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.

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.

Time & Space Complexity

Time Complexity

⏱️

O(n^2)

For each divisor of total sum (at most n), we perform DFS traversal of O(n)

⚠ Quadratic Growth

Space Complexity

O(n)

Recursion stack for DFS and adjacency list storage

⚡ Linearithmic Space

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

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
✓ 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)	O(2^n * n^2)	O(n^2)	Try every possible combination of edge deletions

DFS with Target Sum Enumeration (Optimal) — Algorithm Steps

Calculate total sum of all node values
Try each possible number of components k from 2 to n
For each k, calculate target sum per component (total/k)
Use DFS to traverse tree and count valid edge cuts
An edge can be cut if subtree sum equals target sum
Return maximum cuts found across all valid k values

Visualization

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Step-by-Step Walkthrough

Calculate Total Sum

Sum all node values to get total = 12

Try Component Counts

Try k=2 (target=6), k=3 (target=4), k=4 (target=3)

DFS Traversal

For each target, use DFS to find valid cuts

Cut Decision

Cut edge if subtree sum equals target

Return Maximum

Return maximum cuts found across all valid targets

Code -

solution.c — C

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

typedef struct {
    int sum;
    int cuts;
} DFSResult;

int adj[1000][1000];
int adjSize[1000];
int* nums;

DFSResult dfs(int node, int parent, int target, int n) {
    int currentSum = nums[node];
    int edgesCut = 0;
    
    for (int i = 0; i < adjSize[node]; i++) {
        int child = adj[node][i];
        if (child == parent) continue;
        
        DFSResult childResult = dfs(child, node, target, n);
        
        if (childResult.sum == -1) {
            return (DFSResult){-1, -1}; // Impossible
        } else if (childResult.sum == 0) {
            edgesCut += childResult.cuts + 1;
        } else if (childResult.sum == target) {
            edgesCut += childResult.cuts + 1;
        } else {
            currentSum += childResult.sum;
            edgesCut += childResult.cuts;
        }
    }
    
    if (currentSum == target) {
        return (DFSResult){0, edgesCut};
    } else if (currentSum < target) {
        return (DFSResult){currentSum, edgesCut};
    } else {
        return (DFSResult){-1, -1}; // Impossible
    }
}

int componentValue(int* numsInput, int numsSize, int** edges, int edgesSize, int* edgesColSize) {
    int n = numsSize;
    if (n == 1) return 0;
    
    nums = numsInput;
    
    int totalSum = 0;
    for (int i = 0; i < n; i++) {
        totalSum += nums[i];
    }
    
    // Initialize adjacency list
    memset(adjSize, 0, sizeof(adjSize));
    
    // Build adjacency list
    for (int i = 0; i < edgesSize; i++) {
        int u = edges[i][0], v = edges[i][1];
        adj[u][adjSize[u]++] = v;
        adj[v][adjSize[v]++] = u;
    }
    
    int maxCuts = 0;
    
    // Try each possible number of components
    for (int k = 2; k <= n; k++) {
        if (totalSum % k != 0) continue;
        
        int target = totalSum / k;
        DFSResult result = dfs(0, -1, target, n);
        
        if (result.sum == 0 || result.sum == target) {
            if (result.cuts > maxCuts) {
                maxCuts = result.cuts;
            }
        }
    }
    
    return maxCuts;
}

Time & Space Complexity

Time Complexity

⏱️

O(n^2)

For each divisor of total sum (at most n), we perform DFS traversal of O(n)

⚠ Quadratic Growth

Space Complexity

O(n)

Recursion stack for DFS and adjacency list storage

⚡ Linearithmic Space

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

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Input & Output

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Related Problems

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Common Approaches

DFS with Target Sum Enumeration (Optimal) — Algorithm Steps

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Time & Space Complexity

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