900. RLE Iterator
Medium
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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.
class RLEIterator {
private:
vector<pair<int,int>> Vec;
public:
RLEIterator(vector<int>& A) {
for(int i = 0;i < A.size();i = i + 2){
if(A[i] == 0){continue;}
Vec.push_back(make_pair(A[i + 1],A[i]));
}
}
int next(int n) {
for(auto &a : Vec){
if(a.second == 0){
continue;
}
if(a.second >= n){
a.second = a.second - n;
return a.first;
}
n = n - a.second;
a.second = 0;
}
return -1;
}
};
/**
* Your RLEIterator object will be instantiated and called as such:
* RLEIterator* obj = new RLEIterator(A);
* int param_1 = obj->next(n);
*/
