grid47 Exploring patterns and algorithms
Oct 3, 2024
6 min read
Solution to LeetCode 341: Flatten Nested List Iterator Problem
You are given a nested list of integers nestedList, where each element can either be an integer or a sublist containing integers or other sublists. Implement an iterator to flatten this nested list.
Problem
Approach
Steps
Complexity
Input: The input is a nested list of integers where each element is either an integer or a list of integers.
Example: nestedList = [[2,3],4,[5,6]]
Constraints:
• 1 <= nestedList.length <= 500
• The integer values in the nested list are within the range [-10^6, 10^6].
Output: The output is a list of integers that is the flattened version of the nested list.
Example: [2, 3, 4, 5, 6]
Constraints:
• The flattened list should maintain the order of the elements from the original nested list.
Goal: The goal is to implement an iterator that flattens the nested list efficiently.
Steps:
• Use a stack to keep track of the current position in the nested list.
• For each call to next(), move to the next element in the flattened list.
• For each call to hasNext(), check if there are more elements left to return in the nested list.
Goal: The solution must handle nested lists of size up to 500 and integer values in the range [-10^6, 10^6].
Steps:
• The solution must ensure that it handles nested lists efficiently without excessive recursion or memory usage.
Assumptions:
• The input list may contain nested sublists of various depths.
• Input: nestedList = [[2,3],4,[5,6]]
• Explanation: The input nested list contains sublists and integers. The iterator should flatten this list to return the integers in the order [2, 3, 4, 5, 6].
• Input: nestedList = [1, [2, [3, 4]]]
• Explanation: The input list has nested sublists. The iterator should flatten it to return the integers in the order [1, 2, 3, 4].
Approach: The approach to flattening the nested list is to use a stack to iteratively traverse the elements and flatten them.
Observations:
• The nested list contains integers and sublists of integers.
• A stack can help manage the state of the iterator and allow efficient traversal through the nested list.
Steps:
• Initialize a stack with the beginning and end iterators of the nested list.
• Use the stack to traverse the nested list. For each integer, return it; for each list, dive into its elements.
Empty Inputs:
• If the nested list is empty, the iterator should immediately return false from hasNext() and not produce any output.
Large Inputs:
• For larger nested lists, the solution must avoid excessive recursion and handle deeply nested sublists efficiently.
Special Values:
• Consider cases where the nested list contains only integers or only empty sublists.
Constraints:
• The iterator should not exceed O(n) time complexity, where n is the total number of integers in the flattened list.
NestedIterator(vector<NestedInteger>&nestedList) {
begins.push(nestedList.begin());
ends.push(nestedList.end());
}
intnext() {
hasNext();
return (begins.top()++)->getInteger();
}
boolhasNext() {
while (begins.size()) {
if (begins.top() == ends.top()) {
begins.pop();
ends.pop();
} else {
auto x = begins.top();
if (x->isInteger())
returntrue;
begins.top()++;
begins.push(x->getList().begin());
ends.push(x->getList().end());
}
}
returnfalse;
}
private:stack<vector<NestedInteger>::iterator> begins, ends;
The constructor takes a reference to a vector of `NestedInteger` objects, representing a potentially nested list, and initializes two stacks to keep track of the current and end iterators for each list level.
2 : Push Begin Iterator
begins.push(nestedList.begin());
The begin iterator for the first level of the nested list is pushed onto the `begins` stack.
3 : Push End Iterator
ends.push(nestedList.end());
The end iterator for the first level of the nested list is pushed onto the `ends` stack.
4 : Next Function Declaration
intnext() {
The `next()` function retrieves the next integer in the nested structure, if available.
5 : HasNext Call
hasNext();
Calls `hasNext()` to ensure there is an integer available before proceeding to retrieve it.
6 : Return Integer
return (begins.top()++)->getInteger();
The next integer is retrieved from the top of the `begins` stack by advancing the iterator and calling `getInteger()` on the current `NestedInteger` object.
7 : HasNext Function Declaration
boolhasNext() {
The `hasNext()` function checks if there are more integers available in the nested list.
8 : While Loop Start
while (begins.size()) {
This `while` loop continues as long as there are elements in the `begins` stack, indicating that there are more levels in the nested list to explore.
9 : Check End of Current List
if (begins.top() == ends.top()) {
Checks if the current iterator has reached the end of the current list level.
10 : Pop Iterators
begins.pop();
If the iterator has reached the end, both the `begins` and `ends` iterators for that level are popped off the stacks.
11 : Pop End Iterator
ends.pop();
Pops the end iterator for the current level.
12 : Else Block Start
} else {
If the current iterator has not reached the end, continue processing the current element.
13 : Check if Integer
auto x = begins.top();
Grabs the current `NestedInteger` from the top of the `begins` stack.
14 : Return True if Integer
if (x->isInteger())
Checks if the current `NestedInteger` is an integer. If it is, the function returns `true`, indicating that the next integer is available.
15 : Return True
returntrue;
Returns `true` indicating that the next integer is available.
16 : Increment Iterator
begins.top()++;
Advances the iterator to the next element in the current list.
17 : Push Next Level Begin Iterator
begins.push(x->getList().begin());
If the current `NestedInteger` is a list, pushes the begin iterator of the nested list onto the `begins` stack.
18 : Push Next Level End Iterator
ends.push(x->getList().end());
Pushes the end iterator of the nested list onto the `ends` stack.
19 : Return False
returnfalse;
Returns `false` if there are no more integers available in the nested structure.
Two stacks are declared to keep track of iterators for traversing the nested list: `begins` holds the current iterators, and `ends` holds the end iterators for each level.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear because each integer is visited once.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the stack used for traversal and the result storage.
