Leetcode 1130: Minimum Cost Tree From Leaf Values
You are given an array arr of positive integers. Consider all binary trees such that each node has either 0 or 2 children, the values of arr correspond to the values of each leaf in an in-order traversal of the tree, and the value of each non-leaf node is equal to the product of the largest leaf values in its left and right subtrees. Return the smallest possible sum of the values of the non-leaf nodes among all possible binary trees.
Problem
Approach
Steps
Complexity
Input: You are given an array arr of integers where each integer represents a leaf node value in a binary tree.
Example: Input: arr = [5,3,8]
Constraints:
• 2 <= arr.length <= 40
• 1 <= arr[i] <= 15
• The answer fits into a 32-bit signed integer.
Output: Return the smallest possible sum of the values of each non-leaf node in the tree.
Example: Output: 75
Constraints:
• The sum of non-leaf node values should fit in a 32-bit signed integer.
Goal: The goal is to find the binary tree that minimizes the sum of non-leaf node values, considering all possible binary trees that can be formed from the given leaf values.
Steps:
• Initialize a stack with a value greater than any leaf node.
• Iterate through the array of leaf values, and for each value, check if it can form a valid tree node by combining with the previous elements in the stack.
• For each valid combination, compute the product of the current non-leaf node values and accumulate the result.
Goal: Ensure the solution is efficient even for the largest input sizes.
Steps:
• The array should have between 2 and 40 elements.
• Each element in the array is between 1 and 15.
Assumptions:
• The tree will be a full binary tree with 0 or 2 children per node.
• It is assumed that the values in the array will always be valid positive integers within the specified range.
• Input: Input: arr = [5,3,8]
• Explanation: The goal is to arrange the leaf nodes in such a way that the sum of non-leaf nodes is minimized. The smallest possible non-leaf node sum is 75, as calculated by the optimal tree structure.
• Input: Input: arr = [10,15,12]
• Explanation: The optimal binary tree for this set of leaf nodes produces the minimum sum of non-leaf node values, which is 330.
Approach: The solution involves using a stack to help compute the smallest sum by simulating the process of building a binary tree and evaluating the cost for each non-leaf node.
Observations:
• We need to optimize the tree structure such that the non-leaf node values are minimized.
• Using a stack allows us to easily manage and combine leaf nodes while maintaining an efficient calculation of the product of the largest values from each subtree.
Steps:
• Start with an empty stack and iterate through the leaf values in the array.
• For each leaf node, combine it with previous elements in the stack while ensuring that the product of the non-leaf nodes is minimized.
• Keep track of the sum of the non-leaf nodes and return the final result after processing all leaf values.
Empty Inputs:
• The problem guarantees that the input will have at least 2 elements, so no need to handle empty arrays.
Large Inputs:
• The algorithm should handle inputs with up to 40 leaf nodes efficiently.
Special Values:
• Consider cases where all elements in the array are equal, as this might affect the structure of the binary tree.
Constraints:
• Ensure that the solution runs within the time limits for the largest input sizes.
int mctFromLeafValues(vector<int>& arr) {
int res = 0;
vector<int> stk = { INT_MAX };
for(int a : arr) {
while(stk.back() <= a) {
int mid = stk.back();
stk.pop_back();
res += mid * min(stk.back(), a);
}
stk.push_back(a);
}
for(int i = 2; i < stk.size(); i++) {
res += stk[i] * stk[i - 1];
}
return res;
}
1 : Function Definition
int mctFromLeafValues(vector<int>& arr) {
This function definition begins the implementation of the `mctFromLeafValues` function, which takes a vector of integers `arr` and returns an integer.
2 : Variable Initialization
int res = 0;
This line initializes the result variable `res`, which will store the total cost of the minimum cost tree.
3 : Stack Initialization
vector<int> stk = { INT_MAX };
A stack `stk` is initialized with the maximum possible integer (`INT_MAX`) as a sentinel value to handle edge cases when processing the tree structure.
4 : Loop Setup
for(int a : arr) {
This `for` loop iterates through the elements in the input vector `arr`.
5 : Loop Condition
while(stk.back() <= a) {
This `while` loop checks if the current stack's top value is less than or equal to the current element `a`. If true, it processes the current stack element.
6 : Stack Manipulation
int mid = stk.back();
The top value of the stack is stored in `mid` for further calculations.
7 : Stack Pop
stk.pop_back();
This line removes the top element from the stack after processing it.
8 : Cost Calculation
res += mid * min(stk.back(), a);
The result variable `res` is updated by adding the product of the `mid` value and the minimum of the stack's new top value and the current element `a`.
9 : Stack Push
stk.push_back(a);
After the inner `while` loop, the current element `a` is pushed onto the stack.
10 : Additional Processing
for(int i = 2; i < stk.size(); i++) {
A second `for` loop starts at index 2 to process the remaining stack elements after the main loop finishes.
11 : Cost Calculation
res += stk[i] * stk[i - 1];
This line adds the product of each adjacent pair of stack elements to the result `res`.
12 : Return Statement
return res;
The final result `res` is returned, which contains the minimum cost of the tree.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear in terms of the number of leaf nodes due to the stack-based approach.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the space required for the stack.
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