Class Solution


  • public class Solution
    extends Object
    1529 - Minimum Suffix Flips.

    Medium

    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'.
    • Constructor Detail

      • Solution

        public Solution()
    • Method Detail

      • minFlips

        public int minFlips​(String target)