1234. Replace the sub -string to get a balanced string One question daily
topic：
1234. Replace the sub -string to get a balanced string.md
Thought：
all():if bool(x) For all the values in iterative objects x All for True，Then return True。 if可迭代对象为空，Then return True。
Tongxiang dual pointer，The solution of the spiritual god of this question。 If in this string，These four characters happen just to appear n/4 Second-rate，Then it is one「Balanced string」。 if在待替换子串之外的任意字符的出现Second-rate数都Exceed ，So no matter how you replace it，都无法make这个字符的出现Second-rate数等于m。 on the other hand，if在待替换子串之外的任意字符的出现Second-rate数都不Exceed m，
So it can be replaced ，make s 为Balanced string，即每个字符的出现Second-rate数均为 m。 For this question，The left and right end points of the sub -string are left and right，enumerate right， if子串外的任意字符的出现Second-rate数都不Exceedm，The explanation from left arrive rightThis sub -string can be to replace the sub -string，Length right−left+1 Update the minimum value of the answer，Move to the right left，Sumid sub -string length。
Code：
class Solution:
def balancedString(self, s: str) -> int:
s_c = Counter(s)
n = len(s)
if all(s_c[v] <= n//4 for v in s_c):
return 0
ans, left = inf, 0
# enumerate右端点
for i, j in enumerate(s):
s_c[j] -= 1
while all(s_c[v] <= n // 4 for v in s_c):
ans = min(ans, i - left + 1)
s_c[s[left]] += 1
left += 1
return ansfunc balancedString(s string) int {
cnt, m := ['X']int{}, len(s)/4
for _, c := range s {
cnt[c]++
}
if cnt['Q'] == m && cnt['W'] == m && cnt['E'] == m && cnt['R'] == m {
return 0
}
ans, left := len(s), 0
for right, c := range s {
cnt[c]--
for cnt['Q'] <= m && cnt['W'] <= m && cnt['E'] <= m && cnt['R'] <= m {
ans = min(ans, right-left+1)
cnt[s[left]]++
left++
}
}
return ans
}
func min(a, b int) int { if a > b { return b }; return a }贡献者
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