Leetcode 2583: Kth Largest Sum in a Binary Tree
You are given a 0-indexed array of strings
words and two integers left and right. A string is considered a vowel string if it starts and ends with a vowel character (vowels are ‘a’, ’e’, ‘i’, ‘o’, ‘u’). Your task is to return the number of vowel strings in the array words where the indices fall within the range [left, right].Problem
Approach
Steps
Complexity
Input: The input consists of an array `words` and two integers `left` and `right`.
Example: For example, `words = ["apple", "banana", "oreo"], left = 0, right = 2`.
Constraints:
• 1 <= words.length <= 1000
• 1 <= words[i].length <= 10
• words[i] consists of only lowercase English letters
• 0 <= left <= right < words.length
Output: The output is an integer that represents the number of vowel strings within the specified range of indices.
Example: For input `words = ["apple", "banana", "oreo"], left = 0, right = 2`, the output is `2`.
Constraints:
• The result will always be a valid integer.
Goal: The goal is to find the number of vowel strings in the given range of indices.
Steps:
• 1. Iterate through the array `words` from index `left` to index `right`.
• 2. For each word, check if the first and last characters are vowels.
• 3. Count how many words meet the condition of being a vowel string.
Goal: The length of the array is at most 1000 and each string in the array has a maximum length of 10 characters.
Steps:
• 1 <= words.length <= 1000
• 1 <= words[i].length <= 10
• 0 <= left <= right < words.length
Assumptions:
• The input array is valid, containing only lowercase English letters.
• Input: For `words = ["apple", "banana", "oreo"], left = 0, right = 2`
• Explanation: The valid vowel strings in the range are "apple" and "oreo", so the result is 2.
Approach: To solve the problem, we can loop through the array `words` within the specified range, and for each word, check if it starts and ends with a vowel. This can be achieved using basic string operations.
Observations:
• We can easily check the start and end characters of each word to see if they are vowels.
• By iterating over the range and checking each word, we can count the number of valid vowel strings efficiently.
Steps:
• 1. Start iterating from index `left` to index `right`.
• 2. For each word in this range, check if the first and last characters are vowels.
• 3. Count the number of words that satisfy this condition.
Empty Inputs:
• The problem guarantees that the input will contain at least one word.
Large Inputs:
• The solution should handle arrays with up to 1000 words efficiently.
Special Values:
• If `left` equals `right`, we are checking only one word, which should be handled accordingly.
Constraints:
• The given range [left, right] will always be valid for the input array.
long long kthLargestLevelSum(TreeNode* root, int k) {
priority_queue<long long, vector<long long>, greater<long long>> pq;
if(!root) return 0;
queue<TreeNode*> q;
q.push(root);
while(!q.empty()) {
int sz = q.size();
long long sum = 0;
while(sz--) {
auto it = q.front();
sum += it->val;
q.pop();
if(it->left) q.push(it->left);
if(it->right) q.push(it->right);
}
pq.push(sum);
if(pq.size() > k) pq.pop();
}
if(pq.size() < k) return -1;
return pq.top();
}
1 : Function Definition
long long kthLargestLevelSum(TreeNode* root, int k) {
Define the function `kthLargestLevelSum` which takes a `TreeNode* root` (the root of the binary tree) and an integer `k` (the level number to find the kth largest sum for).
2 : Priority Queue Initialization
priority_queue<long long, vector<long long>, greater<long long>> pq;
Create a min-heap priority queue `pq` to store the sums of levels, which will help track the kth largest sum.
3 : Edge Case Check
if(!root) return 0;
Check if the root is null. If the tree is empty, return 0 since there are no level sums.
4 : Queue Initialization
queue<TreeNode*> q;
Initialize a queue `q` to perform a breadth-first search (BFS) on the binary tree.
5 : Add Root to Queue
q.push(root);
Push the root node into the queue to start the BFS traversal.
6 : BFS Loop
while(!q.empty()) {
Start a while loop that runs until the queue is empty, ensuring that we process each level of the binary tree.
7 : Level Size Calculation
int sz = q.size();
Store the number of nodes at the current level (`sz`) in the variable `sz`.
8 : Level Sum Initialization
long long sum = 0;
Initialize a variable `sum` to store the sum of the node values at the current level.
9 : Process Each Node at the Level
while(sz--) {
Loop through all nodes at the current level.
10 : Node Processing
auto it = q.front();
Get the front node from the queue (`it`).
11 : Sum Accumulation
sum += it->val;
Add the value of the current node (`it->val`) to the `sum`.
12 : Queue Update
q.pop();
Remove the current node from the queue.
13 : Left Child Check
if(it->left) q.push(it->left);
If the current node has a left child, add it to the queue for processing in the next iteration.
14 : Right Child Check
if(it->right) q.push(it->right);
If the current node has a right child, add it to the queue for processing in the next iteration.
15 : Push Sum to Priority Queue
pq.push(sum);
Push the sum of the current level to the priority queue.
16 : Priority Queue Size Check
if(pq.size() > k) pq.pop();
If the priority queue exceeds size `k`, pop the smallest sum to ensure we only keep the top `k` largest sums.
17 : Check for kth Largest
if(pq.size() < k) return -1;
Check if the priority queue has fewer than `k` sums. If so, return -1 as there are not enough levels.
18 : Return kth Largest Sum
return pq.top();
Return the top element of the priority queue, which is the kth largest sum of a level.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) where n is the length of the range [left, right]. For each word, we perform constant time operations to check if the first and last characters are vowels.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1) since we only use a few extra variables for counting and checking the vowels.
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