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 <= 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, $\lceil \frac{|d|}{limit} \rceil$.

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 $nums$.

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

Java Code
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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);
    }
}

C++ Code
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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;
    }
};

Go Code
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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
}

TypeScript Code
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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);
}

Rust Code
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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
    }
}

C Code
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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;
}