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.

900. RLE Iterator
struct RLEIterator {
    prefix: Vec<usize>,
    values: Vec<i32>,
    index: usize,
    size: usize,
}

impl RLEIterator {
    fn new(a: Vec<i32>) -> Self {
        let mut prefix = vec![];
        let mut values = vec![];
        let n = a.len();
        let index = 0;
        let mut prev = 0;
        for i in 0..n / 2 {
            let time = a[i * 2] as usize;
            if time != 0 {
                prev += time;
                prefix.push(prev);
                values.push(a[i * 2 + 1]);
            }
        }
        let size = values.len();
        RLEIterator {
            prefix,
            values,
            index,
            size,
        }
    }

    fn next(&mut self, n: i32) -> i32 {
        self.index += n as usize;
        match self.prefix.binary_search(&self.index) {
            Ok(i) => self.values[i],
            Err(i) => {
                if i < self.size {
                    self.values[i]
                } else {
                    -1
                }
            }
        }
    }
}

#[test]
fn test() {
    let mut obj = RLEIterator::new(vec![3, 8, 0, 9, 2, 5]);
    assert_eq!(obj.next(2), 8);
    assert_eq!(obj.next(1), 8);
    assert_eq!(obj.next(1), 5);
    assert_eq!(obj.next(2), -1);
}