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Which modification to the space-optimized DP code below correctly counts the number of such subsets?

hard🎤 Interviewer Follow-up Q15 of Q15
Subsets & Combinations - Count of Subsets With Sum K
Suppose the problem changes: now you can use each element in the array unlimited times to form subsets summing to K (i.e., unlimited repetitions allowed). Which modification to the space-optimized DP code below correctly counts the number of such subsets?
def count_subsets_unlimited(arr, K):
    dp = [0] * (K + 1)
    dp[0] = 1
    for num in arr:
        for j in range(???):
            dp[j] += dp[j - num]
    return dp[K]
Choose the correct range for the inner loop to handle unlimited usage of elements.
Afor j in range(K, num - 1, -1): # iterate backward
Bfor j in range(K + 1): # iterate forward including invalid indices
Cfor j in range(0, K + 1): # iterate forward from 0
Dfor j in range(num, K + 1): # iterate forward
Step-by-Step Solution
Solution:
  1. Step 1: Understand unlimited usage impact

    When elements can be reused unlimited times, the inner loop must iterate forward to allow multiple counts of the same element.
  2. Step 2: Choose correct iteration range

    Iterating forward from num to K ensures dp[j] accumulates counts including repeated usage of num.
  3. Final Answer:

    Option D -> Option D
  4. Quick Check:

    Forward iteration enables unlimited reuse of elements in subset sums [OK]
Quick Trick: Unlimited usage requires forward iteration to accumulate counts correctly [OK]
Common Mistakes:
MISTAKES
  • Using backward iteration which prevents reuse
  • Starting loop from 0 causing invalid indices
Trap Explanation:
PITFALL
  • Backward iteration is correct for single usage but breaks unlimited usage by not allowing repeated additions.
Interviewer Note:
CONTEXT
  • Tests candidate's understanding of DP iteration order changes for unlimited item usage variants.
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