How do you solve Minimum Increments to Equalize Leaf Paths on LeetCode?

Minimum Increments to Equalize Leaf Paths (LeetCode #3593) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Each path must match the maximum leaf path cost.

What is the time complexity of Minimum Increments to Equalize Leaf Paths?

The optimal solution for Minimum Increments to Equalize Leaf Paths runs in O(n) time and uses O(n) space. DFS visits each node once, leading to linear time complexity.

What is the brute force solution for Minimum Increments to Equalize Leaf Paths?

The brute force approach for Minimum Increments to Equalize Leaf Paths is Brute Force, with O(n²) time complexity and O(1) space complexity. Evaluate every root-to-leaf path and calculate the cost needed to equalize all paths to the maximum cost. This involves checking each node and summing costs repeatedly. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Minimum Increments to Equalize Leaf Paths?

Clarify the tree structure and input format. Discuss edge cases like single-node trees. Explain your thought process while coding.

What are common mistakes when solving Minimum Increments to Equalize Leaf Paths?

Forgetting to check all paths during DFS. Incorrectly calculating the cost increments needed.

#3593

Minimum Increments to Equalize Leaf Paths

Medium

ArrayDynamic ProgrammingTreeDepth-First SearchTree TraversalDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

Use DFS to traverse the tree, calculating the path costs dynamically while keeping track of the maximum leaf cost. Adjust node costs only when necessary.

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Algorithm

3 steps

1Step 1: Perform DFS to calculate the total cost of each root-to-leaf path.
2Step 2: Track the maximum leaf cost found during traversal.
3Step 3: For each node, compute the increments needed to reach the maximum leaf cost and count the nodes that need adjustment.

solution.py19 lines

1def minIncrements(n, edges, cost):
2    from collections import defaultdict
3    tree = defaultdict(list)
4    for u, v in edges:
5        tree[u].append(v)
6        tree[v].append(u)
7    maxLeafCost = 0
8    def dfs(node, parent, currentCost):
9        nonlocal maxLeafCost
10        currentCost += cost[node]
11        isLeaf = True
12        for neighbor in tree[node]:
13            if neighbor != parent:
14                isLeaf = False
15                dfs(neighbor, node, currentCost)
16        if isLeaf:
17            maxLeafCost = max(maxLeafCost, currentCost)
18    dfs(0, -1, 0)
19    return maxLeafCost

ℹ

Complexity note: DFS visits each node once, leading to linear time complexity.

1Each path must match the maximum leaf path cost.
2Only nodes contributing to paths with lesser costs need adjustments.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.