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.
\n
Example 1:
\n\n\nInput: target = "10111"\nOutput: 3\nExplanation: Initially, s = "00000".\nChoose index i = 2: "00000" -> "00111"\nChoose index i = 0: "00111" -> "11000"\nChoose index i = 1: "11000" -> "10111"\nWe need at least 3 flip operations to form target.\n\n\n
Example 2:
\n\n\nInput: target = "101"\nOutput: 3\nExplanation: Initially, s = "000".\nChoose index i = 0: "000" -> "111"\nChoose index i = 1: "111" -> "100"\nChoose index i = 2: "100" -> "101"\nWe need at least 3 flip operations to form target.\n\n\n
Example 3:
\n\n\nInput: target = "00000"\nOutput: 0\nExplanation: We do not need any operations since the initial s already equals target.\n\n\n
\n
Constraints:
\n\nn == target.length1 <= n <= 105target[i] is either '0' or '1'.Compiled with clang 11 using the latest C++ 20 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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