Write an iterator that iterates through a run-length encoded sequence.
The iterator is initialized by
RLEIterator(int[] A)
, whereA
is a run-length encoding of some sequence. More specifically, for all eveni
,A[i]
tells us the number of times that the non-negative integer valueA[i+1]
is repeated in the sequence.The iterator supports one function:
next(int n)
, which exhausts the nextn
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 toRLEIterator.next(int n)
per test case.- Each call to
RLEIterator.next(int n)
will have1 <= n <= 10^9
.
Approach #1: Array. [Java]
class RLEIterator {
int index;
int[] A;
public RLEIterator(int[] A) {
this.A = A;
index = 0;
} public int next(int n) {
while (index < A.length && n > A[index]) {
n = n - A[index];
index += 2;
}
if (index >= A.length) return -1;
A[index] = A[index] - n;
return A[index+1];
}
} /**
* Your RLEIterator object will be instantiated and called as such:
* RLEIterator obj = new RLEIterator(A);
* int param_1 = obj.next(n);
*/
Refereence:
https://leetcode.com/problems/rle-iterator/discuss/168294/Java-Straightforward-Solution-O(n)-time-O(1)-space