equal to the total count of ones.
As the window slides, keep track of how many ones are currently inside the window.
The more ones inside the window, the fewer adjustments are needed since more ones are already together.
Find the position of the window that contains the largest number of ones.
The number of zeros inside this window represents the minimum number of adjustments needed, because these zeros need to be swapped out to make a complete group of ones.

Code Implementation

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

    # If no ones or all ones, no swaps are needed.
    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 window and count initial ones
    for i in range(window_size):
        current_ones_in_window += data[i]

    max_ones_in_window = current_ones_in_window

    # Slide the window to find the max 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 minimum swaps needed.
    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 length of the sequence. Then, it uses a sliding window of fixed size (equal to the total number of ones) to iterate through the sequence. The sliding window operation involves checking the number of ones within the window in O(1) time for each slide because we can maintain a running count. Since the window slides at most n times, this part also takes O(n) time. Thus, the overall time complexity is dominated by the initial count and the sliding window iteration, resulting in O(n) + O(n) which simplifies to O(n).

Space Complexity

O(1) – The algorithm primarily uses a sliding window approach and stores a count of ones within the window, as well as the total count of ones. These are stored in constant space integer variables. The space required does not scale with the input array size, denoted as N. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty or null input array	Return 0 since no swaps are needed for an empty or null array.
Array with all 0s	Return 0 since no swaps are needed as there are no 1s to group.
Array with all 1s	Return 0 since all 1s are already grouped.
Array with only one element	Return 0 since a single element is already grouped.
Array with only two elements, both 0	Return 0 as no swap needed.
Array with only two elements, both 1	Return 0 as no swap needed.
Large array with a few scattered 1s	Sliding window approach should efficiently find the optimal window regardless of array size or 1s distribution.
Integer overflow when calculating the number of ones in a large array (language specific)	Use appropriate data types (e.g., long in Java or C++) to store the count of ones to prevent overflow.

Empty or null input array

How to Handle:

Return 0 since no swaps are needed for an empty or null array.

Array with all 0s

How to Handle:

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

Array with all 1s

How to Handle:

Return 0 since all 1s are already grouped.

Array with only one element

How to Handle:

Return 0 since a single element is already grouped.

Array with only two elements, both 0

How to Handle:

Return 0 as no swap needed.

Array with only two elements, both 1

How to Handle:

Return 0 as no swap needed.

Large array with a few scattered 1s

How to Handle:

Sliding window approach should efficiently find the optimal window regardless of array size or 1s distribution.

Integer overflow when calculating the number of ones in a large array (language specific)

How to Handle:

Use appropriate data types (e.g., long in Java or C++) to store the count of ones to prevent overflow.

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, and what is the expected performance in terms of time and space complexity?
What should I return if the input array is null or empty?
Does the array contain only 0s and 1s as stated, or could there be other integer values present?
If there are no 1s in the array, what should the return value be?
If multiple subarrays containing all the 1's with the same minimum swaps are possible, is any one of them acceptable?

Brute Force Solution

Approach

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

First, figure out how many 1's there are in the entire list.
Then, consider a section of the list that is exactly the same size as the number of 1's we found.
Move this section across the entire list, one position at a time.
At each position, count how many 1's are already in that section.
Calculate the number of swaps needed for each section by subtracting the number of 1's in the section from the total number of 1's in the list.
Keep track of the lowest number of swaps needed that we've seen so far.
After checking all the positions, the lowest number of swaps we kept track of will be the answer.

Code Implementation

def min_swaps_to_group_ones_brute_force(data):
    number_of_ones = sum(data)

    minimum_swaps_needed = float('inf')

    # Iterate through all possible subarrays of length number_of_ones
    for i in range(len(data) - number_of_ones + 1):
        current_window = data[i:i + number_of_ones]

        # Count the number of ones in the current window
        ones_in_current_window = sum(current_window)

        # Calculate swaps needed: total ones - ones in window
        swaps_needed = number_of_ones - ones_in_current_window

        # Update the minimum swaps seen so far
        minimum_swaps_needed = min(minimum_swaps_needed, swaps_needed)

    return minimum_swaps_needed

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, count the total number of ones in the entire sequence. This tells us the size of the ideal group we want to create.
Imagine a window sliding across the sequence, with the window's length equal to the total count of ones.
As the window slides, keep track of how many ones are currently inside the window.
The more ones inside the window, the fewer adjustments are needed since more ones are already together.
Find the position of the window that contains the largest number of ones.
The number of zeros inside this window represents the minimum number of adjustments needed, because these zeros need to be swapped out to make a complete group of ones.

Code Implementation

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

    # If no ones or all ones, no swaps are needed.
    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 window and count initial ones
    for i in range(window_size):
        current_ones_in_window += data[i]

    max_ones_in_window = current_ones_in_window

    # Slide the window to find the max 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 minimum swaps needed.
    return window_size - max_ones_in_window

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty or null input array	Return 0 since no swaps are needed for an empty or null array.
Array with all 0s	Return 0 since no swaps are needed as there are no 1s to group.
Array with all 1s	Return 0 since all 1s are already grouped.
Array with only one element	Return 0 since a single element is already grouped.
Array with only two elements, both 0	Return 0 as no swap needed.
Array with only two elements, both 1	Return 0 as no swap needed.
Large array with a few scattered 1s	Sliding window approach should efficiently find the optimal window regardless of array size or 1s distribution.
Integer overflow when calculating the number of ones in a large array (language specific)	Use appropriate data types (e.g., long in Java or C++) to store the count of ones to prevent overflow.

Empty or null input array

How to Handle:

Return 0 since no swaps are needed for an empty or null array.

Array with all 0s

How to Handle:

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

Array with all 1s

How to Handle:

Return 0 since all 1s are already grouped.

Array with only one element

How to Handle:

Return 0 since a single element is already grouped.

Array with only two elements, both 0

How to Handle:

Return 0 as no swap needed.

Array with only two elements, both 1

How to Handle:

Return 0 as no swap needed.

Large array with a few scattered 1s

How to Handle:

Sliding window approach should efficiently find the optimal window regardless of array size or 1s distribution.

Integer overflow when calculating the number of ones in a large array (language specific)

How to Handle:

Use appropriate data types (e.g., long in Java or C++) to store the count of ones to prevent overflow.