# 926. Flip String to Monotone Increasing

A string of `'0'`s and `'1'`s is monotone increasing if it consists of some number of `'0'`s (possibly 0), followed by some number of `'1'`s (also possibly 0.)

We are given a string `S` of `'0'`s and `'1'`s, and we may flip any `'0'` to a `'1'` or a `'1'` to a `'0'`.

Return the minimum number of flips to make `S` monotone increasing.

Example 1:

``````Input: "00110"
Output: 1
Explanation: We flip the last digit to get 00111.
``````

Example 2:

``````Input: "010110"
Output: 2
Explanation: We flip to get 011111, or alternatively 000111.
``````

Example 3:

``````Input: "00011000"
Output: 2
Explanation: We flip to get 00000000.
``````

Note:

1. `1 <= S.length <= 20000`
2. `S` only consists of `'0'` and `'1'` characters.

``````class Solution {
public:
int minFlipsMonoIncr(string S) {
int n = S.size(), res = INT_MAX;
vector<int> cnt1(n + 1), cnt0(n + 1);
for (int i = 1, j = n - 1; j >= 0; ++i, --j) {
cnt1[i] += cnt1[i - 1] + (S[i - 1] == '0' ? 0 : 1);
cnt0[j] += cnt0[j + 1] + (S[j] == '1' ? 0 : 1);
}
for (int i = 0; i <= n; ++i) res = min(res, cnt1[i] + cnt0[i]);
return res;
}
};
``````

``````class Solution {
public:
int minFlipsMonoIncr(string S) {
int n = S.size(), res = 0, cnt1 = 0;
for (int i = 0; i < n; ++i) {
(S[i] == '0') ? ++res : ++cnt1;
res = min(res, cnt1);
}
return res;
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/926

https://leetcode.com/problems/flip-string-to-monotone-increasing/

https://leetcode.com/problems/flip-string-to-monotone-increasing/discuss/183851/C%2B%2BJava-4-lines-O(n)-or-O(1)-DP

https://leetcode.com/problems/flip-string-to-monotone-increasing/discuss/183896/Prefix-Suffix-Java-O(N)-One-Pass-Solution-Space-O(1)

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