RLE Iterator

Similar Problems:

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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

- Solution:

// Blog link: https://code.dennyzhang.com/rle-iterator // Basic Ideas: linkedin list // Complexity: Time O(n), Space O(n) type RLENode struct { cnt, val int next *RLENode } type RLEIterator struct { head *RLENode } func Constructor(A []int) RLEIterator { head := &RLENode{0, 0, nil} p := head for i:= 0; i < len(A)-1; i+=2 { node := &RLENode{A[i], A[i+1], nil} p.next = node p = p.next } return RLEIterator{head} } func (this *RLEIterator) Next(n int) int { if this.head.next == nil { return -1 } p := this.head.next cnt := p.cnt for cnt < n && p != nil { p = p.next if p != nil { cnt += p.cnt } } res := -1 if cnt < n || p == nil { this.head.next = nil } else { p.cnt = cnt-n this.head.next = p res = p.val } return res } /** * Your RLEIterator object will be instantiated and called as such: * obj := Constructor(A); * param_1 := obj.Next(n); */