firsDuplicate.md

First Duplicate

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Given an array a that contains only numbers in the range from 1 to a.length, find the first duplicate number for which the second occurrence has the minimal index. In other words, if there are more than 1 duplicated numbers, return the number for which the second occurrence has a smaller index than the second occurrence of the other number does. If there are no such elements, return -1.

Example

• For a = [2, 1, 3, 5, 3, 2], the output should be solution(a) = 3.

  There are 2 duplicates: numbers 2 and 3. The second occurrence of 3 has a smaller index than the second occurrence of 2 does, so the answer is 3.

• For a = [2, 2], the output should be solution(a) = 2;

• For a = [2, 4, 3, 5, 1], the output should be solution(a) = -1.

Input/Output

• [execution time limit] 0.5 seconds (cpp)

• [input] array.integer a

  Guaranteed constraints:
  1 ≤ a.length ≤ 10^5^,
  1 ≤ a[i] ≤ a.length.

• [output] integer

  The element in a that occurs in the array more than once and has the minimal index for its second occurrence. If there are no such elements, return -1.

[C++] Syntax Tips

// Prints help message to the console
// Returns a string
string helloWorld(string name) {
    cout << "This prints to the console when you Run Tests" << endl;
    return "Hello, " + name;
}

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Solution

int solution(vector<int> a) {
    bool arr[a.size()];
    for (int i = 0; i < a.size(); i++) {
        arr[i] = false;
    }
    for( int i = 0; i < a.size(); i++ ){
        if(arr[a[i] - 1]){
            return a[i];
        }
        arr[a[i] - 1] = true;
    }
    return -1;
}

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