## 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`

.

## Rust Solution

```
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);
}
```

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