LeetCode 1785. Minimum Elements to Add to Form a Given Sum Solution in Java, C++, Python & More | Explanation + Code

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 <= 105
• 1 <= 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 ^|d|⁄_limit \rceil.

Note that in this problem, the data range of array elements is [-10⁶, 10⁶], the maximum number of elements is 10⁵, 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.

PythonJavaC++GoTypeScriptRustC

class Solution: def minElements(self, nums: List[int], limit: int, goal: int) -> int: d = abs(sum(nums) - goal) return (d + limit - 1) // limit(code-box)

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); } }(code-box)

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; } };(code-box)

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 }(code-box)

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); }(code-box)

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 } }(code-box)

int minElements(int* nums, int numsSize, int limit, int goal) { long long sum = 0; for (int i = 0; i < numsSize; i++) { sum += nums[i]; } long long diff = labs(goal - sum); return (diff + limit - 1) / limit; }(code-box)