LeetCode in Kotlin

2673. Make Costs of Paths Equal in a Binary Tree

Medium

You are given an integer n representing the number of nodes in a perfect binary tree consisting of nodes numbered from 1 to n. The root of the tree is node 1 and each node i in the tree has two children where the left child is the node 2 * i and the right child is 2 * i + 1.

Each node in the tree also has a cost represented by a given 0-indexed integer array cost of size n where cost[i] is the cost of node i + 1. You are allowed to increment the cost of any node by 1 any number of times.

Return the minimum number of increments you need to make the cost of paths from the root to each leaf node equal.

Note:

• A perfect binary tree is a tree where each node, except the leaf nodes, has exactly 2 children.
• The cost of a path is the sum of costs of nodes in the path.

Example 1:

Input: n = 7, cost = [1,5,2,2,3,3,1]

Output: 6

Explanation: We can do the following increments:

• Increase the cost of node 4 one time.
• Increase the cost of node 3 three times.
• Increase the cost of node 7 two times.

Each path from the root to a leaf will have a total cost of 9.

The total increments we did is 1 + 3 + 2 = 6. It can be shown that this is the minimum answer we can achieve.

Example 2:

Input: n = 3, cost = [5,3,3]

Output: 0

Explanation: The two paths already have equal total costs, so no increments are needed.

Constraints:

• 3 <= n <= 105
• n + 1 is a power of 2
• cost.length == n
• 1 <= cost[i] <= 104

Solution

class Solution {
    fun minIncrements(n: Int, cost: IntArray): Int {
        val last = n / 2 - 1
        var res = 0
        for (i in last downTo 0) {
            var abs = cost[2 * i + 1] - cost[2 * i + 2]
            if (abs < 0) abs *= -1
            cost[i] += maxOf(cost[2 * i + 1], cost[2 * i + 2])
            res += abs
        }
        return res
    }
}

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