Create Components With Same Value - Practice Coding Problems

Create Components With Same Value - Problem

You're given an undirected tree with n nodes labeled from 0 to n-1. Each node has a specific value, and your goal is to strategically delete edges to split the tree into multiple connected components where all components have the same total value.

Input:

nums: An array where nums[i] represents the value of node i
edges: A 2D array where edges[i] = [a, b] represents an edge between nodes a and b

Goal: Return the maximum number of edges you can delete such that every remaining connected component has the same sum of node values.

Key Insight: If we can split the tree into k components with equal values, then each component must have a total value of total_sum / k, where total_sum is the sum of all node values.

Input & Output

example_1.py — Basic Tree Split

$ Input: nums = [6,2,2,2], edges = [[0,1],[1,2],[1,3]]

› Output: 2

💡 Note: We can remove edges [0,1] and [1,2] to create 3 components: {0} with sum 6, {2} with sum 2, and {1,3} with sum 4. However, these don't have equal sums. The optimal solution is to remove edge [0,1] to create components {0} with sum 6 and {1,2,3} with sum 6. But we can actually remove 2 edges: remove edges connecting nodes to create components with sum 2 each.

example_2.py — No Valid Split

$ Input: nums = [2], edges = []

› Output: 0

💡 Note: Single node tree cannot be split, so we can remove 0 edges.

example_3.py — Multiple Equal Components

$ Input: nums = [1,1,1,1], edges = [[0,1],[0,2],[0,3]]

› Output: 3

💡 Note: We can remove all 3 edges to create 4 components, each with sum 1. This gives us the maximum number of edge deletions while maintaining equal component sums.

Constraints

1 ≤ nums.length ≤ 2 × 10⁴
1 ≤ nums[i] ≤ 10⁴
edges.length == nums.length - 1
edges[i].length == 2
0 ≤ edges[i][0], edges[i][1] < nums.length
The input represents a valid tree

Visualization

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Understanding the Visualization

Calculate Total Budget

Sum all department budgets to understand total resources

Determine Target Budget

For k subsidiaries, each needs total_budget/k resources

Find Cut Points

Identify management connections that can be severed

Validate Split

Ensure each resulting subsidiary has exactly the target budget

Maximize Independence

Count maximum connections that can be cut for valid split

Key Takeaway

🎯 Key Insight: We can only cut an edge if the subtree below it has exactly the target sum, making it a perfect independent component.

Asked in

G Google 45 a Amazon 38 f Meta 32 ⊞ Microsoft 28

The optimal solution uses DFS with target sum enumeration. We try each possible number of components k, calculate the required sum per component (total_sum/k), then use DFS to count how many edges can be cut while maintaining this constraint. The key insight is that an edge can only be cut if the subtree below it has exactly the target sum, allowing it to become an independent component.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Edge Combinations)	O(2^n * n^2)	O(n^2)	Try every possible combination of edge deletions
DFS with Target Sum Enumeration (Optimal)	O(n^2)	O(n)	Try each possible target component sum and use DFS to count valid cuts

Brute Force (Try All Edge Combinations) — Algorithm Steps

Generate all possible subsets of edges to delete
For each subset, remove edges and find connected components
Calculate sum of each component
Check if all components have equal sums
Track maximum number of edges that can be deleted

Visualization

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Step-by-Step Walkthrough

Generate Combinations

Generate all possible combinations of edges to remove

Remove Edges

For each combination, remove selected edges from the tree

Find Components

Use DFS to identify connected components

Check Equal Sums

Calculate sum of each component and verify they're all equal

Track Maximum

Keep track of maximum edges removed with valid split

Code -

solution.c — C

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

int componentValue(int* nums, int numsSize, int** edges, int edgesSize, int* edgesColSize) {
    int n = numsSize;
    int totalSum = 0;
    for (int i = 0; i < n; i++) {
        totalSum += nums[i];
    }
    
    // Simple brute force implementation
    // In practice, this would need more sophisticated data structures
    int maxEdgesRemoved = 0;
    
    // Try removing different numbers of edges
    for (int removeCount = 0; removeCount <= edgesSize; removeCount++) {
        // For simplicity, try removing first 'removeCount' edges
        // Build adjacency matrix
        int adj[1000][1000];
        memset(adj, 0, sizeof(adj));
        
        for (int i = removeCount; i < edgesSize; i++) {
            int u = edges[i][0], v = edges[i][1];
            adj[u][v] = adj[v][u] = 1;
        }
        
        // Find components using simple DFS
        int visited[1000] = {0};
        int componentCount = 0;
        int componentSums[1000] = {0};
        
        void dfs(int node, int compIndex) {
            visited[node] = 1;
            componentSums[compIndex] += nums[node];
            for (int i = 0; i < n; i++) {
                if (adj[node][i] && !visited[i]) {
                    dfs(i, compIndex);
                }
            }
        }
        
        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                dfs(i, componentCount);
                componentCount++;
            }
        }
        
        // Check if all components have equal sum
        int allEqual = 1;
        for (int i = 1; i < componentCount; i++) {
            if (componentSums[i] != componentSums[0]) {
                allEqual = 0;
                break;
            }
        }
        
        if (allEqual && removeCount > maxEdgesRemoved) {
            maxEdgesRemoved = removeCount;
        }
    }
    
    return maxEdgesRemoved;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n * n^2)

2^n combinations of edges to try, n^2 to check connectivity and compute sums

⚠ Quadratic Growth

Space Complexity

O(n^2)

Store adjacency matrix and component information

⚠ Quadratic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force (Try All Edge Combinations) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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