2025-10-17 20:03:35 +0300 MSK
Maximize the Number of Partitions After Operations
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Code
class Solution:
def maxPartitionsAfterOperations(self, s: str, k: int) -> int:
n = len(s)
left = [[0] * 3 for _ in range(n)]
right = [[0] * 3 for _ in range(n)]
num, mask, count = 0, 0, 0
for i in range(n - 1):
binary = 1 << (ord(s[i]) - ord("a"))
if not (mask & binary):
count += 1
if count <= k:
mask |= binary
else:
num += 1
mask = binary
count = 1
left[i + 1][0] = num
left[i + 1][1] = mask
left[i + 1][2] = count
num, mask, count = 0, 0, 0
for i in range(n - 1, 0, -1):
binary = 1 << (ord(s[i]) - ord("a"))
if not (mask & binary):
count += 1
if count <= k:
mask |= binary
else:
num += 1
mask = binary
count = 1
right[i - 1][0] = num
right[i - 1][1] = mask
right[i - 1][2] = count
max_val = 0
for i in range(n):
seg = left[i][0] + right[i][0] + 2
tot_mask = left[i][1] | right[i][1]
tot_count = bin(tot_mask).count("1")
if left[i][2] == k and right[i][2] == k and tot_count < 26:
seg += 1
elif min(tot_count + 1, 26) <= k:
seg -= 1
max_val = max(max_val, seg)
return max_val