Watch 6 video solutions for Minimum Elements to Add to Form a Given Sum, a medium level problem involving Array, Greedy. This walkthrough by Coding Decoded has 767 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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 <= 1051 <= limit <= 106-limit <= nums[i] <= limit-109 <= goal <= 109Problem Overview: You are given an integer array nums, a maximum allowed absolute value limit, and a target goal. You can append numbers between -limit and limit. The task is to compute the minimum number of elements required so the final array sum becomes exactly goal.
Approach 1: Simple Calculation Using Greedy Strategy (Time: O(n), Space: O(1))
First compute the current sum of the array using a single pass. The difference between the target and current sum is diff = |goal - sum(nums)|. Each number you add can contribute at most limit toward reducing this difference because its absolute value cannot exceed limit. The optimal strategy is greedy: always use the maximum allowed magnitude to close the gap as quickly as possible. The number of elements required is therefore ceil(diff / limit). This works because using smaller values would only increase the count of elements needed. The algorithm only scans the array once to compute the sum and then performs constant-time math.
This method relies on a greedy observation rather than exploring combinations. Since each inserted element contributes independently to reducing the remaining difference, maximizing each contribution minimizes the number of operations. The approach fits naturally into problems tagged with Greedy and basic Array processing.
Approach 2: Binary Search on Number of Added Elements (Time: O(n + log(diff)), Space: O(1))
Instead of computing the answer directly, you can frame the problem as searching for the smallest k such that k * limit ≥ diff. Start by calculating the same difference diff = |goal - sum(nums)|. Then perform Binary Search over the possible number of added elements. For a candidate value k, check if the maximum achievable adjustment (k * limit) is sufficient to cover diff. If it is, try smaller values; otherwise increase k. This approach is more algorithmic but ultimately arrives at the same answer as the greedy formula.
The binary search formulation can help when reasoning about feasibility conditions or when constraints become more complex in follow‑up problems. However, the mathematical greedy calculation is simpler and faster in practice.
Recommended for interviews: The greedy calculation is the expected solution. Interviewers typically want to see the observation that each added number can reduce the difference by at most limit, which immediately leads to ceil(diff / limit). Explaining the reasoning clearly demonstrates stronger problem‑solving skills than implementing a heavier search approach.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Greedy Calculation | O(n) | O(1) | Best general solution. Direct mathematical insight using ceil(diff/limit). |
| Binary Search on Answer | O(n + log(diff)) | O(1) | Useful when modeling the problem as a feasibility check or extending to more complex constraints. |