Python solution O(n) time with flexible set of characters

  • 0
        def firstUniqChar(self, s):
            :type s: str
            :rtype: int
            s_list = list(s)
            c_index = {}
            for i, c in enumerate(s_list):
                if c not in c_index:
                    c_index[c] = [i,1]
                    c_index[c][1] += 1
            index = len(s_list)
            for x in c_index:
                if c_index[x][1] == 1 and c_index[x][0] < index:
                    index = c_index[x][0]
            return -1 if index == len(s_list) else index

  • 0

    why do you return the last statement if the index equals the length. Would you just return the first character if they are all unique?

  • 0

    @ColorfulCodes Nevermind. That is if all are repeating. Thank you for this solution.

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