codility lesson 4 frog river one

function solution(X, A) {
  if (A.length === 1) {
    if (A[0] === 1 && X === 1) return 0; 
    else return -1;
  }
    
    i = -1
    sumOfArr = 0
    sequenceSum = (X * (X+1)) / 2
    arrOfFound = []
    
  do {
    i++;
    if (arrOfFound[A[i]]) continue;
    arrOfFound[A[i]] = true;
    sumOfArr += A[i];
    if (sumOfArr === sequenceSum) break;
  } while (i < A.length)
  
  if (i === A.length) return -1;
    
  return i;
}