Leetcode 2673: Make Costs of Paths Equal in a Binary Tree

Input: The input consists of an integer `n` representing the number of nodes in a perfect binary tree, followed by a list of integers `cost`, where `cost[i]` is the cost of node `i+1`.

Example: Input: n = 5, cost = [2, 4, 3, 2, 5]

Constraints:

• 3 <= n <= 10^5

• n + 1 is a power of 2

• cost.length == n

• 1 <= cost[i] <= 10^4

Output: The output should be a single integer, which represents the minimum number of increments required to make the cost of all root-to-leaf paths equal.

Example: Output: 4

Constraints:

• The result should be a non-negative integer.

Goal: The goal is to minimize the number of increments required to equalize the path costs.

Steps:

• Step 1: Starting from the leaf nodes, propagate the required increments up to the root node, ensuring each path has the same total cost.

• Step 2: For each non-leaf node, compare the costs of its left and right children. Increment the smaller one to match the larger one and count the number of increments.

• Step 3: Repeat the process until you reach the root node.

Goal: The solution must efficiently handle the constraints specified.

Steps:

• Ensure the solution works within O(n) time complexity to handle the largest inputs.

Assumptions:

• The tree is a perfect binary tree, meaning every non-leaf node has exactly two children.

• All nodes have a non-negative cost value.

• Input: Input: n = 5, cost = [2, 4, 3, 2, 5]

• Explanation: In this case, the two paths from root to leaves are [2, 4, 5] and [2, 3, 5]. To equalize the path costs, we increment the cost of node 2 to match node 3's cost. This requires 4 increments, so the total increments required are 4.

Approach: To solve this problem, we will work our way from the leaf nodes upwards to the root, balancing the path costs by making necessary increments.

Observations:

• We need to ensure that the cost of each root-to-leaf path is the same.

• This requires comparing the costs of left and right children of every node and balancing them.

• Starting from the leaf nodes and adjusting the internal nodes' costs seems to be an efficient way to solve this problem.

Steps:

• Step 1: Initialize an array `cost` of size `n`.

• Step 2: Start from the last non-leaf node and move upwards. For each node, compare the costs of its left and right children and increment the smaller cost to match the larger one.

• Step 3: Accumulate the number of increments required as you move upwards towards the root.

Empty Inputs:

• The input will always have at least 3 nodes, ensuring a valid binary tree structure.

Large Inputs:

• The solution must handle up to 10^5 nodes efficiently.

Special Values:

• If the tree already has equal costs for all root-to-leaf paths, the result should be 0 increments.

Constraints:

• Ensure the solution runs in O(n) time complexity to handle the largest inputs.

int minIncrements(int n, vector<int>& A) {
    int res = 0;
    for (int i = n / 2 - 1; i >= 0; --i) {
        int l = i * 2 + 1, r = i * 2 + 2;
        res += abs(A[l] - A[r]);
        A[i] += max(A[l], A[r]);
    }
    return res;
}

1 : Function Definition

int minIncrements(int n, vector<int>& A) {

This is the function definition of 'minIncrements', which takes an integer 'n' (size of the array) and a reference to a vector 'A'. The goal is to compute the minimum number of increments needed to turn the array into a valid heap.

2 : Result Initialization

    int res = 0;

We initialize 'res' to 0. This variable will keep track of the total number of increments made while adjusting the array.

3 : Loop Start

    for (int i = n / 2 - 1; i >= 0; --i) {

We start a loop that iterates over the parent nodes of the binary heap, starting from the last parent node (at index 'n / 2 - 1') and moving upwards.

4 : Child Nodes Calculation

        int l = i * 2 + 1, r = i * 2 + 2;

For each parent node, we calculate the indices of its left ('l') and right ('r') children using the standard binary heap indexing formula.

5 : Difference Calculation

        res += abs(A[l] - A[r]);

We calculate the absolute difference between the values of the left and right children, adding this value to 'res' to account for the required increments to make the children equal.

6 : Adjust Parent Value

        A[i] += max(A[l], A[r]);

To maintain the heap property, we adjust the parent node's value by adding the maximum of its left or right child's value.

7 : Return Result

    return res;

We return the total number of increments made to the array, stored in 'res'.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The time complexity is O(n) as we only need to traverse the tree once, adjusting costs from the leaves to the root.

Best Case: O(n)

Worst Case: O(n)

Description: The space complexity is O(n) due to the space needed for the cost array.

Leetcode 2673: Make Costs of Paths Equal in a Binary Tree

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