## Solution

#### Approach 1: Store Exhausted Position and Quantity

Intuition

We can store an index i and quantity q which represents that q elements of A[i] (repeated A[i+1] times) are exhausted.

For example, if we have A = [1,2,3,4] (mapping to the sequence [2,4,4,4]) then i = 0, q = 0 represents that nothing is exhausted; i = 0, q = 1 represents that [2] is exhausted, i = 2, q = 1 will represent that we have currently exhausted [2, 4], and so on.

Algorithm

Say we want to exhaust n more elements. There are currently D = A[i] - q elements left to exhaust (of value A[i+1]).

If n > D, then we should exhaust all of them and continue: n -= D; i += 2; q = 0.

Otherwise, we should exhaust some of them and return the current element's value: q += D; return A[i+1].

Complexity Analysis

• Time Complexity: , where is the length of A, and is the number of calls to RLEIterator.next.

• Space Complexity: .

Analysis written by: @awice.