1151. Minimum Swaps to Group All 1's Together

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

Example 1:

Input: [1,0,1,0,1]
Output: 1
Explanation: 
There are 3 ways to group all 1's together:
[1,1,1,0,0] using 1 swap.
[0,1,1,1,0] using 2 swaps.
[0,0,1,1,1] using 1 swap.
The minimum is 1.

Example 2:

Input: [0,0,0,1,0]
Output: 0
Explanation: 
Since there is only one 1 in the array, no swaps needed.

Example 3:

Input: [1,0,1,0,1,0,0,1,1,0,1]
Output: 3
Explanation: 
One possible solution that uses 3 swaps is [0,0,0,0,0,1,1,1,1,1,1].

Note:

1. 1 <= data.length <= 10^5
2. 0 <= data[i] <= 1

intuition: the # of 1s that should be grouped together is the # of 1's the whole array has. every subarray of size ones, need several number of swaps to reach, which is the number of zeros in that subarray. 

use sliding window, check all the window with the same length n (# of 1s), find the maximum one which already contains the most 1s. then swap the rest: n-max.

time = O(n), space = O(1)

class Solution {
    public int minSwaps(int[] data) {
        int numOfOnes = 0;
        for(int num : data) {
            if(num == 1) {
                numOfOnes++;
            }
        }
        
        int slow = 0, fast = 0, counter = 0, max = 0;   // max # of 1s in current window
        while(fast < data.length) {
            while(fast < data.length && fast - slow < numOfOnes) {  // window size of numOfOnes
                if(data[fast++] == 1) {
                    counter++;
                }
            }
            max = Math.max(max, counter);
            if(fast == data.length) {
                break;
            }
            
            if(data[slow++] == 1) {
                counter--;
            }
        }
        return numOfOnes - max;
    }
}