# 1529. Minimum Suffix Flips

## Description

You are given a 0-indexed binary string target of length n. You have another binary string s of length n that is initially set to all zeros. You want to make s equal to target.

In one operation, you can pick an index i where 0 <= i < n and flip all bits in the inclusive range [i, n - 1]. Flip means changing '0' to '1' and '1' to '0'.

Return the minimum number of operations needed to make s equal to target.

Example 1:

Input: target = "10111"
Output: 3
Explanation: Initially, s = "00000".
Choose index i = 2: "00000" -> "00111"
Choose index i = 0: "00111" -> "11000"
Choose index i = 1: "11000" -> "10111"
We need at least 3 flip operations to form target.


Example 2:

Input: target = "101"
Output: 3
Explanation: Initially, s = "000".
Choose index i = 0: "000" -> "111"
Choose index i = 1: "111" -> "100"
Choose index i = 2: "100" -> "101"
We need at least 3 flip operations to form target.


Example 3:

Input: target = "00000"
Output: 0
Explanation: We do not need any operations since the initial s already equals target.


Constraints:

• n == target.length
• 1 <= n <= 105
• target[i] is either '0' or '1'.

## Solutions

• class Solution {
public int minFlips(String target) {
int ans = 0;
for (int i = 0; i < target.length(); ++i) {
int v = target.charAt(i) - '0';
if (((ans & 1) ^ v) != 0) {
++ans;
}
}
return ans;
}
}

• class Solution {
public:
int minFlips(string target) {
int ans = 0;
for (char c : target) {
int v = c - '0';
if ((ans & 1) ^ v) {
++ans;
}
}
return ans;
}
};

• class Solution:
def minFlips(self, target: str) -> int:
ans = 0
for v in target:
if (ans & 1) ^ int(v):
ans += 1
return ans


• func minFlips(target string) int {
ans := 0
for _, c := range target {
v := int(c - '0')
if ((ans & 1) ^ v) != 0 {
ans++
}
}
return ans
}