Finding a vector leader faking the usage of a stack

StackBasedLeader.js

function stackBasedLeader(A) {
  var n = A.length;
  var occurrences = {};
  var stackSize = 0;
  var stackElem = -1;
  
  for (let a of A) {
    if (!occurrences[a]) occurrences[a] = 0;
    occurrences[a]++;

    if (stackSize > 0 && a !== stackElem) {
      stackSize--;
      continue;
    }
    if (stackSize === 0) {
      stackSize++;
      stackElem = a;
      continue;
    }
    if (a === stackElem) {
      stackSize++;      
    }
  }
  return (occurrences[stackElem] > n/2) ? stackElem : -1;
}

Finding a vector leader in O(n) time is achieved using a stack to pop couples of not-equal elements and returning either what's left in the stack or -1. As mentioned here, we don't need a real stack, as the stack will always contain elements of the same value.
Therefore we implement the stack using a stackSize and a stackElem.
Accumulator occurrences is needed for the final check.