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Minimum Elements to Add to Form a Given Sum

Med
#1633Time: O(N + abs(goal - sum) / limit), where N is the number of elements in `nums`. The initial summation takes O(N) time. The `while` loop can take a very long time if the difference is large and the limit is small, making this approach inefficient for the given constraints.Space: O(1) extra space, as we only use a few variables to store the sum and the count.1 company
Patterns
Greedy
Data structures
Array
Companies
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Prompt

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 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 <= 105
  • 1 <= limit <= 106
  • -limit <= nums[i] <= limit
  • -109 <= goal <= 109

Approaches

2 approaches with complexity analysis and trade-offs.

This approach simulates the process of adding elements one by one. It first calculates the total difference required to reach the goal. Then, in a loop, it repeatedly subtracts limit from this difference, incrementing a counter each time. This process continues until the difference is covered. This is a direct simulation of a greedy strategy where we add the largest possible value (limit) in each step.

Algorithm

  • Calculate the sum of all elements in the nums array. Use a long data type to prevent potential integer overflow.
  • Determine the difference between the goal and the calculated sum. Let this be diff = goal - sum.
  • Take the absolute value of the difference, let's call it remaining = abs(diff).
  • If remaining is 0, the sum is already at the goal, so return 0.
  • Initialize a counter for the number of added elements, count = 0.
  • Start a loop that continues as long as remaining > 0.
  • Inside the loop, subtract limit from remaining and increment count.
  • This simulates adding one element of the maximum possible magnitude (limit) in each step.
  • Once the loop finishes, count will hold the minimum number of elements required. Return count.

Walkthrough

First, we need to determine the total value we need to add to the array. This is the difference between the goal and the current sum of the array's elements. We calculate the current sum by iterating through the nums array. It's crucial to use a long variable for the sum to avoid potential integer overflow, given the problem's constraints.

Let diff = goal - currentSum. The absolute value of this difference, abs(diff), is the total magnitude we need to add. To minimize the number of elements, we should greedily add elements with the maximum allowed magnitude, which is limit.

We can simulate this process. We initialize a counter for the number of new elements to 0. We then loop as long as the remaining difference is greater than 0. In each iteration, we subtract limit from the remaining difference and increment our counter. This loop effectively counts how many times we need to add an element of magnitude limit to cover the total required difference. The final value of the counter is the minimum number of elements required.

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);        if (diff == 0) {            return 0;        }         int count = 0;        while (diff > 0) {            diff -= limit;            count++;        }        return count;    }}

Complexity

Time

O(N + abs(goal - sum) / limit), where N is the number of elements in `nums`. The initial summation takes O(N) time. The `while` loop can take a very long time if the difference is large and the limit is small, making this approach inefficient for the given constraints.

Space

O(1) extra space, as we only use a few variables to store the sum and the count.

Trade-offs

Pros

  • The logic is straightforward and easy to understand as it directly models the problem statement.

Cons

  • This approach can be very slow if the difference diff is much larger than the limit. The number of iterations is proportional to diff / limit, which can lead to a 'Time Limit Exceeded' error on large inputs.

Solutions

class Solution { public int minElements ( int [] nums , int limit , int goal ) { // long s = Arrays.stream(nums).asLongStream().sum(); long s = 0 ; for ( int v : nums ) { s += v ; } long d = Math . abs ( s - goal ); return ( int ) (( d + limit - 1 ) / limit ); } }

Video walkthrough

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