You are given an integer array nums
and two integers limit
and goal
. The array nums
has an interesting property that abs(nums[i]) <= limit
.
Return the minimum number of elements you need to add to make the sum of the array equal to goal
. The array must maintain its property that abs(nums[i]) <= limit
.
Note that abs(x)
equals x
if x >= 0
, and -x
otherwise.
Example 1:
Input: nums = [1,-1,1], limit = 3, goal = -4 Output: 2 Explanation: You can add -2 and -3, then the sum of the array will be 1 - 1 + 1 - 2 - 3 = -4.
Example 2:
Input: nums = [1,-10,9,1], limit = 100, goal = 0 Output: 1
Constraints:
1 <= nums.length <= 105
1 <= limit <= 106
-limit <= nums[i] <= limit
-109 <= goal <= 109
Solution: Math
Time complexity: O(n)
Space complexity: O(1)
Compute the diff = abs(sum(nums) – goal)
ans = (diff + limit – 1)) / limit
C++
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// Author: Huahua class Solution { public: int minElements(vector<int>& nums, int limit, int goal) { int64_t diff = abs(goal - accumulate(begin(nums), end(nums), 0LL)); return (diff + limit - 1) / limit; } }; |
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