_900.java

package com.fishercoder.solutions;

/**
 * 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:
 *
 * 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.
 */
public class _900 {
  public static class Solution1 {
    public static class RLEIterator {

      int index;
      int[] array;

      public RLEIterator(int[] A) {
        index = 0;
        array = A;
      }

      public int next(int n) {
        int lastElement = -1;
        while (n > 0 && index < array.length) {
          if (array[index] > n) {
            array[index] -= n;
            lastElement = array[index + 1];
            break;
          } else if (array[index] == n) {
            array[index] = 0;
            lastElement = array[index + 1];
            index += 2;
            break;
          } else {
            n -= array[index];
            index += 2;
          }
        }
        return lastElement;
      }

    }
  }
}