Two-pointer solution:

    keyboards.sort()
    drives.sort(reverse=True)
    
    best = -1
    i = j = 0
    
    while i < len(keyboards) and j < len(drives):
        total = keyboards[i] + drives[j]
        if total > b:
            j += 1
        else:
            if total > best:
                best = total
            i += 1
    return best

C# static int getMoneySpent(int[] keyboards, int[] drives, int b) { int max = -1; Array.Sort(keyboards); Array.Sort(drives);

 int i = 0;
 int j = drives.Length - 1;

 while (i <= keyboards.Length - 1 && j >= 0)
 {
     int sum = keyboards[i] + drives[j];

     if (sum > b)
     {
         j--;
     }
     else
     {
         i++;

         if (sum > max)
         {
             max = sum;
         }
     }
 }

 return max;

}

costs = []

for i in keyboards:
    for j in drives:
        costs.append (i+j) if (i+j <= b) else costs.append (0)
return max(costs) if (max(costs) > 0) else -1

This doesn't have a typescript option?

here is python3 solution to the problem sol1: by taking an empty list def getMoneySpent(keyboards, drives, b): l=[] for i in keyboards: for j in drives: if i+j<=b: l.append(i+j) if not l: return -1 else:
return max(l) ` sol2: since list might end up taking too much storage def getMoneySpent(keyboards, drives, b): spent=-1 for i in keyboards: for j in drives: total=i+j if total<=b: spent= max(spent,total) return spent