How do you solve RLE Iterator on LeetCode?

RLE Iterator (LeetCode #900) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding run-length encoding is crucial for this problem.

What is the brute force solution for RLE Iterator?

The brute force approach for RLE Iterator is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves reconstructing the entire sequence from the run-length encoded array each time we need to retrieve elements. This means we will repeatedly generate the sequence and exhaust elements as requested. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve RLE Iterator?

RLE Iterator uses the following concepts: Array, Design, Counting, Iterator. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for RLE Iterator?

Explain your thought process clearly while coding. Consider edge cases and test your code with them. Optimize your solution step-by-step, explaining why each improvement is necessary.

What are common mistakes when solving RLE Iterator?

Reconstructing the entire sequence instead of managing state. Not handling edge cases where there are not enough elements to exhaust.

#900

RLE Iterator

Medium

ArrayDesignCountingIteratorArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution maintains the current position in the encoded array and keeps track of how many elements are left to consume. This avoids reconstructing the entire sequence and allows for efficient element retrieval.

⚙️

Algorithm

3 steps

1Step 1: Initialize the iterator with the encoded array, setting the current index and counts.
2Step 2: In the 'next(n)' method, check if there are enough elements left to exhaust.
3Step 3: If there are enough elements, decrement the count and return the current value; otherwise, return -1.

solution.py21 lines

1class RLEIterator:
2    def __init__(self, encoded):
3        self.encoded = encoded
4        self.index = 0
5        self.current_count = 0
6        self.current_value = 0
7
8    def next(self, n):
9        while n > 0:
10            if self.current_count == 0:
11                if self.index >= len(self.encoded):
12                    return -1
13                self.current_count = self.encoded[self.index]
14                self.current_value = self.encoded[self.index + 1]
15                self.index += 2
16            if n <= self.current_count:
17                self.current_count -= n
18                return self.current_value
19            n -= self.current_count
20            self.current_count = 0
21        return -1

ℹ

Complexity note: This complexity is linear because we only traverse the encoded array once, and we keep track of our position without needing to reconstruct the entire sequence.

1Understanding run-length encoding is crucial for this problem.
2Efficiently managing state (current index and counts) is key to optimizing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.