Introduction

In this blog post, we will explore how to solve the LeetCode problem "341" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, 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

You are given a nested list of integers nestedList. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it. Implement the NestedIterator class: NestedIterator(List nestedList) Initializes the iterator with the nested list nestedList. int next() Returns the next integer in the nested list. boolean hasNext() Returns true if there are still some integers in the nested list and false otherwise. Your code will be tested with the following pseudocode: initialize iterator with nestedList res = [] while iterator.hasNext() append iterator.next() to the end of res return res If res matches the expected flattened list, then your code will be judged as correct. Example 1: Input: nestedList = [[1,1],2,[1,1]] Output: [1,1,2,1,1] Explanation: By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,1,2,1,1]. Example 2: Input: nestedList = [1,[4,[6]]] Output: [1,4,6] Explanation: By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,4,6]. Constraints: 1 <= nestedList.length <= 500 The values of the integers in the nested list is in the range [-106, 106].

Explanation

Here's a breakdown of the solution:

• Stack-Based Iteration: Use a stack to keep track of the nested lists we need to explore. This allows us to efficiently handle the nested structure.
• Pre-processing in hasNext(): Instead of doing the flattening inside next(), which can be called repeatedly, we proactively move the iterator forward in hasNext() until the top of the stack contains an actual integer. This amortizes the cost of finding the next integer.

• Efficient Traversal: Only expand lists when necessary, avoiding unnecessary recursion and memory usage.

• Runtime & Storage Complexity: O(1) average time complexity for next() and hasNext(), O(D) space complexity, where D is the maximum depth of nesting in the input list.

Code

    # """
# This is the interface that allows for creating nested lists.
# You should not implement it, or speculate about its implementation
# """
#class NestedInteger:
#    def isInteger(self) -> bool:
#        """
#        @return True if this NestedInteger holds a single integer, rather than a nested list.
#        """
#
#    def getInteger(self) -> int:
#        """
#        @return the single integer that this NestedInteger holds, if it holds a single integer
#        Return None if this NestedInteger holds a nested list
#        """
#
#    def getList(self) -> list[NestedInteger]:
#        """
#        @return the nested list that this NestedInteger holds, if it holds a nested list
#        Return None if this NestedInteger holds a single integer
#        """

class NestedIterator:
    def __init__(self, nestedList: list[NestedInteger]):
        self.stack = nestedList[::-1] # Initialize stack with reversed list

    def next(self) -> int:
        if not self.hasNext():
            return None  # Or raise an exception, as appropriate
        return self.stack.pop().getInteger()

    def hasNext(self) -> bool:
        while self.stack:
            top = self.stack[-1]
            if top.isInteger():
                return True
            else:
                self.stack.pop()
                nested_list = top.getList()
                self.stack.extend(nested_list[::-1]) # Add to stack in reversed order

        return False