def getMoneySpent(keyboards, drives, b): max_val=0 for i in keyboards: for j in drives: if(i+j<=b and max_val

Here is my c++ solution, you can find the video here : https://youtu.be/yC-TXToDbD0

int getMoneySpent(vector<int> keyboards, vector<int> drives, int b) {
    int ans = -1;
    for(int k : keyboards){
        for(int d : drives){
            if(k + d > ans && k + d <= b) ans = k + d;
        }
    }
    return ans;
}

//Golang Solution func getMoneySpent(keyboards []int32, drives []int32, b int32) int32 {

 var result int32 =-1

 for i := 0  ; i < len(keyboards) ;i++{

     for j :=0 ; j < len(drives);j++{

         if( keyboards[i] + drives[j] > result  && keyboards[i] + drives[j] <=b) {

             result = keyboards[i] + drives[j]
         } 
     }
 }

 return result

}

O(nlogn)+O(mlogm)+O(n+k) solution

The logic is simple, we sort both the arrays, then put pointer 1 on the end of 1 array and pointer 2 at the start of the other. Then just like doing 2 pointer problems, if the sum of value at pointer 1 + pointer 2 is less or equal than our target value, we store the result and increment pointer 2, if its greater then we decrement pointer 1, thereby only traversing both arrays only once.

`int getMoneySpent(vector keyboards, vector drives, int b) { int n{static_cast(keyboards.size())}; int m{static_cast(drives.size())};

sort(keyboards.begin(),keyboards.end()); //O(nlogn)
sort(drives.begin(), drives.end());     //O(mlogm)

int kb_ptr{n-1};
int d_ptr{0};

int result{-1};
while(kb_ptr >=0 && d_ptr<=m ) {                //O(n+m)
    if(keyboards[kb_ptr]+ drives[d_ptr] <= b){
        result=max(result, keyboards[kb_ptr] + drives[d_ptr]);
        ++d_ptr;
    }
    else{
        --kb_ptr;
    }
}
return result;

}