Unable to find what's wrong with my code for solve spoj CWC15
algorithm,dynamic-programming,memoization,subset-sum
Suggestion: The way you are solving will not work because of complexity. Although space complexity will work (limit is 1536MB and space used is around 785MB), but time complexity is too high for time limit of 5s. An estimate is 10^8 could be executed safely within 1 sec. If you...
Calculating Minimal Subset With Given Sum
algorithm,scala,subset,subset-sum
Making a full list of sums is unnecessary, and using BigInt is wasteful regardless--a Long can hold up to 9.2e18, and you are promised that only 1e15 is going to appear. (In fact, I think 1e15 was chosen so that even a Double can hold the answer without loss of...
Is my recurrence relation right for subset sum?
dynamic-programming,recurrence-relation,subset-sum
You are almost correct ( not sure why you used min ). But let dp[i][j] store the answer of whether a subset of arr[0],arr[1],....arr[j] (here arr[] is the array of elements ) can sum upto i. That is dp[i][j] is 1 if answer is yes and 0 if answer is...
Finding subset sum recursively producing incorrect output
c,algorithm,recursion,subset-sum
#include <stdio.h> #define SIZE(a) sizeof(a)/sizeof(a[0]) int binary[100]; void show(int* p, int size) { int j; for (j = 0; j < size; j++) printf("%d\n", p[j]); } void foo(int target, int i, int sum, int *a, int size) { if (sum == target) { // solution found show(binary, size); } else...
Partitioning an array into 3 sets
arrays,algorithm,divide-and-conquer,subset-sum,sub-arrays
here is brute-force java solution you can use, note - complexity of this solution is O(3^N), which is very very slow /** * Returns absolute difference between 3 values in array */ static int getdiff(final int s[]) { return Math.abs(s[0] - s[1]) + Math.abs(s[1] - s[2]) + Math.abs(s[2] - s[0]);...
Find all k-size subsets with sum s of an n-size bag of duplicate unsorted positive integers
c#,algorithm,.net-2.0,dynamic-programming,subset-sum
My previous answer works on the principle of cutting off the number of combinations to check. But this can be improved significantly once you sort the array. The principle is similar, but since the solution is entirely different, I've decided to put it in a separate answer. I was careful...
algorithm to get subset of array with target sum is not giving working
java,arrays,algorithm,subset,subset-sum
Your solution is wrong because it's a greedy approach. It decides if you should add a number or not based on the fact that adding it does not violate the sum, at the moment. However, this greedy approach does not work, with a simple example of the following array: [1,9,6,5]...
Decreasing the run time for my recursive subset sum algorthim
java,algorithm,recursion,big-o,subset-sum
Some minor optimizations if I understand this correctly: public static void recDfs(int[] arr, int k, int sum) { if (sum < 0) return; if (sum == 0) { counter++; return; } if (k == 0) { return; } recDfs(arr, k - 1, sum - arr[k - 1]); recDfs(arr, k -...
Efficient method to filter and add based on certain conditions (3 conditions in this case)
r,data.table,plyr,dplyr,subset-sum
You may try aggregate aggregate(d ~ a + b + c, data = df, sum) # a b c d # 1 1 1 1 500 # 2 1 3 1 0 # 3 1 1 2 600 # 4 1 2 3 300 As noted by @Roland, for bigger...
Reduce Subset Sum to Polyomino Packing
algorithm,np-complete,subset-sum
I would recommend doing this reduction in two steps. First, reduce subset-sum to the partition problem. This problem is related to subset-sum, except that instead of being given a target k, the goal is to split the set into two exactly equal halves. This isn't too hard (I'll leave it...
Algorithm for subset-sum with subtraction
I think your idea is mostly right: generate each combination of the terms, do the sum and see if it's a hit. You can optimize your code though. The problem is that once you generate 1 + 2, you see that it's not a match for your desired sum and...
Python Subset Sum
Based on your solution: def subsetsum(array,num): if num == 0 or num < 1: return None elif len(array) == 0: return None else: if array[0] == num: return [array[0]] else: with_v = subsetsum(array[1:],(num - array[0])) if with_v: return [array[0]] + with_v else: return subsetsum(array[1:],num) ...