Problem Description

This problem asks you to implement an iterator for a run-length encoded (RLE) array. Run-length encoding is a compression technique where consecutive repeated values are stored as a count-value pair.

In the RLE format used here:

• The encoding array has even length
• At every even index i, encoding[i] represents the count (how many times a value appears)
• At every odd index i+1, encoding[i+1] represents the actual value being repeated

For example, the sequence [8,8,8,5,5] can be encoded as [3,8,2,5], meaning "3 occurrences of 8, followed by 2 occurrences of 5".

You need to implement a class RLEIterator with two methods:

1. Constructor RLEIterator(int[] encoded): Initializes the iterator with the run-length encoded array.

2. Method next(int n): This method "exhausts" (consumes) the next n elements from the decoded sequence and returns the last element that was exhausted. If there aren't enough elements left to exhaust n elements, it returns -1.

The key challenge is to efficiently traverse through the encoded array without actually decoding it into the full sequence. When next(n) is called, you need to:

• Skip through n elements in the logical decoded sequence
• Keep track of your position within the encoded array
• Return the value of the nth element you exhausted, or -1 if you run out of elements

For instance, if the encoding is [3,8,2,5] and you call next(2), you exhaust 2 elements from the sequence (both are 8s), so you return 8. If you then call next(1), you exhaust one more 8, returning 8. If you call next(2) again, you exhaust the last two 5s, returning 5.

Intuition

The key insight is that we don't need to actually decode the entire RLE array into its full sequence. Instead, we can navigate through the encoded array directly by keeping track of our position within it.

Think of the encoded array as a series of "blocks" where each block contains multiple copies of the same value. For example, [3,8,2,5] represents two blocks: a block of 3 eights and a block of 2 fives.

When we need to exhaust n elements, we're essentially "consuming" elements from these blocks. The challenge is: what if n is larger than the current block? We'd need to consume the entire current block and move to the next one, continuing until we've exhausted n elements total.

This leads us to use two pointers:

• Pointer i tracks which block we're currently in (it points to the count in the encoding array)
• Pointer j tracks how many elements we've already consumed from the current block

When next(n) is called, we need to handle two cases:

1. Current block has fewer remaining elements than n: If encoding[i] - j < n, we consume all remaining elements in this block (which is encoding[i] - j elements), reduce n by that amount, and move to the next block by advancing i by 2 and resetting j to 0.

2. Current block has enough elements: If encoding[i] - j >= n, we can satisfy the request entirely from the current block. We advance j by n and return the value at encoding[i + 1].

We keep repeating case 1 until we either satisfy the request (case 2) or run out of blocks entirely (return -1).

This approach is efficient because we're jumping through blocks rather than iterating through individual elements, giving us a time complexity proportional to the number of blocks rather than the total number of elements in the decoded sequence.

Solution Approach

The solution uses a two-pointer approach to efficiently traverse the run-length encoded array without decoding it.

Data Structure:

• self.encoding: Stores the run-length encoded array
• self.i: Points to the current count position in the encoding (always at even indices)
• self.j: Tracks how many elements have been consumed from the current block

Implementation Details:

1. Initialization (__init__ method):

   • Store the encoding array
   • Initialize i = 0 to point to the first count
   • Initialize j = 0 to indicate no elements consumed from the first block

2. The next(n) method:

   The method uses a while loop to process blocks until we exhaust n elements:

   while self.i < len(self.encoding):

   Inside the loop, we check if the current block has enough remaining elements:

   • Case 1: Current block insufficient (self.encoding[self.i] - self.j < n):
     • Calculate remaining elements in current block: self.encoding[self.i] - self.j
     • Subtract these from n: n -= self.encoding[self.i] - self.j
     • Move to next block: self.i += 2 (skip both count and value)
     • Reset consumption counter: self.j = 0
     • Continue to next iteration
   • Case 2: Current block sufficient (self.encoding[self.i] - self.j >= n):
     • Consume n elements from current block: self.j += n
     • Return the value of current block: return self.encoding[self.i + 1]

   If the loop completes without returning (meaning we've exhausted all blocks), return -1.

Example Walkthrough:

Given encoding [3,8,2,5]:

• next(2): i=0, j=0. Block has 3 elements, need 2. Update j=2, return 8
• next(1): i=0, j=2. Block has 1 remaining, need 1. Update j=3, return 8
• next(2): i=0, j=3. Block has 0 remaining, need 2. Move to next block (i=2, j=0), then consume 2 from new block, return 5

Time Complexity: O(k) per call where k is the number of blocks we need to traverse Space Complexity: O(1) as we only maintain two pointers regardless of input size

Example Walkthrough

Let's walk through a concrete example with the encoding [3,8,2,5,1,4] which represents the sequence: [8,8,8,5,5,4].

Initial State:

• encoding = [3,8,2,5,1,4]
• i = 0 (pointing to count 3)
• j = 0 (no elements consumed from current block)

Call 1: next(2)

• Current block: 3 eights, with 0 consumed → 3 remaining
• Need to exhaust: 2 elements
• Since 3 >= 2, we can satisfy this from the current block
• Update j = 0 + 2 = 2
• Return encoding[1] = 8
• State after: i=0, j=2

Call 2: next(3)

• Current block: 3 eights, with 2 consumed → 1 remaining
• Need to exhaust: 3 elements
• Since 1 < 3, current block insufficient
  • Exhaust remaining 1 element: n = 3 - 1 = 2
  • Move to next block: i = 0 + 2 = 2, j = 0
• Now at block of 2 fives
• Since 2 >= 2, we can satisfy remaining from this block
• Update j = 0 + 2 = 2
• Return encoding[3] = 5
• State after: i=2, j=2

Call 3: next(2)

• Current block: 2 fives, with 2 consumed → 0 remaining
• Need to exhaust: 2 elements
• Since 0 < 2, current block insufficient
  • Exhaust remaining 0 elements: n = 2 - 0 = 2
  • Move to next block: i = 2 + 2 = 4, j = 0
• Now at block of 1 four
• Since 1 < 2, current block still insufficient
  • Exhaust remaining 1 element: n = 2 - 1 = 1
  • Move to next block: i = 4 + 2 = 6, j = 0
• Now i = 6 >= len(encoding), we've run out of elements
• Return -1

This example demonstrates all key scenarios: consuming within a block, spanning multiple blocks, and running out of elements.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n + q) where n is the length of the run-length encoding array and q is the total number of times next(n) is called.

The initialization takes O(1) time. For the next(n) method, in the worst case, we might traverse through the entire encoding array across all calls to next(). Each element in the encoding array is visited at most once throughout all calls (as we move forward with index i and never go back). Since the encoding has length n, and we make q calls to next(), the total time complexity is O(n + q).

Space Complexity: O(n) where n is the length of the run-length encoding array.

We store the encoding array which takes O(n) space. The additional variables i and j are just integer pointers that take O(1) space. Therefore, the overall space complexity is O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Index Management When Moving to Next Block

Pitfall: A common mistake is incrementing the index by 1 instead of 2 when moving to the next count-value pair, or forgetting to reset the consumed count.

# WRONG - Only increments by 1
if available < n:
    n -= available
    self.current_index += 1  # Should be += 2
    # Forgot to reset consumed_count

Solution: Always remember that the encoding array stores pairs, so you need to skip both the count and value:

# CORRECT
if available < n:
    n -= available
    self.current_index += 2  # Skip both count and value
    self.consumed_count = 0  # Reset for new block

2. Handling Zero-Count Entries

Pitfall: The encoding might contain entries with count 0 (e.g., [0,5,3,8]). If not handled properly, this can cause infinite loops or incorrect results.

# PROBLEMATIC - Can get stuck if encoding[i] is 0
while self.current_index < len(self.encoding):
    available = self.encoding[self.current_index] - self.consumed_count
    if available < n:
        n -= available  # If available is 0, n never decreases
        # ... rest of code

Solution: Skip zero-count entries immediately:

def next(self, n: int) -> int:
    while self.current_index < len(self.encoding):
        # Skip zero-count entries
        if self.encoding[self.current_index] == 0:
            self.current_index += 2
            continue
          
        available = self.encoding[self.current_index] - self.consumed_count
        # ... rest of logic

3. Not Preserving State Between Calls

Pitfall: Resetting the iterator state in each next() call or using local variables instead of instance variables.

# WRONG - Loses state between calls
def next(self, n: int) -> int:
    current_index = 0  # Should use self.current_index
    consumed_count = 0  # Should use self.consumed_count
    # ... rest of code

Solution: Use instance variables to maintain state across multiple next() calls:

# CORRECT
def __init__(self, encoding: List[int]) -> None:
    self.encoding = encoding
    self.current_index = 0  # Instance variable
    self.consumed_count = 0  # Instance variable

4. Off-by-One Error in Available Count Calculation

Pitfall: Incorrectly calculating how many elements are available in the current block.

# WRONG - Might use wrong formula
available = self.encoding[self.current_index] - self.consumed_count - 1
# or
available = self.encoding[self.current_index + 1] - self.consumed_count

Solution: The correct formula is straightforward:

# CORRECT
available = self.encoding[self.current_index] - self.consumed_count

5. Returning Wrong Value When Block Partially Consumed

Pitfall: When exhausting elements from a partially consumed block, accidentally returning the count instead of the value.

# WRONG - Returns count instead of value
if available >= n:
    self.consumed_count += n
    return self.encoding[self.current_index]  # Should be self.current_index + 1

Solution: Always return the value at the odd index (current_index + 1):

# CORRECT
if available >= n:
    self.consumed_count += n
    return self.encoding[self.current_index + 1]  # Value is at odd index