Formatted question description: https://leetcode.ca/all/900.html

# 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:**

`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`

.

**Related Topics**:

Array

## Solution 1.

```
// OJ: https://leetcode.com/problems/rle-iterator/
// Time:
// RLEIterator: O(N)
// next: O(1) amortized
// Space: O(N)
class RLEIterator {
typedef long long LL;
vector<LL> sizes, nums;
LL i = -1, j = 0;
public:
RLEIterator(vector<int>& A) {
LL sum = 0;
for (int i = 0; i + 1 < A.size(); i += 2) {
LL cnt = A[i], num = A[i + 1];
if (cnt == 0) continue;
sum += cnt;
sizes.push_back(sum);
nums.push_back(num);
}
}
int next(int n) {
i += n;
while (j < sizes.size() && i >= sizes[j]) ++j;
if (j >= sizes.size()) return -1;
return nums[j];
}
};
```