LC 900. RLE Iterator

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.

 

 

 

Runtime: 95 ms, faster than 82.10% of Java online submissions for RLE Iterator.

 

class RLEIterator {
  private Queue<Integer> q = new ArrayDeque<>();
  private int cnt = -1;
  private int value = -1;
  public RLEIterator(int[] A) {
    for(int i=0; i<A.length; i++){
      q.add(A[i]);
    }
  }

  public int next(int n) {
    if(cnt == -1){
      if(q.isEmpty()) {
        return -1;
      }else {
        cnt = q.peek();q.poll();
        value = q.peek(); q.poll();
      }
    }
    if(cnt >= n){
      cnt -= n;
      return value;
    }else {
      n -= cnt;
      cnt = -1;
    }
    while(!q.isEmpty()){
      cnt = q.peek(); q.poll();
      value = q.peek(); q.poll();
      if(cnt >= n) {
        cnt -= n;
        return value;
      }else{
        n -= cnt;
        cnt = -1;
      }
    }
    return -1;
  }
}