You have n
boxes. You are given a binary string boxes
of length n
, where boxes[i]
is '0'
if the ith
box is empty, and '1'
if it contains one ball.
In one operation, you can move one ball from a box to an adjacent box. Box i
is adjacent to box j
if abs(i - j) == 1
. Note that after doing so, there may be more than one ball in some boxes.
Return an array answer
of size n
, where answer[i]
is the minimum number of operations needed to move all the balls to the ith
box.
Each answer[i]
is calculated considering the initial state of the boxes.
Example 1:
Input: boxes = "110" Output: [1,1,3] Explanation: The answer for each box is as follows: 1) First box: you will have to move one ball from the second box to the first box in one operation. 2) Second box: you will have to move one ball from the first box to the second box in one operation. 3) Third box: you will have to move one ball from the first box to the third box in two operations, and move one ball from the second box to the third box in one operation.
Example 2:
Input: boxes = "001011" Output: [11,8,5,4,3,4]
Constraints:
n == boxes.length
1 <= n <= 2000
boxes[i]
is either'0'
or'1'
Solution: Prefix Sum + DP
Time complexity: O(n)
Space complexity: O(n)
C++
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 |
// Author: Huahua class Solution { public: vector<int> minOperations(string boxes) { const int n = boxes.size(); vector<int> ans(n); for (int i = 0, c = 0, s = 0; i < n; ++i) { ans[i] += c; c += (s += boxes[i] - '0'); } for (int i = n - 1, c = 0, s = 0; i >= 0; --i) { ans[i] += c; c += (s += boxes[i] - '0'); } return ans; } }; |
请尊重作者的劳动成果,转载请注明出处!花花保留对文章/视频的所有权利。
如果您喜欢这篇文章/视频,欢迎您捐赠花花。
If you like my articles / videos, donations are welcome.
Be First to Comment