Minimum swaps to group all 1s together II

Link: https://leetcode.com/problems/minimum-swaps-to-group-all-1s-together-ii/

Solution:

Topics: sliding window

Intuition
This is essentially the same problem as Minimum swaps to group all 1s together except for the fact that the first are last element are adjacent, so the idea here is to employ 2 passes by wrapping around the array, no extra memory is required.

Implementation

def min_swaps(nums):
	window = sum(nums)
	most_ones = 0
	curr_sum = 0
	l = 0
	for r in range(len(nums)*2):
		curr_sum += nums[r % len(nums)]
		if r >= window:
			curr_sum -= nums[l % len(nums)]
			l += 1
		most_ones = max(most_ones, curr_sum)
	return window - most_ones
 
#time: o(n)
#memory: o(1)

Review 1
Same as Minimum swaps to group all 1s together

review