Create Minimum Elements to Add to Form a Given Sum.md

Author: sreevidya-16

dsa-solutions/lc-solutions/1700-1799/Minimum Elements to Add to Form a Given Sum.md

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+---
+id: minimum-elements-to-add-to-form-a-given-sum
+title: Minimum Elements to Add to Form a Given Sum
+sidebar_label: 1785 - Minimum Elements to Add to Form a Given Sum
+tags: [Greedy, Array, C++]
+description: Solve the problem of finding the minimum number of elements to add to an array to make the sum equal to a given goal while maintaining the property that absolute value of elements does not exceed a specified limit.
+---
+
+## 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 &lt;= `nums.length` &lt;= 10^5
+- 1 &lt;= `limit` &lt;= 10^6
+- -`limit` &lt;= `nums[i]` &lt;= `limit`
+- -10^9 &lt;= `goal` &lt;= 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++
+
+```cpp
+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
+```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 
+```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)
+    }
+}
+```