I know the algorithm to solve the coin change problem for infinite number of denominations but is there any algorithm for finite number of denominations using DP?
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I know the algorithm to solve the coin change problem for infinite number of denominations but is there any algorithm for finite number of denominations using DP?
Yes. Modify the initial algorithm such that, when it’s about to add a coin that would exceed the number of available coins of that denomination, it doesn’t, instead. Then it will only print the valid combos.
Another, more simple way is: run the algorithm without bounds, then filter the output based on what combinations are invalid. Thinking of it this way makes it really obvious that the problem is indeed solvable.