Number of Ways to Assign Edge Weights II - Problem

Array Math Dynamic Programming Bit Manipulation Tree Depth-First Search Hard

There is an undirected tree with n nodes labeled from 1 to n, rooted at node 1. The tree is represented by a 2D integer array edges of length n - 1, where edges[i] = [u_i, v_i] indicates that there is an edge between nodes u_i and v_i.

Initially, all edges have a weight of 0. You must assign each edge a weight of either 1 or 2.

The cost of a path between any two nodes u and v is the total weight of all edges in the path connecting them.

You are given a 2D integer array queries. For each queries[i] = [u_i, v_i], determine the number of ways to assign weights to edges in the path such that the cost of the path between u_i and v_i is odd.

Return an array answer, where answer[i] is the number of valid assignments for queries[i]. Since the answer may be large, apply modulo 10^9 + 7 to each answer[i].

Note: For each query, disregard all edges not in the path between node u_i and v_i.

Input & Output

Example 1 — Basic Tree Path

$ Input: edges = [[1,2],[2,3],[3,4]], queries = [[1,4],[2,3]]

› Output: [4,1]

💡 Note: For query [1,4]: path 1→2→3→4 has 3 edges, so 2^(3-1) = 4 ways to make sum odd. For query [2,3]: path 2→3 has 1 edge, so 2^(1-1) = 1 way to make sum odd.

Example 2 — Adjacent Nodes

$ Input: edges = [[1,2],[2,3]], queries = [[1,2],[2,3],[1,3]]

› Output: [1,1,2]

💡 Note: Query [1,2]: 1 edge, 1 way to make odd (weight=1). Query [2,3]: 1 edge, 1 way to make odd. Query [1,3]: path 1→2→3 has 2 edges, 2^(2-1) = 2 ways to make sum odd.

Example 3 — Same Node Query

$ Input: edges = [[1,2]], queries = [[1,1],[2,2]]

› Output: [0,0]

💡 Note: Same node queries have 0 edges in path, cannot make sum odd (sum is 0), so 0 ways.

Constraints

1 ≤ n ≤ 10⁴
edges.length == n - 1
1 ≤ u_i, v_i ≤ n
1 ≤ queries.length ≤ 10⁴
queries[i].length == 2

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that for a path with k edges, there are exactly 2^(k-1) ways to assign weights (1 or 2) such that the total sum is odd. This is because we need an odd number of edges with weight 1. Best approach uses tree preprocessing with parent pointers for efficient path length queries. Time: O(N + Q), Space: O(N)

Common Approaches

✓ Brute Force Path Finding

⏱️ Time: O(Q × N) Space: O(N)

For each query, use DFS to find the unique path between two nodes in the tree. Then calculate how many ways to assign weights 1 or 2 to make the sum odd.

Optimized with LCA Preprocessing

⏱️ Time: O(N + Q × log N) Space: O(N)

Build the tree structure once with parent pointers and node depths. For each query, find the path length using lowest common ancestor (LCA) approach without full DFS traversal.

Brute Force Path Finding — Algorithm Steps

Step 1: Build adjacency list from edges
Step 2: For each query, use DFS to find path between u and v
Step 3: Count edges in path, calculate 2^(edges-1) mod (10^9+7)

Visualization

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Step-by-Step Walkthrough

Build Graph

Create adjacency list from edges

Find Path

Use DFS to find unique path between query nodes

Calculate Ways

Count edges in path, compute 2^(k-1) for odd sums

Code -

solution.c — C

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

#define MOD 1000000007
#define MAX_N 10005

static int adj[MAX_N][100];
static int adjCount[MAX_N];
static int visited[MAX_N];
static int path[MAX_N];
static int pathSize;

long long powerMod(long long base, long long exp, long long mod) {
    long long result = 1;
    base = base % mod;
    while (exp > 0) {
        if (exp % 2 == 1) {
            result = (result * base) % mod;
        }
        exp = exp >> 1;
        base = (base * base) % mod;
    }
    return result;
}

int findPath(int start, int end) {
    if (start == end) {
        return 1;
    }
    visited[start] = 1;
    
    for (int i = 0; i < adjCount[start]; i++) {
        int neighbor = adj[start][i];
        if (!visited[neighbor]) {
            path[pathSize++] = neighbor;
            if (findPath(neighbor, end)) {
                return 1;
            }
            pathSize--;
        }
    }
    return 0;
}

int* solution(int** edges, int edgesSize, int** queries, int queriesSize, int* returnSize) {
    *returnSize = queriesSize;
    int* result = (int*)malloc(queriesSize * sizeof(int));
    
    int n = edgesSize + 1;
    
    // Build adjacency list
    memset(adjCount, 0, sizeof(adjCount));
    
    for (int i = 0; i < edgesSize; i++) {
        int u = edges[i][0];
        int v = edges[i][1];
        adj[u][adjCount[u]++] = v;
        adj[v][adjCount[v]++] = u;
    }
    
    for (int q = 0; q < queriesSize; q++) {
        int u = queries[q][0];
        int v = queries[q][1];
        
        if (u == v) {
            result[q] = 0;
            continue;
        }
        
        memset(visited, 0, sizeof(visited));
        pathSize = 1;
        path[0] = u;
        
        findPath(u, v);
        
        int edgeCount = pathSize - 1;
        if (edgeCount == 0) {
            result[q] = 0;
        } else {
            // For k edges, 2^(k-1) ways to make sum odd
            long long ways = powerMod(2, edgeCount - 1, MOD);
            result[q] = (int)ways;
        }
    }
    
    return result;
}

void skipWhitespace(char** ptr) {
    while (**ptr && isspace(**ptr)) (*ptr)++;
}

int parseInt(char** ptr) {
    skipWhitespace(ptr);
    int sign = 1;
    if (**ptr == '-') {
        sign = -1;
        (*ptr)++;
    }
    int num = 0;
    while (**ptr && isdigit(**ptr)) {
        num = num * 10 + (**ptr - '0');
        (*ptr)++;
    }
    return num * sign;
}

int** parse2DArray(char* str, int* rows, int* cols) {
    int capacity = 100;
    int** arr = (int**)malloc(capacity * sizeof(int*));
    *rows = 0;
    *cols = 2;
    
    char* ptr = str;
    skipWhitespace(&ptr);
    
    if (*ptr == '[') ptr++;
    
    while (*ptr != '\0' && *ptr != ']') {
        skipWhitespace(&ptr);
        if (*ptr == '[') {
            ptr++;
            
            if (*rows >= capacity) {
                capacity *= 2;
                arr = (int**)realloc(arr, capacity * sizeof(int*));
            }
            
            arr[*rows] = (int*)malloc(2 * sizeof(int));
            arr[*rows][0] = parseInt(&ptr);
            
            skipWhitespace(&ptr);
            if (*ptr == ',') ptr++;
            
            arr[*rows][1] = parseInt(&ptr);
            (*rows)++;
            
            skipWhitespace(&ptr);
            if (*ptr == ']') ptr++;
        }
        
        skipWhitespace(&ptr);
        if (*ptr == ',') ptr++;
    }
    
    return arr;
}

int main() {
    char line[100000];
    
    if (fgets(line, sizeof(line), stdin) == NULL) return 0;
    int edgesRows, edgesCols;
    int** edges = parse2DArray(line, &edgesRows, &edgesCols);
    
    if (fgets(line, sizeof(line), stdin) == NULL) {
        for (int i = 0; i < edgesRows; i++) free(edges[i]);
        free(edges);
        return 0;
    }
    int queriesRows, queriesCols;
    int** queries = parse2DArray(line, &queriesRows, &queriesCols);
    
    int returnSize;
    int* result = solution(edges, edgesRows, queries, queriesRows, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    for (int i = 0; i < edgesRows; i++) free(edges[i]);
    free(edges);
    for (int i = 0; i < queriesRows; i++) free(queries[i]);
    free(queries);
    free(result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(Q × N)

For Q queries, each DFS traversal can visit up to N nodes

✓ Linear Growth

Space Complexity

O(N)

Adjacency list stores N nodes and N-1 edges, plus DFS recursion stack

✓ Linear Space

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Number of Ways to Assign Edge Weights II - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Path Finding — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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