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# knapsack problem

## How can the running time of the knapsack problem be reduced from O(N*M

27 mars 2023 à 13:45:06

I'm presently working on an algorithm assignment for the knapsack issue, and the question is:
The size of the knapsack is M, and there are N items, each with an integer value and an integer weight; we must determine the ideal maximum weight that we can place inside the knapsack, just like in this case.
I'm solving this problem with dynamic programming and saving the results just for the current and previous loops, which consumes M2 space. Although this saves space, the time complexity is still O(MN) because there are two for loops; is there any way to improve this?
Thanks
12 juillet 2023 à 9:02:21

Yes, there are ways to improve the time complexity of the dynamic programming solution to the knapsack problem. One approach is to use a 1D array to store the results instead of a 2D array, which can reduce the space complexity to O(M). This is possible because we only need to know the results from the previous iteration of the loop to calculate the results for the current iteration. For exmple, in bloxd io, you could modify your dynamic programming solution to use a 1D array:

• Initialize an array dp of size M+1 with all elements set to 0.
• For each item i from 1 to N, do the following:
• For each weight j from M down to the weight of item i, do the following:
• If j is greater than or equal to the weight of item i, update dp[j] to be the maximum of dp[j] and dp[j - weight[i]] + value[i].
• Return the last element of the dp array.

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Edité par KeithMoody 12 juillet 2023 à 9:04:13

15 mars 2024 à 8:26:10

To further optimize the time complexity, you can consider using the "Space Optimization" technique. Instead of saving results for the current and previous loops, you can use a single array of size M to store the intermediate results. This way, the space complexity will be reduced to O(M), but the time complexity will remain the same (O(MN)). This technique can be achieved by iteratively updating the array in a backward manner, starting from the maximum weight down to the minimum weight.