Minimum Elements to Add to Form a Given Sum

Problem Statement

Problem 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

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

Intuition

To solve this problem efficiently, we can use a greedy approach. We need to calculate the difference between the current sum of nums and the goal. Based on this difference, we can determine how many elements are required to achieve the desired sum. The key steps are:

1. Calculate the Current Sum: Compute the sum of the nums array.
2. Determine the Difference: Find the absolute difference between the current sum and the goal.
3. Calculate the Minimum Number of Elements: Given that each element added can have a maximum magnitude of limit, compute the minimum number of such elements required to cover the difference.

Time Complexity and Space Complexity Analysis

• Time Complexity:

  • Calculating the sum of the nums array takes $O(n)$, where n is the length of the array.
  • Calculating the minimum number of elements involves a simple arithmetic operation, which is $O(1)$.

  Overall time complexity is $O(n)$.

• Space Complexity:

  • The space complexity is $O(1)$ as we are using a fixed amount of extra space.

Code

C++

class Solution {
public:
    int minElements(vector<int>& nums, int limit, int goal) {
        long long sum = accumulate(nums.begin(), nums.end(), 0LL);
        long long diff = abs(sum - goal);
        return (diff + limit - 1) / limit; // Equivalent to ceil(diff / limit)
    }
};

Python

class Solution:
    def minElements(self, nums: List[int], limit: int, goal: int) -> int:
        current_sum = sum(nums)
        diff = abs(current_sum - goal)
        return (diff + limit - 1) // limit  # Equivalent to ceil(diff / limit)

Java

import java.util.List;

class Solution {
    public int minElements(List<Integer> nums, int limit, int goal) {
        long currentSum = 0;
        for (int num : nums) {
            currentSum += num;
        }
        long diff = Math.abs(currentSum - goal);
        return (int) ((diff + limit - 1) / limit); // Equivalent to ceil(diff / limit)
    }
}