Minimum Number of Operations to Move All Balls to Each Box

/*
Method:-1
class Solution {
public:
    vector<int> minOperations(string boxes) {
        vector <int> left,right;
        int prev_val=0,prev_count=0;
        for(int i=0;i<boxes.size();i++){
            left.push_back(prev_val+prev_count);
            prev_val=prev_val+prev_count;
            if(boxes[i]=='1') prev_count++;
        }
        prev_val=0;
        prev_count=0;
        for(int i=boxes.size()-1;i>=0;i--){
            left[i]+=prev_val+prev_count;
            prev_val=prev_val+prev_count;
            if(boxes[i]=='1') prev_count++;
        }
        return left;
    }
};
*/
/*
   Leetcode:
   First to bring all the balls to 0th position will be the sum of all the balls indexes.
   Assume at ith position,We need n opeation to shift all balls into ith box.
   Let's assume we know all the balls in the left and right of ith position
   as per the initial state.
   so,in case of (i+1)th box,all the balls in the left of (i+1)th box will
   move one more box than the ith box and all the balls in the right of (i+1)th
   including (i+1)th box will move one less boxes.
   so,
   arr[i+1]=arr[i]+left-right(including current box)
*/

class Solution {
public:
    vector<int> minOperations(string boxes) {
        vector <int> ans;
        int cur=0,right=0,left=0;
        for(int i=1;i<boxes.size();i++){
            if(boxes[i]=='1'){ cur+=i; right++;}
        }
        if(boxes[0]=='1') left++;
        ans.push_back(cur);
        for(int i=1;i<boxes.size();i++){
            ans.push_back(ans.back()-right+left);
            if(boxes[i]=='1'){ left++; right--;}
        }
        return ans;
    }
};

#Python
class Solution:
    def minOperations(self, boxes: str) -> List[int]:
        right=0
        ops=0
        for i in range(1,len(boxes)):
            ops+=i if boxes[i]=='1' else 0
            right+=1 if boxes[i]=='1' else 0
        left=1 if boxes[0]=='1' else 0
        ans=[ops]
        for i in range(1,len(boxes)):
            ans.append(ans[i-1]+left-right)
            right-=1 if boxes[i]=='1' else 0
            left+=1 if boxes[i]=='1' else 0
        return ans