900.rle-iterator.java

/*
 * @lc app=leetcode id=900 lang=java
 *
 * [900] RLE Iterator
 *
 * https://leetcode.com/problems/rle-iterator/description/
 *
 * algorithms
 * Medium (50.76%)
 * Total Accepted:    19.4K
 * Total Submissions: 36.9K
 * Testcase Example:  '["RLEIterator","next","next","next","next"]\n[[[3,8,0,9,2,5]],[2],[1],[1],[2]]'
 *
 * 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:
 *
 *
 * 0 <= A.length <= 1000
 * A.length is an even integer.
 * 0 <= A[i] <= 10^9
 * There are at most 1000 calls to RLEIterator.next(int n) per test case.
 * Each call to RLEIterator.next(int n) will have 1 <= n <= 10^9.
 *
 *
 */
class RLEIterator {

   private int cur = 0;
   private int[] A;

   public RLEIterator(int[] A) {
       this.A = A;
   }

   public int next(int n) {
       while (n > 0) {
           // Find the first non-zero counter
           while (cur < A.length && A[cur] == 0) {
               cur += 2;
           }

           // The whole stream has been exhausted
           if (cur >= A.length) {
               return -1;
           }

           int minus = Math.min(A[cur], n);

           A[cur] -= minus;
           n -= minus;
       }

       return A[cur + 1];
   }
}

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