Minimum Swaps to group all 1s

Given an array of 0's and 1's, we need to find the minimum swaps required to group all 1's present in the array together. In a swap operation, we swap two elements at different indexes.

Examples:

Input: arr[] = [1, 0, 1, 0, 1]
Output: 1
Explanation: Only 1 swap is required to group all 1's together. Swapping index 1 with 4 will give arr[] = [1, 1, 1, 0, 0]

Input: arr[] = [1, 1, 0, 1, 0, 1, 1]
Output: 2
Explanation: Only 2 swap is required to group all 1's together. Swapping index 0 with 2 and 1 with 4 will give arr[] = [0, 0, 1, 1, 1, 1, 1]

Input: arr[] = [0, 0, 0]
Output: -1
Explanation: No 1s are present in the array, so return -1.

[Naive Approach] Using Nested loops - O(n^2) Time and O(n) Space

The idea is based on the fact that a subarray with more 1s would require less swaps.

• Count total number of 1’s in the array. To group 1s, we need to know how many 1s are there. Let this count be x.
• Find the subarray of length x with maximum number of 1’s. To group all 1s, we need to choose a subarray that has more 1s as would require less swaps.
• The minimum swaps required will be the number of 0’s in this subarray of length x.

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[Expected Approach] Maintaining Window K(count of 1) - O(n) Time and O(1) Space

This is mainly an optimization over the above approach. Count the total number of 1s (x) in the array, then find a subarray of length x with the maximum 1s using a sliding count. Track the maximum 1s in any such subarray and return x - maxOnes as the minimum swaps needed.

• Given arr = {1, 0, 1, 0, 1}, total number of 1s x = 3, so required window size = 3
• Consider all subarrays of size 3 and count number of 1s in each
• [1, 0, 1] have 2 ones, [0, 1, 0] have 1 one, [1, 0, 1] have 2 ones
• Maximum 1s in any window = maxOnes = 2
• Minimum swaps required = number of 0s in best window = x - maxOnes = 3 - 2 = 1

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