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 iA[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]],,,,]
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 .

.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.

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];
}
};