This is basically a bounded knapsack variant and I knew that going in, but the 'distinct sums' framing threw me for a second because I kept second-guessing whether I needed to track combinations or just reachability.
Model the problem as a bounded knapsack reachability problem, using a boolean array to track achievable sums. Optimize with binary splitting of coin counts or a monotonic queue to handle up to 1 million sums efficiently. Discuss time and space complexity, and address large sum handling with bitsets or memory-conscious techniques.
Pro tip: Mention that Visa cares about scalability and real-world constraints—highlight how your solution avoids O(maxSum * totalCoins) by using binary splitting or monotonic queues, and note that bitset operations can give a 64x speedup in practice.
Restate the problem: given denominations and counts, count distinct positive sums achievable using at most the given number of each coin. Confirm that coins are indistinguishable and order doesn't matter.
Use a boolean DP array where dp[s] indicates if sum s is achievable. Initialize dp[0]=true and iterate through each denomination, updating reachable sums within the allowed count.
Apply binary splitting to convert each coin count into powers of two, turning the problem into 0/1 knapsack. Alternatively, use a monotonic queue to process each denomination in O(maxSum) time.
State time complexity: O(maxSum * sum(log(count_i))) with binary splitting, or O(maxSum * numDenominations) with monotonic queue. Space is O(maxSum). For maxSum ~1e6, mention bitset optimization for speed and memory.
Compare binary splitting vs. monotonic queue: binary splitting is simpler but may be slower for large counts; monotonic queue is optimal but complex. Handle edge cases like zero counts, duplicate denominations, and sums exceeding 1e6.
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