README.md

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.

Solutions (Rust)

1. Solution

struct RLEIterator {
    iterator: std::vec::IntoIter<i32>,
    remain: (i32, i32),
}


/** 
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl RLEIterator {

    fn new(A: Vec<i32>) -> Self {
        Self {
            iterator: A.into_iter(),
            remain: (0, -1),
        }
    }

    fn next(&mut self, mut n: i32) -> i32 {
        while n > self.remain.0 {
            n -= self.remain.0;
            self.remain.0 = 0;
            match self.iterator.next() {
                Some(x) => self.remain = (x, self.iterator.next().unwrap()),
                None => return -1,
            }
        }

        self.remain.0 -= n;

        self.remain.1
    }
}

/**
 * Your RLEIterator object will be instantiated and called as such:
 * let obj = RLEIterator::new(A);
 * let ret_1: i32 = obj.next(n);
 */