Leetcode: RLE Iterator

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:

  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.

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);
 */
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