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1785. Minimum Elements to Add to Form a Given Sum
Description
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 <= 1051 <= limit <= 106-limit <= nums[i] <= limit-109 <= goal <= 109
Solutions
Solution 1: Greedy
First, we calculate the sum of the array elements $s$, and then calculate the difference $d$ between $s$ and $goal$.
| The number of elements to be added is the absolute value of $d$ divided by $limit$ and rounded up, that is, $\lceil \frac{ | d | }{limit} \rceil$. |
Note that in this problem, the data range of array elements is $[-10^6, 10^6]$, the maximum number of elements is $10^5$, the total sum $s$ and the difference $d$ may exceed the range of 32-bit integers, so we need to use 64-bit integers.
The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array $nums$.
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class Solution { public int minElements(int[] nums, int limit, int goal) { // long s = Arrays.stream(nums).asLongStream().sum(); long s = 0; for (int v : nums) { s += v; } long d = Math.abs(s - goal); return (int) ((d + limit - 1) / limit); } } -
class Solution { public: int minElements(vector<int>& nums, int limit, int goal) { long long s = accumulate(nums.begin(), nums.end(), 0ll); long long d = abs(s - goal); return (d + limit - 1) / limit; } }; -
class Solution: def minElements(self, nums: List[int], limit: int, goal: int) -> int: d = abs(sum(nums) - goal) return (d + limit - 1) // limit -
func minElements(nums []int, limit int, goal int) int { s := 0 for _, v := range nums { s += v } d := abs(s - goal) return (d + limit - 1) / limit } func abs(x int) int { if x < 0 { return -x } return x } -
function minElements(nums: number[], limit: number, goal: number): number { const sum = nums.reduce((r, v) => r + v, 0); const diff = Math.abs(goal - sum); return Math.floor((diff + limit - 1) / limit); } -
impl Solution { pub fn min_elements(nums: Vec<i32>, limit: i32, goal: i32) -> i32 { let limit = limit as i64; let goal = goal as i64; let mut sum = 0; for &num in nums.iter() { sum += num as i64; } let diff = (goal - sum).abs(); ((diff + limit - 1) / limit) as i32 } }