1785. Minimum Elements to Add to Form a Given Sum
Problem Description
The problem provides an array of integers nums, where each integer satisfies the condition abs(nums[i]) <= limit. You are also given two integers, limit and goal. The task is to find the minimum number of elements that need to be added to the array so that the sum of the array equals the goal while still maintaining the condition that the absolute value of any element does not exceed limit.
An important thing to note is how absolute values work: abs(x) = x if x >= 0, and abs(x) = -x otherwise. This tells us that the elements we may add can be as large as limit but no larger.
Intuition
The intuition behind the solution is to find the difference d between the current sum of the array sum(nums) and the goal. This difference tells us what the deficit or surplus is relative to the goal. If the current sum is less than the goal, it means we have a deficit and we need to add elements to reach the goal. If it's greater, we need to subtract elements, but since we can only add elements, we think of what we can add to balance the excess (which mathematically is the same as subtracting to reach a lower goal).
After finding the difference d, we consider the maximum value we are allowed to add, which is limit. To minimize the number of elements we add, we should add elements of value limit until we reach or exceed the goal. However, since we might not reach the goal exactly, we might need to add one last element with a value smaller than limit to make the sum exactly equal the goal.
Mathematically, the minimum number of elements we need is the total difference divided by the limit, but since we're dealing with integers and we want to cover any remainder, we have to use the ceiling of the division, which is achieved by adding limit - 1 to the difference before dividing. This is encapsulated in (d + limit - 1) // limit.
So the approach is:
- Calculate the difference
dbetween the current sum and thegoal. - Divide
dbylimitusing the ceiling of the division to account for any remainder (since we can only add whole elements and need to reach or exceedgoal). - The result of this division gives the minimum number of elements needed to be added.
This is precisely what the implementation does in an efficient and concise manner.
Learn more about Greedy patterns.
How would you design a stack which has a function min that returns the minimum element in the stack, in addition to push and pop? All push, pop, min should have running time O(1).
Solution Approach
The implementation of the solution is straightforward and takes advantage of the mathematical foundation laid out in the intuition.
Here's the step-by-step breakdown of the algorithm using the provided Python code:
- Calculate the difference
dbetween the current sum of elements innumsand thegoal. This is achieved using thesumfunction andabsto ensure the difference is positive regardless of whether the sum is below or above the goal:
1d = abs(sum(nums) - goal)
The abs function is crucial here because it ensures that the difference is treated as a positive value, which aligns with our need to either add or subtract (handled as adding a negative value when the sum is above the goal) to reach the exact goal value.
- Compute the minimum number of elements to add by dividing the difference
dbylimit. Since we have to deal with the possibility of having a non-zero remainder, we aim for the ceiling of the division by addinglimit - 1before performing integer division:
1(d + limit - 1) // limit
The // operator in Python indicates floor division, which would normally take the floor of the division result. However, the trick of adding limit - 1 effectively changes the division result to a ceiling operation for positive numbers, ensuring that if there's a remainder, we count an additional element.
No specific data structures are used apart from the basics provided by Python, and the pattern applied here is purely mathematical. The algorithm's complexity is O(n) due to the summation of the elements in the array, but the actual calculation of the result is done in constant time, O(1). This makes the solution very efficient for inputs where n, the number of elements in nums, is not excessively large.
What is the worst case running time for finding an element in a binary tree (not necessarily binary search tree) of size n?
Example Walkthrough
Let's apply the solution approach to a small example for clarification. Suppose we have an array nums = [1, 2, 3], a limit of 3, and a goal of 10. We want to find out how many minimum additional elements we need to add to nums so that the sum of the array equals goal, without adding any element with an absolute value greater than limit.
- First, we calculate the current sum of the array:
sum(nums) = 1 + 2 + 3 = 6. - We find the difference
dbetween the current sum and thegoal:d = abs(6 - 10) = 4, since the sum is below the goal. - Now, we determine the smallest number of elements of value up to the
limitthat need to be added to reach the goal. Since we want to add as few elements as possible, we will add elements with the maximum value allowed, which islimit(3 in this case). - We calculate this by dividing
dbylimitand using the ceiling of the division to get the minimum number of additional elements:- We add
limit - 1todbefore the division to account for any remainder:d + limit - 1 = 4 + 3 - 1 = 6. - Now, we perform the integer division by
limit:(6) // 3 = 2.
- We add
So, we need to add at least 2 elements to reach the goal. The elements we can add are [3, 1], where we're adding the maximum value limit (3) first and then topping it off with a final element (1) to reach the exact goal. After adding these numbers to our original array, nums now looks like [1, 2, 3, 3, 1] and the sum is 10, which is equal to the goal.
Therefore, using the solution approach outlined, we have determined that the minimum number of elements we need to add to the nums array to reach the goal is 2.
Solution Implementation
1from typing import List
2
3class Solution:
4 def minElements(self, nums: List[int], limit: int, goal: int) -> int:
5 # Calculate the difference between the sum of the array and the goal
6 difference = abs(sum(nums) - goal)
7
8 # The number of elements required is the ceiling of 'difference / limit'
9 # which is computed using (difference + limit - 1) // limit to avoid floating point division
10 min_elements_needed = (difference + limit - 1) // limit
11
12 return min_elements_needed
131class Solution {
2
3 /**
4 * Finds the minimum number of elements with value 'limit' that
5 * must be added to the array to reach the 'goal' sum.
6 *
7 * @param nums The array of integers.
8 * @param limit The maximum value that could be added to or subtracted from the sum.
9 * @param goal The target sum.
10 * @return The minimum number of elements needed to reach the goal.
11 */
12 public int minElements(int[] nums, int limit, int goal) {
13 // Variable to store sum of the elements in the array.
14 long sum = 0;
15
16 // Loop to calculate the cumulative sum of the array elements.
17 for (int number : nums) {
18 sum += number;
19 }
20
21 // Calculate the difference between current sum and goal, using absolute value
22 // because we can add or subtract elements to reach the goal.
23 long difference = Math.abs(sum - goal);
24
25 // Compute the minimum number of elements needed with value 'limit' to cover the difference.
26 // The addition of 'limit - 1' is for upward rounding without using Math.ceil().
27 return (int) ((difference + limit - 1) / limit);
28 }
29}
301#include <vector>
2#include <numeric> // For accumulate
3
4class Solution {
5public:
6 // This function returns the minimum number of elements with the
7 // value 'limit' that must be added to the array to reach the goal
8 int minElements(vector<int>& nums, int limit, int goal) {
9 // Calculate the current sum of the array using long long for large sums
10 long long currentSum = accumulate(nums.begin(), nums.end(), 0ll);
11
12 // Calculate the absolute difference needed to reach the goal
13 long long differenceToGoal = abs(goal - currentSum);
14
15 // Calculate the minimum number of elements needed by dividing the difference
16 // by the limit and taking the ceiling of that value.
17 // (The ceil is implicitly done by adding limit - 1 before division)
18 // This is because we need to round up to make sure any remaining part of
19 // the difference is covered, even if it's less than the limit value.
20 int minElementsNeeded = (differenceToGoal + limit - 1) / limit;
21
22 // Return the calculated number of minimum elements needed
23 return minElementsNeeded;
24 }
25};
261/**
2 * Calculates the minimum number of elements with the given limit to add to or subtract
3 * from an array to achieve a specific goal sum.
4 *
5 * @param nums - The array of numbers representing the current elements.
6 * @param limit - The limit of the absolute value of each element that can be added.
7 * @param goal - The desired sum to be reached.
8 * @returns The minimum number of elements required to reach the goal.
9 */
10function minElements(nums: number[], limit: number, goal: number): number {
11 // Calculate the current sum of the array.
12 const currentSum = nums.reduce((accumulator, value) => accumulator + value, 0);
13
14 // Determine the absolute difference between current sum and goal.
15 const differenceToGoal = Math.abs(goal - currentSum);
16
17 // Calculate the minimum number of elements required to reach or surpass
18 // the absolute difference by dividing by the limit and rounding up.
19 // We subtract one before dividing to handle cases where the difference
20 // is an exact multiple of the limit.
21 const minimumElementsRequired = Math.ceil(differenceToGoal / limit);
22
23 return minimumElementsRequired;
24}
25Which two pointer technique does Quick Sort use?
Time and Space Complexity
The time complexity of the provided code is O(n), where n is the length of the nums array. This is because the code must sum up all elements in the array which takes O(n) time.
The space complexity of the code is O(1), as the additional space used by the function does not depend on the input size and is limited to a fixed number of integer variables (d and the return value).
Learn more about how to find time and space complexity quickly using problem constraints.
Which of the following is equvalent to O(3*2^n + n^3 + n!+ log n)?
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