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What is the time complexity of the backtracking solution with frequency counting for the Combination Sum (Reuse Allowed) problem, where N is the number of candidates, T is the target, and M is the minimum candidate value?

medium🪤 Complexity Trap Q13 of Q15
Subsets & Combinations - Combination Sum (Reuse Allowed)
What is the time complexity of the backtracking solution with frequency counting for the Combination Sum (Reuse Allowed) problem, where N is the number of candidates, T is the target, and M is the minimum candidate value?
AO(N * T) because each candidate is tried once for each target value.
BO(2^N) because each candidate can be either included or excluded once.
CO(N^(T/M + 1)) because the recursion tree branches exponentially with depth proportional to T/M.
DO(T^N) because for each target value, all candidates are tried in all combinations.
Step-by-Step Solution

  1. Step 1: Identify recursion depth and branching factor

    The recursion depth is roughly T/M because the smallest candidate can be used up to T/M times. At each level, up to N candidates are considered with multiple counts.

  2. Step 2: Calculate total number of recursive calls

    Each candidate can be chosen 0 to max_use times, leading to branching factor roughly N^(T/M + 1). This exponential complexity dominates.

  3. Final Answer:

    Option C -> Option C

  4. Quick Check:

    Exponential branching with depth proportional to T/M matches O(N^(T/M + 1)) because the recursion tree branches exponentially with depth proportional to T/M. [OK]
Quick Trick: Exponential in target divided by smallest candidate [OK]
Common Mistakes:
MISTAKES
  • Confusing DP complexity with backtracking
  • Ignoring exponential branching due to reuse
Trap Explanation:
PITFALL
  • Option A looks like DP complexity but backtracking explores many combinations, causing exponential growth.
Interviewer Note:
CONTEXT
  • Tests understanding of recursion tree size and why complexity is exponential, not polynomial.
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