Leetcode 1785: Minimum Elements to Add to Form a Given Sum
You are given an integer array nums and two integers limit and goal. The elements of the array nums have a property such that abs(nums[i]) <= limit. Your task is to determine the minimum number of elements you need to add to the array in order to make the sum equal to goal. The added elements must also satisfy the condition that their absolute value is less than or equal to limit.
Problem
Approach
Steps
Complexity
Input: You are given an integer array nums and two integers, limit and goal.
Example: nums = [2, -5, 7], limit = 6, goal = -4
Constraints:
• 1 <= nums.length <= 10^5
• 1 <= limit <= 10^6
• -limit <= nums[i] <= limit
• -10^9 <= goal <= 10^9
Output: Return the minimum number of elements needed to be added to the array in order to achieve the goal sum. The added elements must also maintain the constraint abs(x) <= limit.
Example: Input: nums = [1, -3, 5], limit = 10, goal = 0, Output: 1
Constraints:
Goal: The goal is to determine how many elements need to be added to the array such that the sum of the array equals the target goal, and each added element must respect the limit condition.
Steps:
• 1. Calculate the sum of the array nums.
• 2. Find the difference between the goal and the current sum.
• 3. Determine how many elements with a value of abs(limit) are required to bridge the difference. This can be calculated as diff / limit (rounded up).
Goal: Ensure that the number of elements in nums is within the given constraints, and the values in nums should satisfy the abs(nums[i]) <= limit condition.
Steps:
• 1 <= nums.length <= 10^5
• 1 <= limit <= 10^6
• -limit <= nums[i] <= limit
• -10^9 <= goal <= 10^9
Assumptions:
• The array nums is provided with the condition abs(nums[i]) <= limit.
• The goal is achievable by adding values within the limit.
• Input: Input: nums = [2, -5, 7], limit = 6, goal = -4
• Explanation: The initial sum of the array is 4, and to make it equal to -4, we need to reduce the sum by 8. Adding -6 and -6 will achieve this.
• Input: Input: nums = [1, -3, 5], limit = 10, goal = 0
• Explanation: The initial sum is 3, and to reach 0, we need to subtract 3. Adding -3 achieves this goal.
Approach: The approach involves calculating the difference between the current sum of the array and the goal. Then, we calculate how many elements with the largest allowed value (abs(limit)) are needed to close this difference.
Observations:
• The sum of the array must be compared with the goal to compute the difference.
• The largest element we can add or subtract is the value of limit.
• To minimize the number of elements added, we should use the maximum allowed value (limit) for each addition.
Steps:
• 1. Calculate the sum of the array nums.
• 2. Compute the difference between the sum and the goal.
• 3. Calculate the minimum number of elements needed to close this difference by dividing it by limit and rounding up.
Empty Inputs:
• The array nums will not be empty as per the constraints.
Large Inputs:
• Ensure that the solution handles large arrays efficiently (up to 100,000 elements).
Special Values:
• The solution should correctly handle cases where the goal is already equal to the sum of the array.
Constraints:
• The solution should ensure that all elements added to nums respect the condition abs(x) <= limit.
int minElements(vector<int>& nums, int limit, int goal) {
long sum = 0;
for(int x: nums)
sum += x;
long diff = abs(goal - sum);
if(diff % limit == 0) return diff/limit;
else return diff/limit + 1;
}
1 : Function Definition
int minElements(vector<int>& nums, int limit, int goal) {
Define the function `minElements`, which takes a vector of integers `nums`, an integer `limit`, and an integer `goal`, and returns the minimum number of operations required to reach the goal.
2 : Variable Initialization
long sum = 0;
Initialize a variable `sum` to 0, which will hold the sum of elements in the `nums` array.
3 : Looping through Array
for(int x: nums)
Loop through each element `x` in the `nums` array to calculate the sum of its elements.
4 : Accumulate Sum
sum += x;
Add the value of each element `x` to the `sum` variable.
5 : Calculate Difference
long diff = abs(goal - sum);
Calculate the absolute difference between the `goal` and the current `sum` of the array elements.
6 : Condition Check
if(diff % limit == 0) return diff/limit;
Check if the difference `diff` is divisible by the `limit`. If it is, return the result of `diff / limit`, which represents the minimum number of operations needed.
7 : Condition Else
else return diff/limit + 1;
If the difference `diff` is not divisible by the `limit`, return `diff / limit + 1`, accounting for the extra operation required.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) due to iterating over the array to calculate the sum.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1) as we only need a few variables to compute the sum and the difference.
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