Python solution using bit manipulation


  • 0
    G
            bitvec = 0
            for char in s:
                bitvec ^= (1 << ord(char))
            return bitvec == bitvec & (-bitvec)
    

    Using a bitvector to maintain each letter's odd/even status, and a bit trick to determine if there are more than one set bit in the bit vector (also returns true for 0 set bits).


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