Marc Lester Tan

Most problems like this one can be solved by using the character count map. This approach is very effective as it can give you O(n) time complexity.

class Solution {
    fun firstUniqChar(s: String): Int {
        if(s.length == 1) return 0
        
        var array = IntArray(26)
        
        for(x in 0 until s.length){
            array[s[x] - 'a'] = array[s[x] - 'a'] + 1
        }
        
        for(x in 0 until s.length){
            if(array[s[x] - 'a'] == 1) return x
        }
        
        return -1
    }
}

Runtime: 216 ms, faster than 89.52% of Kotlin online submissions for First Unique Character in a String.
Memory Usage: 35.9 MB, less than 100.00% of Kotlin online submissions for First Unique Character in a String.

The idea is to store a character count map and then go through the characters again and if you find a single-occuring letter, then return it’s index.