JavaScript solution using Array.find


  • 2
    T
    /**
     * @param {number[]} nums
     * @return {number}
     */
    var singleNonDuplicate = function(nums) {
        return nums.find(n => nums.indexOf(n) === nums.lastIndexOf(n))
    }
    

    edit: switched out "filter" for "find" since we don't need an array


  • 0
    L

    It is not O(log n), but it is a really good solution.


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