2279. Maximum number of backpacks filled with stones

class Solution {
public:
    int maximumBags(vector<int>& capacity, vector<int>& rocks, int additionalRocks) {
      //  The number of items that can be put into each backpack , The same type of put number can only be saved and put once 
      vector<int> spaceVec;
      
      // key：spaceVec The value in 
      // value :  The number of backpacks put in this number  
      map<int, int> map;
      
      int ans = 0;
      
      //  Traverse all backpacks , Record empty remainder 
      for(int i = 0; i < capacity.size(); i++){
        //  Backpack size 
        int cap = capacity[i];
  
        //  The number of backpacks has been put  
        int roc = rocks[i];
        
        if(cap > roc){
          //  The backpack can also be put in quantity 
          int tmp = cap - roc;
          
          auto it = map.find(tmp);
          if(it == end(map)){
            //  Not recorded 
            spaceVec.push_back(tmp);
            map.emplace(tmp, 1);
          }else{
            //  It is recorded that only the quantity is increased by one 
            map[tmp] = map[tmp] + 1;
          }
        }else if(roc == cap){
          //  The backpack itself is full 
          ans++;
        }
      }
      
      //  In order to get the maximum number of full backpacks 
      //  There will be   Number that can be put in   Sort by increments 
      sort(begin(spaceVec), end(spaceVec));
      
      //  Number of objects available 
      int avail = additionalRocks;
      
      //  From small to large   Number that can be put in   Containers 
      for(int i = 0; i < spaceVec.size(); i++){
        //  Number that can be put in 
        int roc = spaceVec[i];

        //  Change the number of occurrences of the number that can be put in 
        int times = map[roc];

        while(times-- > 0){
          //  When the utilization object is smaller than 0 yes , Directly return the current result 
          if(0 > (avail -= roc)){
            return ans;
          }

          //  If it is satisfied, the result will be accumulated 
          ans++;
        }
      }
      
      return ans;
    }
};