Create Components With Same Value

Create Components With Same Value

What is this problem about?

The Create Components With Same Value interview question is a "Hard" tree problem. You are given a tree where each node has a value. You can delete any number of edges to split the tree into several connected components. The goal is to maximize the number of edges deleted such that all resulting components have the same sum of node values.

Why is this asked in interviews?

Sprinklr and other high-growth startups use this tree interview pattern to test a candidate's mastery of graph theory and number theory. It evaluates whether you can identify that the "target component sum" must be a divisor of the total sum of all node values. It tests optimization skills and the ability to implement recursive checks efficiently.

Algorithmic pattern used

The problem uses Divisor Enumeration and DFS.

1. Calculate the totalSum of all node values.
2. Find all divisors of totalSum.
3. For each divisor d (starting from the largest number of components, which means the smallest d):
   • Perform a DFS to see if the tree can be partitioned into components each with sum d.
   • In DFS, a node returns its own value plus the sum of its subtrees. If a subtree's sum reaches exactly d, "cut" the edge (return 0 to the parent) and increment the component count.
4. The first divisor that successfully partitions the tree gives the maximum edges deleted (count - 1).

Example explanation

Tree: 1 (val 2) -- 2 (val 2) -- 3 (val 2). totalSum = 6. Divisors of 6: 1, 2, 3, 6.

1. Try d = 2:
   • Node 3 has sum 2. Cut! Return 0 to 2.
   • Node 2 has sum 2. Cut! Return 0 to 1.
   • Node 1 has sum 2.
   • 3 components found. Edges deleted = 2. (Valid!)
2. Try d = 3:
   • Node 3 returns 2.
   • Node 2 returns 2 + 2 = 4. (Exceeds d without reaching it). Invalid.

Common mistakes candidates make

• Trying all edge combinations: There are $2^{(n-1)}$ ways to delete edges, which is exponential. Using divisors reduces this to O(Divisors * N).
• Wrong DFS return: Not correctly handling the "carry-over" sum from a node to its parent when a component is not yet full.
• Ignoring the total sum: Failing to realize that the target sum must be a factor of the total.

Interview preparation tip

Master the "Post-order DFS for tree partitioning." It's a common technique for problems where you need to make local decisions (like cutting an edge) that satisfy a global constraint.