LeetCode 1785 - Minimum Elements to Add to Form a Given Sum

LeetCode Problem 1785

Difficulty: 🟡 Medium
Topics: Array, Greedy

Solution

Problem Understanding

The problem gives us an integer array nums, a maximum allowed absolute value limit, and a target sum called goal.

Every element in the array already satisfies the condition:

abs(nums[i]) <= limit

We are allowed to add new integers to the array, but every newly added integer must also satisfy the same restriction:

abs(x) <= limit

Our task is to determine the minimum number of elements that must be added so that the final sum of the array becomes exactly equal to goal.

In simpler terms, we first compute the current sum of the array. If the current sum is already equal to the goal, then no elements need to be added. Otherwise, there is some difference between the current sum and the target. We must cover that difference using as few numbers as possible, while ensuring each added number stays within the range [-limit, limit].

The constraints are important:

• nums.length can be as large as 10^5
• limit can be as large as 10^6
• goal can be as large as 10^9 in magnitude

These constraints immediately tell us that we need a very efficient solution. Any exponential search or dynamic programming over sums would be far too slow. We need something close to linear time.

There are several important edge cases to think about:

• The array sum may already equal goal
• The difference between the current sum and the goal may be negative
• The difference may be very large, requiring multiple added elements
• limit may be small, forcing us to use many elements
• The values involved can exceed 32 bit integer ranges in some languages, so careful integer handling matters

The problem guarantees that every original element already satisfies the limit restriction, so we only need to worry about choosing valid added elements.

Approaches

Brute Force Approach

A brute force approach would try different combinations of numbers to add until the array reaches the target sum.

Suppose the difference between the current sum and the goal is diff. One naive strategy would recursively attempt all possible values from -limit to limit, exploring every possible sequence of additions until the remaining difference becomes zero.

This approach is correct because it eventually explores every valid combination of added numbers. However, it is completely impractical. The branching factor is enormous because each added element can take many possible values, and the depth can also become large.

Even a dynamic programming approach based on reachable sums would be infeasible because the target range can reach billions.

The brute force idea helps reveal the key observation: to minimize the number of added elements, each added element should contribute as much as possible toward closing the remaining difference.

Key Insight

Let:

current_sum = sum(nums)
diff = abs(goal - current_sum)

Each added number can contribute at most limit toward reducing this difference, because:

abs(x) <= limit

Therefore, the optimal strategy is always to use the largest possible contribution each time.

If we still need to cover a remaining difference of diff, then:

• One element can cover at most limit
• Two elements can cover at most 2 * limit
• Three elements can cover at most 3 * limit

So the minimum number of elements needed is simply:

ceil(diff / limit)

This greedy reasoning is optimal because using anything smaller than the maximum allowed contribution can only increase the number of elements required.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	Exponential	Exponential	Tries all possible added values recursively
Optimal	O(n)	O(1)	Compute difference and use greedy math

Algorithm Walkthrough

1. Compute the current sum of the array using all elements in nums.

This tells us where the array currently stands relative to the target. 2. Compute the absolute difference between the current sum and goal.

diff = abs(goal - current_sum)

We use the absolute value because it does not matter whether we need to increase or decrease the sum. The number of required elements depends only on the magnitude of the gap. 3. Determine how much one added element can contribute.

Since every added number must satisfy:

abs(x) <= limit

the maximum contribution from one element is exactly limit. 4. Divide the remaining difference by limit.

diff / limit

This tells us how many full maximum sized contributions we need. 5. Round up the division result.

If the difference is not perfectly divisible by limit, we need one extra element to cover the remaining amount.

Mathematically:

answer = ceil(diff / limit)

6. Return the result.

Why it works

The algorithm works because every added element has a hard upper bound on how much it can change the total sum. No element can contribute more than limit toward the remaining difference. Therefore, the best possible strategy is always to use elements with the maximum allowed magnitude until the gap is closed.

If k elements are used, the largest total adjustment possible is:

k * limit

So we need the smallest k such that:

k * limit >= diff

This is exactly:

ceil(diff / limit)

which proves the greedy formula is optimal.

Python Solution

from typing import List

class Solution:
    def minElements(self, nums: List[int], limit: int, goal: int) -> int:
        current_sum = sum(nums)

        difference = abs(goal - current_sum)

        return (difference + limit - 1) // limit

The implementation starts by computing the sum of all elements in nums. This gives the current array total.

Next, we compute the absolute difference between the current sum and the desired goal. Using the absolute value simplifies the problem because we only care about how large the gap is, not whether it is positive or negative.

Finally, we perform ceiling division using:

(difference + limit - 1) // limit

This is a standard integer arithmetic trick for computing:

ceil(difference / limit)

without using floating point operations.

The algorithm uses only a few variables and performs a single pass through the array to compute the sum.

Go Solution

func minElements(nums []int, limit int, goal int) int {
	currentSum := 0

	for _, num := range nums {
		currentSum += num
	}

	difference := goal - currentSum
	if difference < 0 {
		difference = -difference
	}

	return (difference + limit - 1) / limit
}

The Go implementation follows exactly the same logic as the Python version.

One important Go specific detail is handling absolute values manually for integers:

if difference < 0 {
    difference = -difference
}

Go does not provide a built in absolute value function for integers in the standard library.

Another consideration is integer overflow. In this problem, the constraints safely fit within Go's standard int type on LeetCode environments, so no special handling is needed.

Worked Examples

Example 1

Input: nums = [1, -1, 1], limit = 3, goal = -4

First compute the current sum:

Element	Running Sum
1	1
-1	0
1	1

So:

current_sum = 1

Now compute the difference:

difference = abs(-4 - 1)
           = abs(-5)
           = 5

Each added element can contribute at most 3.

Now compute:

ceil(5 / 3) = 2

Using integer arithmetic:

(5 + 3 - 1) // 3
= 7 // 3
= 2

Answer:

2

One valid construction is adding -3 and -2.

Example 2

Input: nums = [1, -10, 9, 1], limit = 100, goal = 0

Compute the current sum:

Element	Running Sum
1	1
-10	-9
9	0
1	1

So:

current_sum = 1

Difference:

difference = abs(0 - 1)
           = 1

Each element can contribute up to 100.

Therefore:

ceil(1 / 100) = 1

Answer:

1

We can simply add -1.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	We traverse the array once to compute the sum
Space	O(1)	Only a few extra variables are used

The dominant operation is computing the sum of the array, which requires visiting every element exactly once. All remaining operations are constant time arithmetic computations.

The algorithm does not allocate any auxiliary data structures proportional to input size, so the space complexity remains constant.

Test Cases

from typing import List

class Solution:
    def minElements(self, nums: List[int], limit: int, goal: int) -> int:
        current_sum = sum(nums)
        difference = abs(goal - current_sum)
        return (difference + limit - 1) // limit

sol = Solution()

assert sol.minElements([1, -1, 1], 3, -4) == 2  # provided example 1
assert sol.minElements([1, -10, 9, 1], 100, 0) == 1  # provided example 2

assert sol.minElements([1, 2, 3], 10, 6) == 0  # already equals goal
assert sol.minElements([1], 1, 10) == 9  # many additions required
assert sol.minElements([-1, -1, -1], 2, 5) == 4  # positive target from negative sum
assert sol.minElements([5], 5, -5) == 2  # exact multiple of limit
assert sol.minElements([0], 1000000, 1000000000) == 1000  # large values
assert sol.minElements([2, 2], 3, -8) == 4  # negative direction
assert sol.minElements([1], 2, 2) == 1  # small remainder
assert sol.minElements([1], 3, 10) == 3  # ceiling division required

Test	Why
[1, -1, 1], limit=3, goal=-4	Validates provided example
[1, -10, 9, 1], limit=100, goal=0	Validates provided example
[1,2,3], goal=6	Tests already satisfied target
[1], limit=1, goal=10	Tests many required additions
[-1,-1,-1], goal=5	Tests large positive adjustment
[5], limit=5, goal=-5	Tests exact divisibility
[0], limit=1000000, goal=1000000000	Tests large constraints
[2,2], limit=3, goal=-8	Tests negative target movement
[1], limit=2, goal=2	Tests small remaining difference
[1], limit=3, goal=10	Tests ceiling division behavior

Edge Cases

One important edge case occurs when the array sum already equals the goal. In this situation, the difference becomes zero, and the correct answer is also zero. A buggy implementation might accidentally return one because of incorrect ceiling division logic. This implementation handles it correctly because:

(0 + limit - 1) // limit

evaluates to zero.

Another important edge case occurs when the required adjustment is negative. For example, the current sum may be much larger than the goal. Since we use:

abs(goal - current_sum)

the algorithm treats positive and negative adjustments symmetrically. This prevents sign related bugs.

A third important edge case involves differences that are not divisible by limit. Suppose the remaining difference is 10 and limit is 3. Three elements can contribute at most 9, so a fourth element is necessary. The ceiling division formula correctly handles this:

(10 + 3 - 1) // 3
= 12 // 3
= 4

Without ceiling division, an implementation using normal integer division would incorrectly return 3.

A final edge case involves very large values near the problem limits. The difference can become extremely large, especially when goal is near 10^9. Since the algorithm only performs simple arithmetic and does not allocate memory proportional to the target size, it remains efficient and safe even for maximum constraints.