LeetCode-in-Java
3884. First Matching Character From Both Ends
Easy
You are given a string s of length n consisting of lowercase English letters.
Return the smallest index i such that s[i] == s[n - i - 1].
If no such index exists, return -1.
Example 1:
Input: s = “abcacbd”
Output: 1
Explanation:
At index i = 1, s[1] and s[5] are both 'b'.
No smaller index satisfies the condition, so the answer is 1.
Example 2:
Input: s = “abc”
Output: 1
Explanation:
At index i = 1, the two compared positions coincide, so both characters are 'b'.
No smaller index satisfies the condition, so the answer is 1.
Example 3:
Input: s = “abcdab”
Output: -1
Explanation:
For every index i, the characters at positions i and n - i - 1 are different.
Therefore, no valid index exists, so the answer is -1.
Constraints:
1 <= n == s.length <= 100sconsists of lowercase English letters.
Solution
public class Solution {
public int firstMatchingIndex(String s) {
int l = 0;
int h = s.length() - 1;
while (l <= h) {
if (s.charAt(l) == s.charAt(h)) {
return l;
}
l++;
h--;
}
return -1;
}
}