Problem Description

Given an integer array nums, an integer limit, and an integer goal, your task is to determine the minimum number of elements you need to add (with each new element's absolute value not exceeding limit) so that the sum of the array equals goal.

• You may add any integers you like, as long as their absolute value is at most limit.
• You can add the same or different values multiple times.
• Each added number is independent; you do not need to use elements from nums.
• There is always at least one valid solution.

The goal is to find the minimum number of additions required to make the sum of nums equal to goal.

Thought Process

The first step is to realize that the only thing that matters is the difference between the current sum of nums and the desired goal. We don't care about the actual values in nums, only their sum.

The brute-force approach would be to try all possible combinations of numbers (with absolute value ≤ limit) to reach the difference. However, this is highly inefficient, especially for large differences.

Instead, we can optimize by using the largest possible number (in absolute value) each time. This is like trying to pay off a debt using the largest available denomination of coins.

Thus, the minimum number of elements required is the ceiling of the absolute difference divided by limit.

Solution Approach

• Step 1: Calculate the current sum of nums.
• Step 2: Find the difference between goal and the current sum.
  • Let’s call this diff = abs(goal - sum(nums)).
• Step 3: To minimize the number of additions, always add (or subtract) the maximum allowed value, which is limit.
• Step 4: The minimum number of elements needed is ceil(diff / limit).
  • This is because each added element can "cover" at most limit of the difference.
• Step 5: Return this number as the answer.

We use the absolute value because the difference could be positive or negative, and we can add either positive or negative numbers to reach the goal.

Example Walkthrough

Example:
nums = [1, -1, 1], limit = 3, goal = -4

1. Sum of nums is 1 + (-1) + 1 = 1.
2. Difference to goal is abs(-4 - 1) = 5.
3. Each added element can be at most 3 or -3.
4. We need to "cover" 5 using numbers with absolute value at most 3.
5. First addition: add -3 (sum is now 1 + (-3) = -2).
6. Second addition: add -3 (sum is now -2 + (-3) = -5).
7. But now we've gone past the goal. Instead, we can add -3 and -2 (since 2 ≤ limit).
8. So, the answer is 2 additions.
9. Mathematically: ceil(5 / 3) = 2.

Time and Space Complexity

• Brute-Force: Would require trying all combinations of additions to reach the goal, which is exponential in the worst case (O(limit^k) where k is the number of elements added).
• Optimized Approach:
  • Time Complexity: O(n) for calculating the sum of nums, where n is the length of nums.
  • Space Complexity: O(1) since we only use a few variables.

The optimized solution is highly efficient and suitable for large inputs.

Summary

The key insight is that the problem reduces to finding the minimal number of steps to reach a target sum, using additions (or subtractions) bounded by limit. By always using the largest possible value, we guarantee the fewest steps. This greedy strategy makes the solution both elegant and efficient, requiring only a simple calculation after summing the input array.

Code Implementation

import math

class Solution:
    def minElements(self, nums, limit, goal):
        diff = abs(goal - sum(nums))
        return math.ceil(diff / limit)
      

#include <vector>
#include <cmath>
#include <numeric>

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

class Solution {
    public int minElements(int[] nums, int limit, int goal) {
        long sum = 0;
        for (int num : nums) sum += num;
        long diff = Math.abs(goal - sum);
        return (int)((diff + limit - 1) / limit);
    }
}
      

var minElements = function(nums, limit, goal) {
    let sum = nums.reduce((a, b) => a + b, 0);
    let diff = Math.abs(goal - sum);
    return Math.ceil(diff / limit);
};