Minimize the Total Price of the Trips - Practice Coding Problems

Minimize the Total Price of the Trips - Problem

Minimize the Total Price of Tree Trips

Imagine you're planning multiple trips through a network of connected cities arranged in a tree structure (no cycles, all cities connected). Each city has a toll price you must pay when passing through it.

You're given:
• An undirected tree with n nodes (cities) indexed from 0 to n-1
• edges array defining connections between cities
• price array where price[i] is the toll for city i
• trips array where each trip goes from start to end city

The Challenge: Before starting your trips, you can choose some non-adjacent cities and halve their toll prices. Your goal is to minimize the total cost of all trips.

Key Rules:
• The cost of a trip is the sum of tolls for all cities on the path
• You can only halve prices of cities that are not directly connected
• Each trip takes the unique path between start and end (since it's a tree)

Return the minimum possible total cost for all trips combined.

Input & Output

example_1.py — Basic Tree

$ Input: n = 4, edges = [[0,1],[1,2],[1,3]], price = [2,2,10,6], trips = [[0,3],[2,1],[2,3]]

› Output: 23

💡 Note: The tree looks like: 0-1-2 and 1-3. Trip paths are: (0,3): 0→1→3, (2,1): 2→1, (2,3): 2→1→3. Node frequencies: 0:1, 1:3, 2:2, 3:2. Optimal strategy: discount nodes 0 and 2 (non-adjacent). Cost = 1×(2/2) + 3×2 + 2×(10/2) + 2×6 = 1 + 6 + 10 + 12 = 29. Wait, let me recalculate... Actually, discount node 1: Cost = 1×2 + 3×(2/2) + 2×10 + 2×6 = 2 + 3 + 20 + 12 = 37. Best is discount nodes 0,2: 1×1 + 3×2 + 2×5 + 2×6 = 1+6+10+12 = 29. Actually optimal is 23.

example_2.py — Linear Tree

$ Input: n = 2, edges = [[0,1]], price = [2,2], trips = [[0,0]]

› Output: 1

💡 Note: Simple case: only one node (0) is used once. We can discount it, so cost = 1×(2/2) = 1.

example_3.py — Star Tree

$ Input: n = 3, edges = [[0,1],[0,2]], price = [1,10,10], trips = [[1,2],[1,2]]

› Output: 7

💡 Note: Star tree with center 0. Both trips go 1→0→2. Node frequencies: 0:2, 1:2, 2:2. Without discount: 2×1 + 2×10 + 2×10 = 42. Discount nodes 1,2 (non-adjacent): 2×1 + 2×5 + 2×5 = 22. Actually optimal is to discount center node 0: 2×0.5 + 2×10 + 2×10 = 41. Wait, that's worse. Let me recalculate properly... Discount 1,2: cost = 22. But can't discount adjacent nodes to 0 simultaneously. Best single discount is node 0: 22+1=23. Actually need to solve this systematically.

Constraints

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

Visualization

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Understanding the Visualization

Map the Highway Network

Build the tree structure from edges - this represents our toll booth network where every booth is connected but no cycles exist.

Count Traffic Volume

For each delivery route (trip), find the unique path and count how many times each toll booth will be used. High-traffic booths are prime candidates for discounts.

Dynamic Discount Strategy

Use tree DP to decide which booths to discount. Each booth can be in two states: discounted (50% off) or full price. Adjacent booths can't both offer discounts.

Optimal Cost Calculation

The DP ensures we maximize savings while respecting the adjacency constraint, giving us the minimum possible total cost for all trips.

Key Takeaway

🎯 Key Insight: This problem brilliantly combines tree traversal, frequency analysis, and dynamic programming. The 'adjacency constraint' transforms a simple optimization into a classic tree DP pattern, similar to the House Robber problem but with weighted nodes based on usage frequency.

Asked in

G Google 28 a Amazon 15 f Meta 12 Apple 8

The optimal solution uses Tree Dynamic Programming combined with frequency counting. First, count how many times each node is used across all trips by finding paths in the tree. Then, apply tree DP where each node has two states: discounted or not discounted. The DP ensures that adjacent nodes cannot both be discounted while maximizing total savings. Time complexity: O(m×n) where m is trips and n is nodes.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Combinations)	O(2^n * m * n)	O(n)	Generate all possible non-adjacent node combinations, calculate total cost for each
Tree DP with Frequency Count (Optimal)	O(m × n + n)	O(n)	Count node usage frequency, then use tree DP to optimally place discounts

Brute Force (Try All Combinations) — Algorithm Steps

Build adjacency list from edges
Generate all possible independent sets of nodes
For each independent set, apply 50% discount to those nodes
Calculate total cost of all trips with current discount configuration
Track and return the minimum total cost found

Visualization

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Step-by-Step Walkthrough

Generate Subsets

Create all possible node combinations using bit masking

Check Independence

Verify no two selected nodes are adjacent in the tree

Calculate Costs

For each valid combination, compute total trip cost with discounts

Find Minimum

Track and return the lowest total cost found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <stdbool.h>

typedef struct {
    int* nodes;
    int size;
    int capacity;
} IntList;

IntList* createList() {
    IntList* list = malloc(sizeof(IntList));
    list->capacity = 10;
    list->nodes = malloc(list->capacity * sizeof(int));
    list->size = 0;
    return list;
}

void addToList(IntList* list, int val) {
    if (list->size == list->capacity) {
        list->capacity *= 2;
        list->nodes = realloc(list->nodes, list->capacity * sizeof(int));
    }
    list->nodes[list->size++] = val;
}

int minimumTotalPrice(int n, int** edges, int edgesSize, int* edgesColSize, int* price, int priceSize, int** trips, int tripsSize, int* tripsColSize) {
    // Build adjacency list
    IntList* adj = malloc(n * sizeof(IntList));
    for (int i = 0; i < n; i++) {
        adj[i] = *createList();
    }
    
    for (int i = 0; i < edgesSize; i++) {
        addToList(&adj[edges[i][0]], edges[i][1]);
        addToList(&adj[edges[i][1]], edges[i][0]);
    }
    
    // Helper function to find path (simplified for brevity)
    int minCost = INT_MAX;
    
    // Try all possible subsets (simplified implementation)
    for (int mask = 0; mask < (1 << n); mask++) {
        // Check if current mask represents valid independent set
        bool valid = true;
        for (int i = 0; i < n && valid; i++) {
            if (mask & (1 << i)) {
                for (int j = 0; j < adj[i].size; j++) {
                    if (mask & (1 << adj[i].nodes[j])) {
                        valid = false;
                        break;
                    }
                }
            }
        }
        
        if (valid) {
            // Calculate cost for this configuration
            int totalCost = 0;
            // Implementation simplified - would need full path finding
            if (totalCost < minCost) {
                minCost = totalCost;
            }
        }
    }
    
    // Cleanup
    for (int i = 0; i < n; i++) {
        free(adj[i].nodes);
    }
    free(adj);
    
    return minCost;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n * m * n)

2^n possible subsets, m trips, n nodes per path on average

⚠ Quadratic Growth

Space Complexity

O(n)

Space for recursion stack and adjacency list

⚡ Linearithmic Space

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Medium Frequency

~35 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Combinations) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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