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1888. Minimum Number of Flips to Make the Binary String Alternating

Description

You are given a binary string s. You are allowed to perform two types of operations on the string in any sequence:

  • Type-1: Remove the character at the start of the string s and append it to the end of the string.
  • Type-2: Pick any character in s and flip its value, i.e., if its value is '0' it becomes '1' and vice-versa.

Return the minimum number of type-2 operations you need to perform such that s becomes alternating.

The string is called alternating if no two adjacent characters are equal.

  • For example, the strings "010" and "1010" are alternating, while the string "0100" is not.

 

Example 1:

Input: s = "111000"
Output: 2
Explanation: Use the first operation two times to make s = "100011".
Then, use the second operation on the third and sixth elements to make s = "101010".

Example 2:

Input: s = "010"
Output: 0
Explanation: The string is already alternating.

Example 3:

Input: s = "1110"
Output: 1
Explanation: Use the second operation on the second element to make s = "1010".

 

Constraints:

  • 1 <= s.length <= 105
  • s[i] is either '0' or '1'.

Solutions

  • class Solution {
    public int minFlips(String s) {
        int n = s.length();
        String target = "01";
        int cnt = 0;
        for (int i = 0; i < n; ++i) {
            if (s.charAt(i) != target.charAt(i & 1)) {
                ++cnt;
            }
        }
        int ans = Math.min(cnt, n - cnt);
        for (int i = 0; i < n; ++i) {
            if (s.charAt(i) != target.charAt(i & 1)) {
                --cnt;
            }
            if (s.charAt(i) != target.charAt((i + n) & 1)) {
                ++cnt;
            }
            ans = Math.min(ans, Math.min(cnt, n - cnt));
        }
        return ans;
    }
}

  • class Solution {
public:
    int minFlips(string s) {
        int n = s.size();
        string target = "01";
        int cnt = 0;
        for (int i = 0; i < n; ++i) {
            if (s[i] != target[i & 1]) {
                ++cnt;
            }
        }
        int ans = min(cnt, n - cnt);
        for (int i = 0; i < n; ++i) {
            if (s[i] != target[i & 1]) {
                --cnt;
            }
            if (s[i] != target[(i + n) & 1]) {
                ++cnt;
            }
            ans = min({ans, cnt, n - cnt});
        }
        return ans;
    }
};

  • class Solution:
    def minFlips(self, s: str) -> int:
        n = len(s)
        target = "01"
        cnt = sum(c != target[i & 1] for i, c in enumerate(s))
        ans = min(cnt, n - cnt)
        for i in range(n):
            cnt -= s[i] != target[i & 1]
            cnt += s[i] != target[(i + n) & 1]
            ans = min(ans, cnt, n - cnt)
        return ans


  • func minFlips(s string) int {
	n := len(s)
	target := "01"
	cnt := 0
	for i := range s {
		if s[i] != target[i&1] {
			cnt++
		}
	}
	ans := min(cnt, n-cnt)
	for i := range s {
		if s[i] != target[i&1] {
			cnt--
		}
		if s[i] != target[(i+n)&1] {
			cnt++
		}
		ans = min(ans, min(cnt, n-cnt))
	}
	return ans
}

  • function minFlips(s: string): number {
    const n = s.length;
    const target = '01';
    let cnt = 0;
    for (let i = 0; i < n; ++i) {
        if (s[i] !== target[i & 1]) {
            ++cnt;
        }
    }
    let ans = Math.min(cnt, n - cnt);
    for (let i = 0; i < n; ++i) {
        if (s[i] !== target[i & 1]) {
            --cnt;
        }
        if (s[i] !== target[(i + n) & 1]) {
            ++cnt;
        }
        ans = Math.min(ans, cnt, n - cnt);
    }
    return ans;
}

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