Assume your target sum is quite small say <1e6. For every number you can memorize the best solution, instead of calling bestsum() recursively. (you call the same target sum multiple times and it grows somehow exponentially).

Denote $$$<x,y>$$$ where $$$x$$$ to be the minimum length of array , $$$y$$$ to be the previous number makes your array minimal, Say $$$<5,100>$$$ means the array has 5 elements and the first 4 elements comes from the optimal solution in $$$100$$$

assume input is 10 3 1 3 4, we calculate the best solution from 0 to your desired target.

target sum 0 : []

target sum 1 : 1 + target sum 0 / 3 + target sum -2 / 4 + target sum -3 = 1 + target sum 0 = <1,0>

target sum 2 : <2,1> = [1,1]

target sum 3 : 1 + <2,1> / 3 + <0,0> = <1,0> = [3]

target sum 4 : 1 + <1,3> / 3 + <1,1> / 4 + <0,0> = <1,0> = [4]

target sum 5 : 1 + <1,4> / 3 + <2,1> / 4 + <1,1> = <2,4> / <2,1> = [1,4]

target sum 6 : 1 + <2,5> / 3 + <1,3> / 4 + <2,2> = <2,3> = [3,3]

then 10 = <3,6> = [4] + optimal solution for 6 = [4] + [3] + optimal solution for 3 = [4,3,3]

bool bestsum(int n,int m,int a[]){
    int minlen[n+1];
    int par[n+1];
    memset(minlen,0,sizeof(minlen));
    memset(par,0,sizeof(par));
    for(int i=1;i<=n;i++){
        int l=-1; // if -1 then no solution
        int p=0;
        for(int j=0;j<m;j++){
            int r=i-a[j];
            if(r>=0&&minlen[r]>=0){ // check r has solution or not
                if(l==-1||minlen[r]<l){ // check is optimal or not
                    l=minlen[r]+1;
                    p=r;
                }
            }
        }
        minlen[i]=l;
        par[i]=p;
    }
    if(minlen[n]>=0){
        int r=par[n];
        while(n>0){
            r=par[n];
            save.push_back(n-r);
            n=r;
        }
        flag = true;
    }
    else flag = false;
}

For anyone that is still trying to find a pretty simple solution, here you go: ```

include <bits/stdc++.h>

using namespace std;

unordered_map<int, vector> memo;

vector bestSum(int targetSum, const vector& numbers) { if (memo.find(targetSum) != memo.end()) { return memo[targetSum]; }

if (targetSum == 0) {
    return {};
}

if (targetSum < 0) {
    return {INT_MIN};  // Return a sentinel value to represent null
}

vector<int> shortestCombination = {INT_MIN};

for (int num : numbers) {
    int remainder = targetSum - num;
    vector<int> remainderCombination = bestSum(remainder, numbers);

    if (remainderCombination != vector<int>{INT_MIN}) {
        vector<int> combination = remainderCombination;
        combination.push_back(num);

        if (shortestCombination == vector<int>{INT_MIN} || combination.size() < shortestCombination.size()) {
            shortestCombination = combination;
        }
    }
}

memo[targetSum] = shortestCombination;
return shortestCombination;

}

int main() { vector numbers = {1, 2, 5, 25}; int targetSum = 100;

vector<int> result = bestSum(targetSum, numbers);

if (result != vector<int>{INT_MIN}) {
    cout << "Shortest combination: ";
    for (int num : result) {
        cout << num << " ";
    }
    cout << endl;
} else {
    cout << "No valid combination found." << endl;
}

return 0;

} ```