Formatted question description: https://leetcode.ca/all/1785.html

# 1785. Minimum Elements to Add to Form a Given Sum

## Level

Medium

## 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 <= 10^5`

`1 <= limit <= 10^6`

`-limit <= nums[i] <= limit`

`-10^9 <= goal <= 10^9`

## Solution

First, calculate `sum`

as the sum of all elements in `nums`

. Next, calculate the absolute difference between `goal`

and `sum`

, and calculate the minimum number of elements to add using a greedy approach, which means consider adding `limit`

or `-limit`

as many as possible and adding one more element if the remaining absolute difference is less than `limit`

.

```
class Solution {
public int minElements(int[] nums, int limit, int goal) {
long sum = 0;
for (int num : nums)
sum += (long) num;
long difference = (long) goal - sum;
if (difference == 0)
return 0;
else {
difference = Math.abs(difference);
long minAdd = difference / limit;
if (difference % limit != 0)
minAdd++;
return (int) minAdd;
}
}
}
```