Let count1 be the total number of 1s in data. In the final arrangement, all 1s must occupy one contiguous block of length count1, because there are exactly count1 ones to place.

So the problem becomes: among all windows of length count1, which window is best to turn into that block? A swap is only needed for each 0 inside the chosen window (swap it with a 1 outside), so we want the window with the fewest 0s, or equivalently the most 1s.

Use a fixed-size sliding window. Keep total as the number of 1s in the current window. Initialize it for the first count1 elements, then slide right by adding the new element and removing the element that left the window. For every window, required swaps are count1 - total. Track the minimum of this value.

This gives an O(n) solution with O(1) extra space.

Implementation

def minSwaps(self, data):
    count1 = data.count(1)
    total = 0
    for i in range(count1): total += data[i]
    swaps = count1-total
    for r in range(count1, len(data)):
        total += data[r]
        total -= data[r-count1]
        swaps = min(swaps, count1-total)
    return swaps