# Rle Iterator Problem

## Description

LeetCode Problem 900.

We can use run-length encoding (i.e., RLE) to encode a sequence of integers. In a run-length encoded array of even length encoding (0-indexed), for all even i, encoding[i] tells us the number of times that the non-negative integer value encoding[i + 1] is repeated in the sequence.

• For example, the sequence arr = [8,8,8,5,5] can be encoded to be encoding = [3,8,2,5]. encoding = [3,8,0,9,2,5] and encoding = [2,8,1,8,2,5] are also valid RLE of arr.

Given a run-length encoded array, design an iterator that iterates through it.

Implement the RLEIterator class:

• RLEIterator(int[] encoded) Initializes the object with the encoded array encoded.
• int next(int n) Exhausts the next n elements and returns the last element exhausted in this way. If there is no element left to exhaust, return -1 instead.

Example 1:

``````1
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Input
["RLEIterator", "next", "next", "next", "next"]
[[[3, 8, 0, 9, 2, 5]], , , , ]
Output
[null, 8, 8, 5, -1]
Explanation
RLEIterator rLEIterator = new RLEIterator([3, 8, 0, 9, 2, 5]); // This maps to the sequence [8,8,8,5,5].
rLEIterator.next(2); // exhausts 2 terms of the sequence, returning 8. The remaining sequence is now [8, 5, 5].
rLEIterator.next(1); // exhausts 1 term of the sequence, returning 8. The remaining sequence is now [5, 5].
rLEIterator.next(1); // exhausts 1 term of the sequence, returning 5. The remaining sequence is now .
rLEIterator.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.
``````

Constraints:

• 2 <= encoding.length <= 1000
• encoding.length is even.
• 0 <= encoding[i] <= 10^9
• 1 <= n <= 10^9
• At most 1000 calls will be made to next.

## Sample C++ Code

``````1
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class RLEIterator {
public:
int curInd;
vector<int> seq;

RLEIterator(vector<int>& A) {
seq = A;
curInd = 0;
}

int next(int n) {
while (curInd < seq.size()) {
if (seq[curInd] >= n) {
seq[curInd] -= n;
return seq[curInd + 1];
} else {
n -= seq[curInd];
curInd += 2;
}
}
return -1;
}
};

/**
* Your RLEIterator object will be instantiated and called as such:
* RLEIterator* obj = new RLEIterator(encoding);
* int param_1 = obj->next(n);
*/
``````