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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 }