Tuesday, September 22, 2020

LeetCode [900] RLE Iterator

 900. RLE Iterator

Medium

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.
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class RLEIterator {
    int left = 0;//left numbers
    int index = -2;//even index on A
    int sz;
    int[] A;
    public RLEIterator(int[] A) {
        this.sz = A.length;
        this.A = A;
    }
    
    public int next(int n) {
        int v = 0;
        while(n>0){
            if(left==0){
                index += 2;
                if(index<sz) left = A[index];
                else break;
            }
            v = A[index+1];
            int d = Math.min(n, left);
            n -= d;
            left -= d;
        }

        if(n>0) return -1;
        else return v;
    }
}


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

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