N-Repeated Element in Size 2N Array The task
is taken from leetcode


  In a array A of size 2N, there are N+1 unique elements, and exactly
  one of these elements is repeated N times.
  
  Return the element repeated N times.
  
  Example 1:
  
  Input: [1,2,3,3] Output: 3 
  
  Example 2:
  
  Input: [2,1,2,5,3,2] Output: 2 
  
  Example 3:
  
  Input: [5,1,5,2,5,3,5,4] Output: 5  
  
  Note:
  
  4 &lt;= A.length &lt;= 10000 0 &lt;= A[i] &lt; 10000 A.length is even


My solution

var repeatedNTimes = function(A) {
    const set = new Set();
    for (const n of A) {
        if (set.has(n)) { return n; }
        set.add(n);
    }
};


I didn't do much with the information size 2N, there are N+1 unique elements, and exactly one of these elements is repeated N times. I think these information is not needed to solve the problem efficiently. I also wonder whether there is a bitwise or purely math based solution to the task.
In the worst case (the repeated element occupies the second half of the array), the set will accommodate all N non-repeated elements, therefore the space complexity of the solution is \$O(N)\$.

Your intuition is correct; the solution doesn't use the important information. It is used to find out how far apart the repeated elements are. If a distance between them is at least \$d\$, the array would be at least \$d\cdot(N-1) + 1\$ long. Since we know that it is \$2N\$ long, we can conclude that \$d \le \dfrac{2N-1}{N-1}\$, which is effectively 2 for N > 2. It is enough to compare each a[i] with a[i+1] and a[i+2]. The space complexity is now constant.