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Minimum Adjacent Swaps for K Consecutive Ones

Hard

Asked by:

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Topics:

ArraysSliding WindowsGreedy Algorithms

You are given an integer array, nums, and an integer k. nums comprises of only 0's and 1's. In one move, you can choose two adjacent indices and swap their values.

Return the minimum number of moves required so that nums has k consecutive 1's.

Example 1:

Input: nums = [1,0,0,1,0,1], k = 2
Output: 1
Explanation: In 1 move, nums could be [1,0,0,0,1,1] and have 2 consecutive 1's.

Example 2:

Input: nums = [1,0,0,0,0,0,1,1], k = 3
Output: 5
Explanation: In 5 moves, the leftmost 1 can be shifted right until nums = [0,0,0,0,0,1,1,1].

Example 3:

Input: nums = [1,1,0,1], k = 2
Output: 0
Explanation: nums already has 2 consecutive 1's.

Constraints:

1 <= nums.length <= 10⁵
nums[i] is 0 or 1.
1 <= k <= sum(nums)

Try coding it in LeetCode

Minimum Adjacent Swaps for K Consecutive Ones

Hard

Asked by:

63 views

Topics:

ArraysSliding WindowsGreedy Algorithms

You are given an integer array, nums, and an integer k. nums comprises of only 0's and 1's. In one move, you can choose two adjacent indices and swap their values.

Return the minimum number of moves required so that nums has k consecutive 1's.

Example 1:

Input: nums = [1,0,0,1,0,1], k = 2
Output: 1
Explanation: In 1 move, nums could be [1,0,0,0,1,1] and have 2 consecutive 1's.

Example 2:

Input: nums = [1,0,0,0,0,0,1,1], k = 3
Output: 5
Explanation: In 5 moves, the leftmost 1 can be shifted right until nums = [0,0,0,0,0,1,1,1].

Example 3:

Input: nums = [1,1,0,1], k = 2
Output: 0
Explanation: nums already has 2 consecutive 1's.

Constraints:

1 <= nums.length <= 10⁵
nums[i] is 0 or 1.
1 <= k <= sum(nums)

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 are the possible values for elements in the `nums` array, and what is the maximum size of the `nums` array?
Can `k` be zero, or larger than the number of ones present in the array?
If it's impossible to find `k` consecutive ones, what should I return?
Is the order of the elements in the input array significant beyond the need for consecutive ones?
Are there any specific constraints on the range of the number of swaps I can return (e.g., should I be concerned about integer overflow)?
Are the ones already sorted to some extent?
In the case of multiple possible k consecutive ones, which one should I consider to be the 'leftmost' k consecutive ones?

Brute Force Solution

Approach

The goal is to find the fewest moves needed to group a certain number of ones together in a row. The brute force approach is to try every possible grouping of ones, and then count how many moves it would take to bring them together.

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

First, identify all the locations where ones exist.
Then, consider every possible consecutive sequence of the desired length from these locations.
For each of these sequences, figure out where the middle spot would be.
Calculate the number of moves needed to bring all the ones in that sequence to be as close as possible to that middle spot.
Remember the total number of moves needed for each sequence.
Finally, compare all the move counts, and choose the smallest one. That's the solution!

Code Implementation

def min_moves_k_consecutive_ones_brute_force(numbers, k_consecutive):

    one_indices = [i for i, number in enumerate(numbers) if number == 1]
    number_of_ones = len(one_indices)

    if number_of_ones < k_consecutive:
        return 0

    minimum_swaps = float('inf')

    # Iterate through all possible consecutive sequences of ones
    for i in range(number_of_ones - k_consecutive + 1):
        current_sequence = one_indices[i:i + k_consecutive]

        # Calculate the median index of the current sequence
        median_index = k_consecutive // 2
        median_value = current_sequence[median_index]

        current_swaps = 0
        # Calculate the number of swaps needed for current sequence
        for j in range(k_consecutive):
            current_swaps += abs(current_sequence[j] - (median_value - median_index + j))

        minimum_swaps = min(minimum_swaps, current_swaps)

    return minimum_swaps

Big(O) Analysis

Time Complexity

O(n*k) – First, we iterate through the input array of size n to identify the positions of all ones. This takes O(n) time. Then, we consider all possible consecutive sequences of length k from these one positions. In the worst case, there could be O(n) such sequences. For each sequence of length k, we calculate the moves needed to group them around the median position within the sequence. Calculating moves for each sequence of k elements takes O(k) time. Therefore, the overall time complexity is approximately O(n * k) because we are calculating the moves for each of the roughly O(n) sequences of length k. The initial O(n) operation of finding 1s is subsumed by this n*k complexity.

Space Complexity

O(N) – The space complexity is primarily determined by the need to store the indices of all the ones in the input array. In the first step, we identify all locations where ones exist, and these locations are implicitly stored in a list of size at most N, where N is the length of the input array. This list drives the space usage. Therefore, the auxiliary space used is O(N).

Optimal Solution

Approach

The key to efficiently solving this problem is to realize that the optimal arrangement has the ones clustered as closely as possible around their middle. We find the positions of the ones and then calculate the cost of moving them to form a consecutive group centered around the median.

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

First, identify the positions where the ones are located in the input.
Consider only the positions of the ones, and imagine you want to create a consecutive sequence of a specific length (K) of ones.
The best place to center this sequence is around the middle one. Find the 'middle' one in the positions of the ones.
Calculate how many moves it would take to bring all the ones in the K-length sequence next to each other, centered around that 'middle' one. Note: to avoid re-calculating, adjust the positions relative to each other.
To do this, consider pairs of ones on either side of the middle one, and find the number of moves to bring them adjacent.
Repeat for different starting points. The best result will be the one that requires the fewest moves.

Code Implementation

def minSwaps(binary_string, k):
    ones_positions = []
    for i, char in enumerate(binary_string):
        if char == '1':
            ones_positions.append(i)

    number_of_ones = len(ones_positions)
    if number_of_ones < k:
        return 0

    minimum_swaps = float('inf')

    for i in range(number_of_ones - k + 1):
        #Consider each possible window of k consecutive ones.
        current_window = ones_positions[i:i+k]
        median_index = k // 2
        median_value = current_window[median_index]
        swaps = 0
        #Adjust positions relative to the median.
        adjusted_positions = [position - median_value for position in current_window]

        current_swaps = 0
        for j in range(k):
            current_swaps += abs(adjusted_positions[j] - (j - median_index))

        minimum_swaps = min(minimum_swaps, current_swaps)

    return minimum_swaps

Big(O) Analysis

Time Complexity

O(n) – The algorithm first identifies the positions of ones, which takes O(n) time where n is the length of the input array. Then, it iterates through the identified positions of ones to find the minimum swaps. Calculating the swaps for each possible K-length sequence involves iterating through K elements, but since K is a constant (as it’s given in the question and does not change as n grows), it's a fixed number of steps. Thus, the dominant operation is finding the ones' positions initially. Therefore, the time complexity is O(n).

Space Complexity

O(N) – The primary auxiliary space usage comes from storing the indices of the ones in the input array. In the worst-case scenario, the input array consists entirely of ones, resulting in a list of size N to hold their positions. The algorithm also uses a few variables to keep track of indices and intermediate sums, which take constant space. Therefore, the dominant space complexity is determined by the list of one positions, which is O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0, as there are no elements and thus 0 swaps are needed to have K consecutive ones.
Input array contains no ones	Return -1 if it's impossible to obtain K consecutive ones (i.e., count of 1s is less than K), as no arrangement will satisfy the constraint.
Input array already has K consecutive ones in their original positions	Return 0 as no swaps are required.
K is larger than the number of ones in the array	Return -1, as it is impossible to have K consecutive ones if the number of ones is less than K.
Large input array with many ones scattered far apart.	Ensure the solution uses an efficient algorithm (e.g., prefix sum with sliding window) to avoid quadratic time complexity.
K is 1 and there is at least one '1' in the array.	Return 0, as any single '1' already satisfies the K=1 constraint.
Integer overflow potential when calculating prefix sums or distances, especially with large indices and array sizes.	Use appropriate data types (e.g., long) to prevent integer overflow.
Input array contains leading/trailing zeros with a relatively small number of ones.	Ensure that the sliding window considers these zeros correctly when minimizing the number of swaps.

Null or empty input array

How to Handle:

Return 0, as there are no elements and thus 0 swaps are needed to have K consecutive ones.

Input array contains no ones

How to Handle:

Return -1 if it's impossible to obtain K consecutive ones (i.e., count of 1s is less than K), as no arrangement will satisfy the constraint.

Input array already has K consecutive ones in their original positions

How to Handle:

Return 0 as no swaps are required.

K is larger than the number of ones in the array

How to Handle:

Return -1, as it is impossible to have K consecutive ones if the number of ones is less than K.

Large input array with many ones scattered far apart.

How to Handle:

Ensure the solution uses an efficient algorithm (e.g., prefix sum with sliding window) to avoid quadratic time complexity.

K is 1 and there is at least one '1' in the array.

How to Handle:

Return 0, as any single '1' already satisfies the K=1 constraint.

Integer overflow potential when calculating prefix sums or distances, especially with large indices and array sizes.

How to Handle:

Use appropriate data types (e.g., long) to prevent integer overflow.

Input array contains leading/trailing zeros with a relatively small number of ones.

How to Handle:

Ensure that the sliding window considers these zeros correctly when minimizing the number of swaps.

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 are the possible values for elements in the `nums` array, and what is the maximum size of the `nums` array?
Can `k` be zero, or larger than the number of ones present in the array?
If it's impossible to find `k` consecutive ones, what should I return?
Is the order of the elements in the input array significant beyond the need for consecutive ones?
Are there any specific constraints on the range of the number of swaps I can return (e.g., should I be concerned about integer overflow)?
Are the ones already sorted to some extent?
In the case of multiple possible k consecutive ones, which one should I consider to be the 'leftmost' k consecutive ones?

Brute Force Solution

Approach

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

First, identify all the locations where ones exist.
Then, consider every possible consecutive sequence of the desired length from these locations.
For each of these sequences, figure out where the middle spot would be.
Calculate the number of moves needed to bring all the ones in that sequence to be as close as possible to that middle spot.
Remember the total number of moves needed for each sequence.
Finally, compare all the move counts, and choose the smallest one. That's the solution!

Code Implementation

def min_moves_k_consecutive_ones_brute_force(numbers, k_consecutive):

    one_indices = [i for i, number in enumerate(numbers) if number == 1]
    number_of_ones = len(one_indices)

    if number_of_ones < k_consecutive:
        return 0

    minimum_swaps = float('inf')

    # Iterate through all possible consecutive sequences of ones
    for i in range(number_of_ones - k_consecutive + 1):
        current_sequence = one_indices[i:i + k_consecutive]

        # Calculate the median index of the current sequence
        median_index = k_consecutive // 2
        median_value = current_sequence[median_index]

        current_swaps = 0
        # Calculate the number of swaps needed for current sequence
        for j in range(k_consecutive):
            current_swaps += abs(current_sequence[j] - (median_value - median_index + j))

        minimum_swaps = min(minimum_swaps, current_swaps)

    return minimum_swaps

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, identify the positions where the ones are located in the input.
Consider only the positions of the ones, and imagine you want to create a consecutive sequence of a specific length (K) of ones.
The best place to center this sequence is around the middle one. Find the 'middle' one in the positions of the ones.
Calculate how many moves it would take to bring all the ones in the K-length sequence next to each other, centered around that 'middle' one. Note: to avoid re-calculating, adjust the positions relative to each other.
To do this, consider pairs of ones on either side of the middle one, and find the number of moves to bring them adjacent.
Repeat for different starting points. The best result will be the one that requires the fewest moves.

Code Implementation

def minSwaps(binary_string, k):
    ones_positions = []
    for i, char in enumerate(binary_string):
        if char == '1':
            ones_positions.append(i)

    number_of_ones = len(ones_positions)
    if number_of_ones < k:
        return 0

    minimum_swaps = float('inf')

    for i in range(number_of_ones - k + 1):
        #Consider each possible window of k consecutive ones.
        current_window = ones_positions[i:i+k]
        median_index = k // 2
        median_value = current_window[median_index]
        swaps = 0
        #Adjust positions relative to the median.
        adjusted_positions = [position - median_value for position in current_window]

        current_swaps = 0
        for j in range(k):
            current_swaps += abs(adjusted_positions[j] - (j - median_index))

        minimum_swaps = min(minimum_swaps, current_swaps)

    return minimum_swaps

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0, as there are no elements and thus 0 swaps are needed to have K consecutive ones.
Input array contains no ones	Return -1 if it's impossible to obtain K consecutive ones (i.e., count of 1s is less than K), as no arrangement will satisfy the constraint.
Input array already has K consecutive ones in their original positions	Return 0 as no swaps are required.
K is larger than the number of ones in the array	Return -1, as it is impossible to have K consecutive ones if the number of ones is less than K.
Large input array with many ones scattered far apart.	Ensure the solution uses an efficient algorithm (e.g., prefix sum with sliding window) to avoid quadratic time complexity.
K is 1 and there is at least one '1' in the array.	Return 0, as any single '1' already satisfies the K=1 constraint.
Integer overflow potential when calculating prefix sums or distances, especially with large indices and array sizes.	Use appropriate data types (e.g., long) to prevent integer overflow.
Input array contains leading/trailing zeros with a relatively small number of ones.	Ensure that the sliding window considers these zeros correctly when minimizing the number of swaps.

Null or empty input array

How to Handle:

Return 0, as there are no elements and thus 0 swaps are needed to have K consecutive ones.

Input array contains no ones

How to Handle:

Return -1 if it's impossible to obtain K consecutive ones (i.e., count of 1s is less than K), as no arrangement will satisfy the constraint.

Input array already has K consecutive ones in their original positions

How to Handle:

Return 0 as no swaps are required.

K is larger than the number of ones in the array

How to Handle:

Return -1, as it is impossible to have K consecutive ones if the number of ones is less than K.

Large input array with many ones scattered far apart.

How to Handle:

Ensure the solution uses an efficient algorithm (e.g., prefix sum with sliding window) to avoid quadratic time complexity.

K is 1 and there is at least one '1' in the array.

How to Handle:

Return 0, as any single '1' already satisfies the K=1 constraint.

Integer overflow potential when calculating prefix sums or distances, especially with large indices and array sizes.

How to Handle:

Use appropriate data types (e.g., long) to prevent integer overflow.

Input array contains leading/trailing zeros with a relatively small number of ones.

How to Handle:

Ensure that the sliding window considers these zeros correctly when minimizing the number of swaps.