# Solving Leetcode Interviews in Seconds with AI: RLE Iterator


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "900" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> We can use run-length encoding (i.e., RLE) to encode a sequence of integers. In a run-length encoded array of even length encoding (0-indexed), for all even i, encoding[i] tells us the number of times that the non-negative integer value encoding[i + 1] is repeated in the sequence.  For example, the sequence arr = [8,8,8,5,5] can be encoded to be encoding = [3,8,2,5]. encoding = [3,8,0,9,2,5] and encoding = [2,8,1,8,2,5] are also valid RLE of arr.  Given a run-length encoded array, design an iterator that iterates through it. Implement the RLEIterator class:  RLEIterator(int[] encoded) Initializes the object with the encoded array encoded. int next(int n) Exhausts the next n elements and returns the last element exhausted in this way. If there is no element left to exhaust, return -1 instead.    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 rLEIterator = new RLEIterator([3, 8, 0, 9, 2, 5]); // This maps to the sequence [8,8,8,5,5]. rLEIterator.next(2); // exhausts 2 terms of the sequence, returning 8. The remaining sequence is now [8, 5, 5]. rLEIterator.next(1); // exhausts 1 term of the sequence, returning 8. The remaining sequence is now [5, 5]. rLEIterator.next(1); // exhausts 1 term of the sequence, returning 5. The remaining sequence is now [5]. rLEIterator.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.    Constraints:  2 <= encoding.length <= 1000 encoding.length is even. 0 <= encoding[i] <= 109 1 <= n <= 109 At most 1000 calls will be made to next.  

	# Explanation
	Here's a breakdown of the approach, complexities, and the Python code:

*   **Key Approach:**
    *   Maintain an index to track the current position in the `encoded` array.
    *   In the `next()` function, iterate through the `encoded` array, decrementing the counts (encoding[i]) by `n`.
    *   Return the corresponding value (encoding[i+1]) when `n` is fully exhausted. If we reach the end of the array before `n` is exhausted, return -1.

*   **Complexity:**
    *   Runtime: O(k), where k is the number of pairs in the encoded array. In the worst case, we iterate through the entire encoded array in the next function.
    *   Storage: O(1), as we only store a single index.

	
	# Code
	```python
	class RLEIterator:

    def __init__(self, encoding: list[int]):
        self.encoding = encoding
        self.index = 0

    def next(self, n: int) -> int:
        while self.index < len(self.encoding):
            if self.encoding[self.index] >= n:
                self.encoding[self.index] -= n
                return self.encoding[self.index + 1]
            else:
                n -= self.encoding[self.index]
                self.index += 2
        return -1
	```