Maximum Weighted K-Edge Path

Maximum Weighted K-Edge Path - Problem

Maximum Weighted K-Edge Path in DAG

You're an explorer navigating through a Directed Acyclic Graph (DAG) representing a network of cities connected by weighted roads. Your mission is to find the most valuable route that satisfies specific constraints.

Given:

A DAG with n nodes (cities) labeled from 0 to n-1
An array edges where edges[i] = [u_i, v_i, w_i] represents a directed road from city u_i to city v_i with value w_i
Two integers: k (exact number of roads to travel) and t (strict upper bound on total value)

Your Goal: Find the maximum possible sum of edge weights for any path that:

Contains exactly k edges
Has total weight strictly less than t

Return the maximum sum, or -1 if no valid path exists.

Example: If you need exactly 3 roads with total value < 100, and you find paths with values [85, 92, 78], return 92 as it's the highest valid sum.

Input & Output

example_1.py — Basic Path Finding

$ Input: n = 4, edges = [[0,1,5],[1,2,3],[2,3,2]], k = 2, t = 10

› Output: 8

💡 Note: The path 0→1→2 uses exactly 2 edges with total weight 5+3=8, which is less than t=10. This is the maximum possible sum meeting both constraints.

example_2.py — No Valid Path

$ Input: n = 3, edges = [[0,1,10],[1,2,15]], k = 2, t = 20

› Output: -1

💡 Note: The only 2-edge path 0→1→2 has weight 10+15=25, which is not less than t=20. No valid path exists.

example_3.py — Multiple Valid Paths

$ Input: n = 4, edges = [[0,1,3],[0,2,7],[1,3,4],[2,3,1]], k = 2, t = 10

› Output: 8

💡 Note: Two valid paths exist: 0→1→3 (weight=7) and 0→2→3 (weight=8). Return 8 as it's the maximum.

Constraints

2 ≤ n ≤ 1000 (number of nodes)
0 ≤ edges.length ≤ 2000 (number of edges)
edges[i].length == 3 (each edge has source, destination, weight)
0 ≤ u_i, v_i ≤ n - 1 (valid node indices)
1 ≤ w_i, t ≤ 10⁴ (positive weights and threshold)
1 ≤ k ≤ 1000 (path length constraint)
The graph is a Directed Acyclic Graph (DAG)

Visualization

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Asked in

G Google 45 a Amazon 38 f Meta 29 ⊞ Microsoft 22

The optimal approach uses Dynamic Programming with a 2D table where dp[node][i] stores the maximum weight sum achievable when reaching a node using exactly i edges. Build the solution incrementally by extending paths layer by layer, ensuring all sums stay strictly below threshold t. Time complexity: O(E × k), Space: O(n × k).

Common Approaches

✓ Dynamic Programming with Memoization

⏱️ Time: O(E * k) Space: O(n * k)

Create a 2D DP table where dp[node][edges] represents the maximum weight sum achievable when reaching 'node' using exactly 'edges' number of edges. Build the solution bottom-up by extending paths incrementally.

Dynamic Programming with Memoization — Algorithm Steps

Initialize DP table: dp[node][0] = 0 for all starting nodes
For each edge count from 1 to k:
For each node, try extending from all incoming edges
Update dp[node][edges] with maximum valid extension
Return maximum value from dp[node][k] where sum < t

Visualization

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Step-by-Step Walkthrough

Initialize

Set dp[node][0] = 0 for all nodes (can start anywhere)

Layer 1

Fill dp[node][1] by extending 0-edge paths with single edges

Layer i

Fill dp[node][i] by extending (i-1)-edge paths

Result

Find maximum value in dp[node][k] column where sum < t

Code -

solution.c — C

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

typedef struct {
    int from, weight;
} InEdge;

typedef struct {
    InEdge* edges;
    int size, capacity;
} InAdjList;

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int maxWeightedPath(int n, int edges[][3], int edgeCount, int k, int t) {
    // Build reverse adjacency list (incoming edges)
    InAdjList* incoming = malloc(n * sizeof(InAdjList));
    for (int i = 0; i < n; i++) {
        incoming[i].edges = malloc(10 * sizeof(InEdge));
        incoming[i].size = 0;
        incoming[i].capacity = 10;
    }
    
    for (int i = 0; i < edgeCount; i++) {
        int to = edges[i][1];
        incoming[to].edges[incoming[to].size++] = (InEdge){edges[i][0], edges[i][2]};
    }
    
    // dp[node][i] = maximum weight using exactly i edges to reach node
    int** dp = malloc(n * sizeof(int*));
    for (int i = 0; i < n; i++) {
        dp[i] = malloc((k + 1) * sizeof(int));
        for (int j = 0; j <= k; j++) {
            dp[i][j] = INT_MIN;
        }
    }
    
    // Base case: can start at any node with 0 edges (weight 0)
    for (int i = 0; i < n; i++) {
        dp[i][0] = 0;
    }
    
    // Fill DP table
    for (int edgeCount = 1; edgeCount <= k; edgeCount++) {
        for (int node = 0; node < n; node++) {
            // Try extending from each incoming edge
            for (int i = 0; i < incoming[node].size; i++) {
                InEdge inEdge = incoming[node].edges[i];
                int prevNode = inEdge.from;
                int weight = inEdge.weight;
                
                if (dp[prevNode][edgeCount - 1] != INT_MIN) {
                    int newWeight = dp[prevNode][edgeCount - 1] + weight;
                    if (dp[node][edgeCount] == INT_MIN || newWeight > dp[node][edgeCount]) {
                        dp[node][edgeCount] = newWeight;
                    }
                }
            }
        }
    }
    
    // Find maximum valid result
    int maxSum = -1;
    for (int node = 0; node < n; node++) {
        if (dp[node][k] != INT_MIN && dp[node][k] < t) {
            if (maxSum == -1 || dp[node][k] > maxSum) {
                maxSum = dp[node][k];
            }
        }
    }
    
    // Cleanup
    for (int i = 0; i < n; i++) {
        free(dp[i]);
        free(incoming[i].edges);
    }
    free(dp);
    free(incoming);
    
    return maxSum;
}

int main() {
    int edges[][3] = {{0,1,5},{1,2,3},{2,3,2}};
    printf("%d\n", maxWeightedPath(4, edges, 3, 2, 10));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(E * k)

For each of k edge counts, we process each edge once, where E is the number of edges

✓ Linear Growth

Space Complexity

O(n * k)

DP table stores maximum weight for each node at each edge count

⚡ Linearithmic Space

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Maximum Weighted K-Edge Path in DAG

Input & Output

Constraints

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Related Problems

Common Approaches

Dynamic Programming with Memoization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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