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//Extra space O(1) Runtime O(nlogn)
    List<Integer> getFibonacciNumber(int[] nums) {
        List<Integer> fibonacciNumbers = new ArrayList<>();
        Arrays.sort(nums);
        int fib1 = 1, fib2 = 1;
        for(int i = 0; i < nums.length;) {
            if(nums[i] < fib2) {
                i++;
            } else if(nums[i] == fib2) {
                fibonacciNumbers.add(nums[i]);
                i++;
            } else {
                int fib3 = fib1 + fib2;
                fib1 = fib2;
                fib2 = fib3;
            }
        }
        return fibonacciNumbers;
    }


    //Math Solution: Extra space O(1) Runtime O(n)
    List<Integer> getFibonacciNumbers(int[] nums) {
        List<Integer> fibonacciNumbers = new ArrayList<>();
        for(int n: nums) {
            if(isFibonacci(n)) fibonacciNumbers.add(n);
        }
        return fibonacciNumbers;
    }
    boolean isFibonacci(int n)
    {
        // n is Fibonacci if one of 5*n*n + 4 or 5*n*n - 4 or both
        // is a perfect square
        return isPerfectSquare(5*n*n + 4) ||
                isPerfectSquare(5*n*n - 4);
    }

    boolean isPerfectSquare(int x)
    {
        int s = (int) Math.sqrt(x);
        return (s*s == x);
    }

Iterate through the list and check if 5(Num)2 +4 OR 5(Num)2 - 4 is a prefect square .
5*n*n + 4 or 5*n*n - 4
add to list if true .