Problem
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.length1 <= n <= 10^5target[i]is either'0'or'1'.
Solution (Java)
class Solution {
public int minFlips(String target) {
int flipCount = target.charAt(0) - 48;
char prev = target.charAt(0);
for (char ch : target.toCharArray()) {
if (ch != prev) {
flipCount++;
prev = ch;
}
}
return flipCount;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).