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1529. Minimum Suffix Flips

Difficulty Topics

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'.

Solution

minimum-suffix-flips.cpp
class Solution {
public:
    int minFlips(string target) {
        int n = target.size();
        int flips = 0;
        char status = '0';

        for (int i = 0; i < n; i++){

            if (target[i] != status){
                flips++;

                status = status == '0' ? '1' : '0';
            }

        }

        return flips;


    }
};