900. RLE Iterator

T: O(n), n is next() input n

S: O(encoding size)

class RLEIterator {

    int[] nums;
    int idx = 0;
    public RLEIterator(int[] encoding) {
        this.nums = encoding;
    }
    
    public int next(int n) {
        while (idx < nums.length) {
            if (n > nums[idx]) {
                n -= nums[idx];
                idx += 2;
            } else {
                nums[idx] -= n;
                return nums[idx+1];
            }
        }
        return -1;
    }
}

/**
 * Your RLEIterator object will be instantiated and called as such:
 * RLEIterator obj = new RLEIterator(encoding);
 * int param_1 = obj.next(n);
 
 
 even i: encoding[i] tells times of
 encoding[i + 1] repeated
 
 arr = [8,8,8,5,5] can be encoded to be 
 encoding = [3,8,2,5]. 
 encoding = [3,8,0,9,2,5] and 
 encoding = [2,8,1,8,2,5] 
 
 4
[3, 8, 0, 9, 2, 5]

n -= arr[idx]
idx+=2

arr[idx]-= n
return arr[idx+1]



 */