LeetCode #341 — MEDIUM

Flatten Nested List Iterator

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

The Problem

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<NestedInteger> 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].

Step 01

Brute Force Baseline

Problem summary: 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<NestedInteger> 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.

Baseline thinking

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

Pattern signal: Stack · Tree · Design

Example 1

[[1,1],2,[1,1]]

Example 2

[1,[4,[6]]]

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 #341: Flatten Nested List Iterator
/**
 * // This is the interface that allows for creating nested lists.
 * // You should not implement it, or speculate about its implementation
 * public interface NestedInteger {
 *
 *     // @return true if this NestedInteger holds a single integer, rather than a nested list.
 *     public boolean isInteger();
 *
 *     // @return the single integer that this NestedInteger holds, if it holds a single integer
 *     // Return null if this NestedInteger holds a nested list
 *     public Integer getInteger();
 *
 *     // @return the nested list that this NestedInteger holds, if it holds a nested list
 *     // Return empty list if this NestedInteger holds a single integer
 *     public List<NestedInteger> getList();
 * }
 */
public class NestedIterator implements Iterator<Integer> {
    private List<Integer> nums = new ArrayList<>();
    private int i = -1;

    public NestedIterator(List<NestedInteger> nestedList) {
        dfs(nestedList);
    }

    @Override
    public Integer next() {
        return nums.get(++i);
    }

    @Override
    public boolean hasNext() {
        return i + 1 < nums.size();
    }

    private void dfs(List<NestedInteger> ls) {
        for (var x : ls) {
            if (x.isInteger()) {
                nums.add(x.getInteger());
            } else {
                dfs(x.getList());
            }
        }
    }
}

/**
 * Your NestedIterator object will be instantiated and called as such:
 * NestedIterator i = new NestedIterator(nestedList);
 * while (i.hasNext()) v[f()] = i.next();
 */

// Accepted solution for LeetCode #341: Flatten Nested List Iterator
/**
 * // This is the interface that allows for creating nested lists.
 * // You should not implement it, or speculate about its implementation
 * type NestedInteger struct {
 * }
 *
 * // Return true if this NestedInteger holds a single integer, rather than a nested list.
 * func (this NestedInteger) IsInteger() bool {}
 *
 * // Return the single integer that this NestedInteger holds, if it holds a single integer
 * // The result is undefined if this NestedInteger holds a nested list
 * // So before calling this method, you should have a check
 * func (this NestedInteger) GetInteger() int {}
 *
 * // Set this NestedInteger to hold a single integer.
 * func (n *NestedInteger) SetInteger(value int) {}
 *
 * // Set this NestedInteger to hold a nested list and adds a nested integer to it.
 * func (this *NestedInteger) Add(elem NestedInteger) {}
 *
 * // Return the nested list that this NestedInteger holds, if it holds a nested list
 * // The list length is zero if this NestedInteger holds a single integer
 * // You can access NestedInteger's List element directly if you want to modify it
 * func (this NestedInteger) GetList() []*NestedInteger {}
 */

type NestedIterator struct {
	nums []int
	i    int
}

func Constructor(nestedList []*NestedInteger) *NestedIterator {
	var dfs func([]*NestedInteger)
	nums := []int{}
	i := -1
	dfs = func(ls []*NestedInteger) {
		for _, x := range ls {
			if x.IsInteger() {
				nums = append(nums, x.GetInteger())
			} else {
				dfs(x.GetList())
			}
		}
	}
	dfs(nestedList)
	return &NestedIterator{nums, i}
}

func (this *NestedIterator) Next() int {
	this.i++
	return this.nums[this.i]
}

func (this *NestedIterator) HasNext() bool {
	return this.i+1 < len(this.nums)
}

# Accepted solution for LeetCode #341: Flatten Nested List Iterator
# """
# 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) -> [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: [NestedInteger]):
        def dfs(ls):
            for x in ls:
                if x.isInteger():
                    self.nums.append(x.getInteger())
                else:
                    dfs(x.getList())

        self.nums = []
        self.i = -1
        dfs(nestedList)

    def next(self) -> int:
        self.i += 1
        return self.nums[self.i]

    def hasNext(self) -> bool:
        return self.i + 1 < len(self.nums)


# Your NestedIterator object will be instantiated and called as such:
# i, v = NestedIterator(nestedList), []
# while i.hasNext(): v.append(i.next())

// Accepted solution for LeetCode #341: Flatten Nested List Iterator
// #[derive(Debug, PartialEq, Eq)]
// pub enum NestedInteger {
//   Int(i32),
//   List(Vec<NestedInteger>)
// }
struct NestedIterator {
    nums: Vec<i32>,
    i: usize,
}

/**
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl NestedIterator {
    fn new(nested_list: Vec<NestedInteger>) -> Self {
        let mut nums = Vec::new();
        Self::dfs(&nested_list, &mut nums);
        NestedIterator { nums, i: 0 }
    }

    fn next(&mut self) -> i32 {
        let result = self.nums[self.i];
        self.i += 1;
        result
    }

    fn has_next(&self) -> bool {
        self.i < self.nums.len()
    }

    fn dfs(nested_list: &Vec<NestedInteger>, nums: &mut Vec<i32>) {
        for ni in nested_list {
            match ni {
                NestedInteger::Int(x) => nums.push(*x),
                NestedInteger::List(list) => Self::dfs(list, nums),
            }
        }
    }
}

// Accepted solution for LeetCode #341: Flatten Nested List Iterator
/**
 * // This is the interface that allows for creating nested lists.
 * // You should not implement it, or speculate about its implementation
 * class NestedInteger {
 *     If value is provided, then it holds a single integer
 *     Otherwise it holds an empty nested list
 *     constructor(value?: number) {
 *         ...
 *     };
 *
 *     Return true if this NestedInteger holds a single integer, rather than a nested list.
 *     isInteger(): boolean {
 *         ...
 *     };
 *
 *     Return the single integer that this NestedInteger holds, if it holds a single integer
 *     Return null if this NestedInteger holds a nested list
 *     getInteger(): number | null {
 *         ...
 *     };
 *
 *     Set this NestedInteger to hold a single integer equal to value.
 *     setInteger(value: number) {
 *         ...
 *     };
 *
 *     Set this NestedInteger to hold a nested list and adds a nested integer elem to it.
 *     add(elem: NestedInteger) {
 *         ...
 *     };
 *
 *     Return the nested list that this NestedInteger holds,
 *     or an empty list if this NestedInteger holds a single integer
 *     getList(): NestedInteger[] {
 *         ...
 *     };
 * };
 */

class NestedIterator {
    private nums: number[] = [];
    private i = -1;
    constructor(nestedList: NestedInteger[]) {
        const dfs = (ls: NestedInteger[]) => {
            for (const x of ls) {
                if (x.isInteger()) {
                    this.nums.push(x.getInteger());
                } else {
                    dfs(x.getList());
                }
            }
        };
        dfs(nestedList);
    }

    hasNext(): boolean {
        return this.i + 1 < this.nums.length;
    }

    next(): number {
        return this.nums[++this.i];
    }
}

/**
 * Your ParkingSystem object will be instantiated and called as such:
 * var obj = new NestedIterator(nestedList)
 * var a: number[] = []
 * while (obj.hasNext()) a.push(obj.next());
 */

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)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.