3003. Maximize the Number of Partitions After Operations

# Time:  O(n)
# Space: O(n)

# prefix sum, greedy
class Solution(object):
    def maxPartitionsAfterOperations(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: int
        """
        def popcount(n):
            n = (n & 0x55555555) + ((n >> 1) & 0x55555555)
            n = (n & 0x33333333) + ((n >> 2) & 0x33333333)
            n = (n & 0x0F0F0F0F) + ((n >> 4) & 0x0F0F0F0F)
            n = (n & 0x00FF00FF) + ((n >> 8) & 0x00FF00FF)
            n = (n & 0x0000FFFF) + ((n >> 16) & 0x0000FFFF)
            return n

        left = [0]*(len(s)+1)
        left_mask = [0]*(len(s)+1)
        cnt = mask = 0
        for i in xrange(len(s)):
            mask |= 1<<(ord(s[i])-ord('a'))
            if popcount(mask) > k:
                cnt += 1
                mask = 1<<(ord(s[i])-ord('a'))
            left[i+1] = cnt
            left_mask[i+1] = mask
        right = [0]*(len(s)+1)
        right_mask = [0]*(len(s)+1)
        cnt = mask = 0
        for i in reversed(xrange(len(s))):
            mask |= 1<<(ord(s[i])-ord('a'))
            if popcount(mask) > k:
                cnt += 1
                mask = 1<<(ord(s[i])-ord('a'))
            right[i] = cnt
            right_mask[i] = mask

        result = 0
        for i in xrange(len(s)):
            curr = left[i]+right[i+1]
            mask = left_mask[i]|right_mask[i+1]
            if popcount(left_mask[i]) == popcount(right_mask[i+1]) == k and popcount(mask) != 26:
                curr += 3
            elif popcount(mask)+int(popcount(mask) != 26) > k:  # test case: s = "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz", k = 26
                curr += 2
            else:
                curr += 1
            result = max(result, curr)
        return result

Solution from kamyu104/LeetCode-Solutions · MIT