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  1. Lonely Integer
  2. Discussions

Lonely Integer

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242 Discussions

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  • nehangma_offici1
     + 0 comments

    Shortest and Fastest solution is to use but manipulation XOR properties: a ^ a = 0 → XOR of a number with itself is 0

    a ^ 0 = a → XOR of a number with 0 is the number

    XOR is commutative and associative, so the order doesn't matter

    public static int lonelyinteger(List a) { int result = 0; for (int num : a) { result ^= num; } return result; }

  • cbiggin
     + 0 comments

    Swift:

    func lonelyinteger(a: [Int]) -> Int {
    var dict = [Int: Bool]()
    
    a.forEach { value in
        // will nil the dictionary entry if more than 1 occurrence
        if dict[value] != nil {
            dict[value] = nil
        } else {
            dict[value] = true
        }
    }
    
    // should only be 1 entry left
    return dict.keys.first ?? -1
}

  • juanp766
     + 0 comments

    Java using two hash sets which operations are O(1), therefore it is only needed to iterate whole array once.

    public static int lonelyinteger(List<Integer> arr) {
        HashSet<Integer> set = new HashSet<>();
        HashSet<Integer> uniqueSet = new HashSet<>();
        
        Integer unique = arr.get(0);
        Integer lastUnique = arr.get(0);
        
        for(Integer element: arr){
            if(!set.contains(element)){
                set.add(element);
                uniqueSet.add(element);
            }else{
                uniqueSet.remove(element);
            }
        }
        
        return uniqueSet.stream().findFirst().get();
    }

  • villamilluis123
     + 0 comments
    public static int lonelyinteger(List<int> a)
    {
        int size = a.Count();
        if(size == 1) return a[0];
        int item = a[0];
        int count = 1;
        List<int> buffer = new List<int>{};
        for(int i = 0; i < size; i++){
            
            if(a[i] < -1 || a[i] > 101) throw new Exception("Invalid items");
            
            count = 1;
            if(buffer.Contains(a[i])) {
                count++;
                continue;
            }
            for(int j = i+1; j <= size - 1; j++){ 
                if(a[i] == a[j]){
                    count++;
                    buffer.Add(a[i]);
                    break;
                }
            }
            if(count < 2) item = a[i];
        }
        return item;
    }

  • robertbielicki
     + 0 comments
    def lonelyinteger(a):
    counter = dict()
    for num in a:
        counter[num] = counter.get(num, 0) +1
    return sorted(counter.items(), key= lambda x: x[1])[0][0]