Description#
A binary string is monotone increasing if it consists of some number of 0's (possibly none), followed by some number of 1's (also possibly none).
You are given a binary string s. You can flip s[i] changing it from 0 to 1 or from 1 to 0.
Return the minimum number of flips to make s monotone increasing.
Example 1:
Input: s = "00110"
Output: 1
Explanation: We flip the last digit to get 00111.
Example 2:
Input: s = "010110"
Output: 2
Explanation: We flip to get 011111, or alternatively 000111.
Example 3:
Input: s = "00011000"
Output: 2
Explanation: We flip to get 00000000.
Constraints:
1 <= s.length <= 105s[i] is either '0' or '1'.
Solutions#
Solution 1#
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| class Solution:
def minFlipsMonoIncr(self, s: str) -> int:
n = len(s)
left, right = [0] * (n + 1), [0] * (n + 1)
ans = 0x3F3F3F3F
for i in range(1, n + 1):
left[i] = left[i - 1] + (1 if s[i - 1] == '1' else 0)
for i in range(n - 1, -1, -1):
right[i] = right[i + 1] + (1 if s[i] == '0' else 0)
for i in range(0, n + 1):
ans = min(ans, left[i] + right[i])
return ans
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| class Solution {
public int minFlipsMonoIncr(String s) {
int n = s.length();
int[] left = new int[n + 1];
int[] right = new int[n + 1];
int ans = Integer.MAX_VALUE;
for (int i = 1; i <= n; i++) {
left[i] = left[i - 1] + (s.charAt(i - 1) == '1' ? 1 : 0);
}
for (int i = n - 1; i >= 0; i--) {
right[i] = right[i + 1] + (s.charAt(i) == '0' ? 1 : 0);
}
for (int i = 0; i <= n; i++) {
ans = Math.min(ans, left[i] + right[i]);
}
return ans;
}
}
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| class Solution {
public:
int minFlipsMonoIncr(string s) {
int n = s.size();
vector<int> left(n + 1, 0), right(n + 1, 0);
int ans = INT_MAX;
for (int i = 1; i <= n; ++i) {
left[i] = left[i - 1] + (s[i - 1] == '1');
}
for (int i = n - 1; i >= 0; --i) {
right[i] = right[i + 1] + (s[i] == '0');
}
for (int i = 0; i <= n; i++) {
ans = min(ans, left[i] + right[i]);
}
return ans;
}
};
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| func minFlipsMonoIncr(s string) int {
n := len(s)
left, right := make([]int, n+1), make([]int, n+1)
ans := math.MaxInt32
for i := 1; i <= n; i++ {
left[i] = left[i-1]
if s[i-1] == '1' {
left[i]++
}
}
for i := n - 1; i >= 0; i-- {
right[i] = right[i+1]
if s[i] == '0' {
right[i]++
}
}
for i := 0; i <= n; i++ {
ans = min(ans, left[i]+right[i])
}
return ans
}
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| /**
* @param {string} s
* @return {number}
*/
var minFlipsMonoIncr = function (s) {
const n = s.length;
let presum = new Array(n + 1).fill(0);
for (let i = 0; i < n; ++i) {
presum[i + 1] = presum[i] + (s[i] == '1');
}
let ans = presum[n];
for (let i = 0; i < n; ++i) {
ans = Math.min(ans, presum[i] + n - i - (presum[n] - presum[i]));
}
return ans;
};
|
Solution 2#
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| class Solution:
def minFlipsMonoIncr(self, s: str) -> int:
n = len(s)
presum = [0] * (n + 1)
for i, c in enumerate(s):
presum[i + 1] = presum[i] + int(c)
ans = presum[-1]
for i in range(n):
ans = min(ans, presum[i] + n - i - (presum[-1] - presum[i]))
return ans
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| class Solution {
public int minFlipsMonoIncr(String s) {
int n = s.length();
int[] presum = new int[n + 1];
for (int i = 0; i < n; ++i) {
presum[i + 1] = presum[i] + (s.charAt(i) - '0');
}
int ans = presum[n];
for (int i = 0; i < n; ++i) {
ans = Math.min(ans, presum[i] + n - i - (presum[n] - presum[i]));
}
return ans;
}
}
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| class Solution {
public:
int minFlipsMonoIncr(string s) {
int n = s.size();
vector<int> presum(n + 1);
for (int i = 0; i < n; ++i) presum[i + 1] = presum[i] + (s[i] == '1');
int ans = presum[n];
for (int i = 0; i < n; ++i) ans = min(ans, presum[i] + n - i - (presum[n] - presum[i]));
return ans;
}
};
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| func minFlipsMonoIncr(s string) int {
n := len(s)
presum := make([]int, n+1)
for i, c := range s {
presum[i+1] = presum[i] + int(c-'0')
}
ans := presum[n]
for i := range s {
ans = min(ans, presum[i]+n-i-(presum[n]-presum[i]))
}
return ans
}
|
