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  1. Lonely Integer
  2. Discussions

Lonely Integer

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  • sanskar_agarwal3
     + 0 comments
    public static int lonelyinteger(List<Integer> a) {
    // Write your code here
    Collections.sort(a);
    System.out.println(a);
    if(a.size()==1){
        return a.get(0);
    }
    for(int i = 0 ; i < a.size() ; i++){
        if(i%2==0 && i<(a.size()-2)){
            if(a.get(i)==a.get(i+1)){
                i++;
            }
            else{
                return a.get(i);
            }
        }
    }
    return a.get(a.size()-1);

    }

  • inq_daniel_eliz1
     + 0 comments

    I decided to reduce the time complexity using a set of a.

    a_set = set(a)
    for i in a_set:
        if a.count(i) == 1:
            return i

  • minh_anm_nguyen
     + 0 comments

    Time: O(N), Space: O(1)

    int lonelyinteger(vector a) { int ret {0};

    for (int num : a)
    ret ^= num;

return ret;

    }

  • m_hassan90
     + 0 comments
    return a.GroupBy(x => x)
                .Where(g => g.Count() == 1)
                .Select(g => g.Key)
                .FirstOrDefault();

  • muhumukip
     + 0 comments

    Python 3 Solution.

    Using list comprehensions.

    • Time complexity of O(n^2)
    • Space Complexity of O(1)
    def lonelyinteger(a):
    loner = [value for value in a if a.count(value) == 1]
    return loner[0]