LeetCode #2646 — HARD

Minimize the Total Price of the Trips

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

There exists an undirected and unrooted tree with n nodes indexed from 0 to n - 1. You are given the integer n and a 2D integer array edges of length n - 1, where edges[i] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i in the tree.

Each node has an associated price. You are given an integer array price, where price[i] is the price of the i^th node.

The price sum of a given path is the sum of the prices of all nodes lying on that path.

Additionally, you are given a 2D integer array trips, where trips[i] = [start_i, end_i] indicates that you start the i^th trip from the node start_i and travel to the node end_i by any path you like.

Before performing your first trip, you can choose some non-adjacent nodes and halve the prices.

Return the minimum total price sum to perform all the given trips.

Example 1:

Input: n = 4, edges = [[0,1],[1,2],[1,3]], price = [2,2,10,6], trips = [[0,3],[2,1],[2,3]]
Output: 23
Explanation: The diagram above denotes the tree after rooting it at node 2. The first part shows the initial tree and the second part shows the tree after choosing nodes 0, 2, and 3, and making their price half.
For the 1^st trip, we choose path [0,1,3]. The price sum of that path is 1 + 2 + 3 = 6.
For the 2^nd trip, we choose path [2,1]. The price sum of that path is 2 + 5 = 7.
For the 3^rd trip, we choose path [2,1,3]. The price sum of that path is 5 + 2 + 3 = 10.
The total price sum of all trips is 6 + 7 + 10 = 23.
It can be proven, that 23 is the minimum answer that we can achieve.

Example 2:

Input: n = 2, edges = [[0,1]], price = [2,2], trips = [[0,0]]
Output: 1
Explanation: The diagram above denotes the tree after rooting it at node 0. The first part shows the initial tree and the second part shows the tree after choosing node 0, and making its price half.
For the 1^st trip, we choose path [0]. The price sum of that path is 1.
The total price sum of all trips is 1. It can be proven, that 1 is the minimum answer that we can achieve.

Constraints:

1 <= n <= 50
edges.length == n - 1
0 <= a_i, b_i <= n - 1
edges represents a valid tree.
price.length == n
price[i] is an even integer.
1 <= price[i] <= 1000
1 <= trips.length <= 100
0 <= start_i, end_i <= n - 1

Step 01

Brute Force Baseline

Problem summary: There exists an undirected and unrooted tree with n nodes indexed from 0 to n - 1. You are given the integer n and a 2D integer array edges of length n - 1, where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree. Each node has an associated price. You are given an integer array price, where price[i] is the price of the ith node. The price sum of a given path is the sum of the prices of all nodes lying on that path. Additionally, you are given a 2D integer array trips, where trips[i] = [starti, endi] indicates that you start the ith trip from the node starti and travel to the node endi by any path you like. Before performing your first trip, you can choose some non-adjacent nodes and halve the prices. Return the minimum total price sum to perform all the given trips.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Tree

Example 1

4
[[0,1],[1,2],[1,3]]
[2,2,10,6]
[[0,3],[2,1],[2,3]]

Example 2

2
[[0,1]]
[2,2]
[[0,0]]

Step 02

Core Insight

What unlocks the optimal approach

The final answer is the price[i] * freq[i], where freq[i] is the number of times node i was visited during the trip, and price[i] is the final price.
To find freq[i] we will use dfs or bfs for each trip and update every node on the path start and end.
Finally, to find the final price[i] we will use dynamic programming on the tree. Let dp(v, 0/1) denote the minimum total price with the node v’s price being halved or not.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2646: Minimize the Total Price of the Trips
class Solution {
    private List<Integer>[] g;
    private int[] price;
    private int[] cnt;

    public int minimumTotalPrice(int n, int[][] edges, int[] price, int[][] trips) {
        this.price = price;
        cnt = new int[n];
        g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        for (var t : trips) {
            int start = t[0], end = t[1];
            dfs(start, -1, end);
        }
        int[] ans = dfs2(0, -1);
        return Math.min(ans[0], ans[1]);
    }

    private boolean dfs(int i, int fa, int k) {
        ++cnt[i];
        if (i == k) {
            return true;
        }
        boolean ok = false;
        for (int j : g[i]) {
            if (j != fa) {
                ok = dfs(j, i, k);
                if (ok) {
                    break;
                }
            }
        }
        if (!ok) {
            --cnt[i];
        }
        return ok;
    }

    private int[] dfs2(int i, int fa) {
        int a = cnt[i] * price[i];
        int b = a >> 1;
        for (int j : g[i]) {
            if (j != fa) {
                var t = dfs2(j, i);
                a += Math.min(t[0], t[1]);
                b += t[0];
            }
        }
        return new int[] {a, b};
    }
}

// Accepted solution for LeetCode #2646: Minimize the Total Price of the Trips
func minimumTotalPrice(n int, edges [][]int, price []int, trips [][]int) int {
	g := make([][]int, n)
	for _, e := range edges {
		a, b := e[0], e[1]
		g[a] = append(g[a], b)
		g[b] = append(g[b], a)
	}
	cnt := make([]int, n)
	var dfs func(int, int, int) bool
	dfs = func(i, fa, k int) bool {
		cnt[i]++
		if i == k {
			return true
		}
		ok := false
		for _, j := range g[i] {
			if j != fa {
				ok = dfs(j, i, k)
				if ok {
					break
				}
			}
		}
		if !ok {
			cnt[i]--
		}
		return ok
	}
	for _, t := range trips {
		start, end := t[0], t[1]
		dfs(start, -1, end)
	}
	var dfs2 func(int, int) (int, int)
	dfs2 = func(i, fa int) (int, int) {
		a := price[i] * cnt[i]
		b := a >> 1
		for _, j := range g[i] {
			if j != fa {
				x, y := dfs2(j, i)
				a += min(x, y)
				b += x
			}
		}
		return a, b
	}
	a, b := dfs2(0, -1)
	return min(a, b)
}

# Accepted solution for LeetCode #2646: Minimize the Total Price of the Trips
class Solution:
    def minimumTotalPrice(
        self, n: int, edges: List[List[int]], price: List[int], trips: List[List[int]]
    ) -> int:
        def dfs(i: int, fa: int, k: int) -> bool:
            cnt[i] += 1
            if i == k:
                return True
            ok = any(j != fa and dfs(j, i, k) for j in g[i])
            if not ok:
                cnt[i] -= 1
            return ok

        def dfs2(i: int, fa: int) -> (int, int):
            a = cnt[i] * price[i]
            b = a // 2
            for j in g[i]:
                if j != fa:
                    x, y = dfs2(j, i)
                    a += min(x, y)
                    b += x
            return a, b

        g = [[] for _ in range(n)]
        for a, b in edges:
            g[a].append(b)
            g[b].append(a)
        cnt = Counter()
        for start, end in trips:
            dfs(start, -1, end)
        return min(dfs2(0, -1))

// Accepted solution for LeetCode #2646: Minimize the Total Price of the Trips
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2646: Minimize the Total Price of the Trips
// class Solution {
//     private List<Integer>[] g;
//     private int[] price;
//     private int[] cnt;
// 
//     public int minimumTotalPrice(int n, int[][] edges, int[] price, int[][] trips) {
//         this.price = price;
//         cnt = new int[n];
//         g = new List[n];
//         Arrays.setAll(g, k -> new ArrayList<>());
//         for (var e : edges) {
//             int a = e[0], b = e[1];
//             g[a].add(b);
//             g[b].add(a);
//         }
//         for (var t : trips) {
//             int start = t[0], end = t[1];
//             dfs(start, -1, end);
//         }
//         int[] ans = dfs2(0, -1);
//         return Math.min(ans[0], ans[1]);
//     }
// 
//     private boolean dfs(int i, int fa, int k) {
//         ++cnt[i];
//         if (i == k) {
//             return true;
//         }
//         boolean ok = false;
//         for (int j : g[i]) {
//             if (j != fa) {
//                 ok = dfs(j, i, k);
//                 if (ok) {
//                     break;
//                 }
//             }
//         }
//         if (!ok) {
//             --cnt[i];
//         }
//         return ok;
//     }
// 
//     private int[] dfs2(int i, int fa) {
//         int a = cnt[i] * price[i];
//         int b = a >> 1;
//         for (int j : g[i]) {
//             if (j != fa) {
//                 var t = dfs2(j, i);
//                 a += Math.min(t[0], t[1]);
//                 b += t[0];
//             }
//         }
//         return new int[] {a, b};
//     }
// }

// Accepted solution for LeetCode #2646: Minimize the Total Price of the Trips
function minimumTotalPrice(
    n: number,
    edges: number[][],
    price: number[],
    trips: number[][],
): number {
    const g: number[][] = Array.from({ length: n }, () => []);
    for (const [a, b] of edges) {
        g[a].push(b);
        g[b].push(a);
    }
    const cnt: number[] = new Array(n).fill(0);
    const dfs = (i: number, fa: number, k: number): boolean => {
        ++cnt[i];
        if (i === k) {
            return true;
        }
        let ok = false;
        for (const j of g[i]) {
            if (j !== fa) {
                ok = dfs(j, i, k);
                if (ok) {
                    break;
                }
            }
        }
        if (!ok) {
            --cnt[i];
        }
        return ok;
    };
    for (const [start, end] of trips) {
        dfs(start, -1, end);
    }
    const dfs2 = (i: number, fa: number): number[] => {
        let a: number = price[i] * cnt[i];
        let b: number = a >> 1;
        for (const j of g[i]) {
            if (j !== fa) {
                const [x, y] = dfs2(j, i);
                a += Math.min(x, y);
                b += x;
            }
        }
        return [a, b];
    };
    const [a, b] = dfs2(0, -1);
    return Math.min(a, b);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.