Given a binary string s, return true if the longest contiguous segment of 1s is strictly longer than the longest contiguous segment of 0s in s. Return false otherwise.
- For example, in
s = "110100010"the longest contiguous segment of1s has length2, and the longest contiguous segment of0s has length3.
Note that if there are no 0s, then the longest contiguous segment of 0s is considered to have length 0. The same applies if there are no 1s.
Example 1:
Input: s = "1101" Output: true Explanation: The longest contiguous segment of 1s has length 2: "1101" The longest contiguous segment of 0s has length 1: "1101" The segment of 1s is longer, so return true.
Example 2:
Input: s = "111000" Output: false Explanation: The longest contiguous segment of 1s has length 3: "111000" The longest contiguous segment of 0s has length 3: "111000" The segment of 1s is not longer, so return false.
Example 3:
Input: s = "110100010" Output: false Explanation: The longest contiguous segment of 1s has length 2: "110100010" The longest contiguous segment of 0s has length 3: "110100010" The segment of 1s is not longer, so return false.
Constraints:
1 <= s.length <= 100s[i]is either'0'or'1'.
Solution: Brute Force
Write a function count to count longest contiguous segment of m, return count(‘1’) > count(‘0’)
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: bool checkZeroOnes(string s) { auto count = [&](char m) { int ans = 0; int l = 0; for (char c : s) if (c != m) l = 0; else ans = max(ans, ++l); return ans; }; return count('1') > count('0'); } }; |