RLE Iterator

Description

Write an iterator that iterates through a run-length encoded sequence.

The iterator is initialized by RLEIterator(int[] A), where A is a run-length encoding of some sequence.  More specifically, for all even i, A[i] tells us the number of times that the non-negative integer value A[i+1] is repeated in the sequence.

The iterator supports one function: next(int n), which exhausts the next n elements (n >= 1) and returns the last element exhausted in this way.  If there is no element left to exhaust, next returns -1 instead.

For example, we start with A = [3,8,0,9,2,5], which is a run-length encoding of the sequence [8,8,8,5,5].  This is because the sequence can be read as "three eights, zero nines, two fives".

 

Example 1:

Input: ["RLEIterator","next","next","next","next"], [[[3,8,0,9,2,5]],[2],[1],[1],[2]]
Output: [null,8,8,5,-1]
Explanation: 
RLEIterator is initialized with RLEIterator([3,8,0,9,2,5]).
This maps to the sequence [8,8,8,5,5].
RLEIterator.next is then called 4 times:
.next(2) exhausts 2 terms of the sequence, returning 8.  The remaining sequence is now [8, 5, 5].
.next(1) exhausts 1 term of the sequence, returning 8.  The remaining sequence is now [5, 5].
.next(1) exhausts 1 term of the sequence, returning 5.  The remaining sequence is now [5].
.next(2) exhausts 2 terms, returning -1.  This is because the first term exhausted was 5,
but the second term did not exist.  Since the last term exhausted does not exist, we return -1.

Note:

1. 0 <= A.length <= 1000
2. A.length is an even integer.
3. 0 <= A[i] <= 10^9
4. There are at most 1000 calls to RLEIterator.next(int n) per test case.
5. Each call to RLEIterator.next(int n) will have 1 <= n <= 10^9.

Solution(javascript)

/**
 * @param {number[]} A
 */
var RLEIterator = function(A) {
    let E = A;
    let i = 0, q = 0;

    this.next = function(n) {
        while(i < E.length){
            if(q + n > E[i]){
                n -= E[i] - q;
                q = 0;
                i += 2;
            } else {
                q += n;
                return E[i+1];
            }
        }

        return -1;
    };
}
/** 
 * Your RLEIterator object will be instantiated and called as such:
 * var obj = new RLEIterator(A)
 * var param_1 = obj.next(n)
 */