LeetCode #900 — MEDIUM

RLE Iterator

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Design

Example 1

["RLEIterator","next","next","next","next"]
[[[3,8,0,9,2,5]],[2],[1],[1],[2]]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #900: RLE Iterator
class RLEIterator {
    private int[] encoding;
    private int i;
    private int j;

    public RLEIterator(int[] encoding) {
        this.encoding = encoding;
    }

    public int next(int n) {
        while (i < encoding.length) {
            if (encoding[i] - j < n) {
                n -= (encoding[i] - j);
                i += 2;
                j = 0;
            } else {
                j += 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);
 */

// Accepted solution for LeetCode #900: RLE Iterator
type RLEIterator struct {
	encoding []int
	i, j     int
}

func Constructor(encoding []int) RLEIterator {
	return RLEIterator{encoding, 0, 0}
}

func (this *RLEIterator) Next(n int) int {
	for this.i < len(this.encoding) {
		if this.encoding[this.i]-this.j < n {
			n -= (this.encoding[this.i] - this.j)
			this.i += 2
			this.j = 0
		} else {
			this.j += 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);
 */

# Accepted solution for LeetCode #900: RLE Iterator
class RLEIterator:
    def __init__(self, encoding: List[int]):
        self.encoding = encoding
        self.i = 0
        self.j = 0

    def next(self, n: int) -> int:
        while self.i < len(self.encoding):
            if self.encoding[self.i] - self.j < n:
                n -= self.encoding[self.i] - self.j
                self.i += 2
                self.j = 0
            else:
                self.j += 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)

// Accepted solution for LeetCode #900: RLE Iterator
struct RLEIterator {
    prefix: Vec<usize>,
    values: Vec<i32>,
    index: usize,
    size: usize,
}

impl RLEIterator {
    fn new(a: Vec<i32>) -> Self {
        let mut prefix = vec![];
        let mut values = vec![];
        let n = a.len();
        let index = 0;
        let mut prev = 0;
        for i in 0..n / 2 {
            let time = a[i * 2] as usize;
            if time != 0 {
                prev += time;
                prefix.push(prev);
                values.push(a[i * 2 + 1]);
            }
        }
        let size = values.len();
        RLEIterator {
            prefix,
            values,
            index,
            size,
        }
    }

    fn next(&mut self, n: i32) -> i32 {
        self.index += n as usize;
        match self.prefix.binary_search(&self.index) {
            Ok(i) => self.values[i],
            Err(i) => {
                if i < self.size {
                    self.values[i]
                } else {
                    -1
                }
            }
        }
    }
}

#[test]
fn test() {
    let mut obj = RLEIterator::new(vec![3, 8, 0, 9, 2, 5]);
    assert_eq!(obj.next(2), 8);
    assert_eq!(obj.next(1), 8);
    assert_eq!(obj.next(1), 5);
    assert_eq!(obj.next(2), -1);
}

// Accepted solution for LeetCode #900: RLE Iterator
class RLEIterator {
    private encoding: number[];
    private i: number;
    private j: number;

    constructor(encoding: number[]) {
        this.encoding = encoding;
        this.i = 0;
        this.j = 0;
    }

    next(n: number): number {
        while (this.i < this.encoding.length) {
            if (this.encoding[this.i] - this.j < n) {
                n -= this.encoding[this.i] - this.j;
                this.i += 2;
                this.j = 0;
            } else {
                this.j += n;
                return this.encoding[this.i + 1];
            }
        }
        return -1;
    }
}

/**
 * Your RLEIterator object will be instantiated and called as such:
 * var obj = new RLEIterator(encoding)
 * var param_1 = obj.next(n)
 */

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n + q)

Space

O(n)

Approach Breakdown

NAIVE

O(n) per op time

O(n) space

Use a simple list or array for storage. Each operation (get, put, remove) requires a linear scan to find the target element — O(n) per operation. Space is O(n) to store the data. The linear search makes this impractical for frequent operations.

OPTIMIZED DESIGN

O(1) per op time

O(n) space

Design problems target O(1) amortized per operation by combining data structures (hash map + doubly-linked list for LRU, stack + min-tracking for MinStack). Space is always at least O(n) to store the data. The challenge is achieving constant-time operations through clever structure composition.

Shortcut: Combine two data structures to get O(1) for each operation type. Space is always O(n).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.