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Formatted question description: https://leetcode.ca/all/900.html

900. RLE Iterator (Medium)

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 iA[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.

Related Topics:
Array

Solution 1.

  • class RLEIterator {
        private int[] encoding;
        private int curr;
        private int i;
    
        public RLEIterator(int[] encoding) {
            this.encoding = encoding;
            curr = 0;
            i = 0;
        }
    
        public int next(int n) {
            while (i < encoding.length) {
                if (curr + n > encoding[i]) {
                    n -= encoding[i] - curr;
                    i += 2;
                    curr = 0;
                } else {
                    curr += n;
                    return encoding[i + 1];
                }
            }
            return -1;
        }
    }
    
    /**
     * Your RLEIterator object will be instantiated and called as such:
     * RLEIterator obj = new RLEIterator(encoding);
     * int param_1 = obj.next(n);
     */
    
  • class RLEIterator {
    public:
        vector<int> encoding;
        int curr;
        int i;
    
        RLEIterator(vector<int>& encoding) {
            this->encoding = encoding;
            this->curr = 0;
            this->i = 0;
        }
    
        int next(int n) {
            while (i < encoding.size()) {
                if (curr + n > encoding[i]) {
                    n -= encoding[i] - curr;
                    curr = 0;
                    i += 2;
                } else {
                    curr += n;
                    return encoding[i + 1];
                }
            }
            return -1;
        }
    };
    
    /**
     * Your RLEIterator object will be instantiated and called as such:
     * RLEIterator* obj = new RLEIterator(encoding);
     * int param_1 = obj->next(n);
     */
    
  • class RLEIterator:
        def __init__(self, encoding: List[int]):
            self.encoding = encoding
            self.i = 0
            self.curr = 0
    
        def next(self, n: int) -> int:
            while self.i < len(self.encoding):
                if self.curr + n > self.encoding[self.i]:
                    n -= self.encoding[self.i] - self.curr
                    self.curr = 0
                    self.i += 2
                else:
                    self.curr += n
                    return self.encoding[self.i + 1]
            return -1
    
    
    # Your RLEIterator object will be instantiated and called as such:
    # obj = RLEIterator(encoding)
    # param_1 = obj.next(n)
    
    
  • type RLEIterator struct {
    	encoding []int
    	curr     int
    	i        int
    }
    
    func Constructor(encoding []int) RLEIterator {
    	return RLEIterator{encoding: encoding, curr: 0, i: 0}
    }
    
    func (this *RLEIterator) Next(n int) int {
    	for this.i < len(this.encoding) {
    		if this.curr+n > this.encoding[this.i] {
    			n -= this.encoding[this.i] - this.curr
    			this.curr = 0
    			this.i += 2
    		} else {
    			this.curr += n
    			return this.encoding[this.i+1]
    		}
    	}
    	return -1
    }
    
    /**
     * Your RLEIterator object will be instantiated and called as such:
     * obj := Constructor(encoding);
     * param_1 := obj.Next(n);
     */
    

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