Problem Description

You have an integer array nums where each element satisfies the constraint abs(nums[i]) <= limit. You also have two integers: limit and goal.

Your task is to find the minimum number of elements you need to add to the array so that the sum of all elements equals exactly goal. Any elements you add must also satisfy the constraint that their absolute value is at most limit.

For example, if you have nums = [1, -1, 1], limit = 3, and goal = -4, the current sum is 1 + (-1) + 1 = 1. To reach the goal of -4, you need a difference of -5. You could add two elements: -3 and -2 (both satisfy abs(x) <= 3), making the new sum 1 + (-3) + (-2) = -4. So the answer would be 2.

The key insight is that to minimize the number of elements added, you should add elements with the maximum allowed absolute value (limit or -limit) to cover the difference as quickly as possible. The solution calculates the absolute difference between the current sum and the goal, then determines how many maximum-value elements are needed to cover this difference, which is ⌈|difference|/limit⌉.

Intuition

Let's think about this problem step by step. We have a current sum of the array, and we need to reach a target goal. The difference between them is what we need to make up by adding new elements.

Since we want to minimize the number of elements we add, we should be greedy - each element we add should contribute as much as possible toward closing the gap. Given the constraint that abs(nums[i]) <= limit, the maximum contribution any single element can make is limit (if we need to increase the sum) or -limit (if we need to decrease the sum).

For example, if our current sum is 10 and our goal is 25, we need to add 15 more to the sum. If limit = 7, we should add elements with value 7 to maximize each element's contribution. We'd need ⌈15/7⌉ = 3 elements: two 7s and one 1.

Similarly, if our current sum is 10 and our goal is -5, we need to subtract 15 from the sum. We'd add elements with value -7 to maximize the contribution in the negative direction. Again, we'd need ⌈15/7⌉ = 3 elements.

The beauty of this approach is that it doesn't matter whether we need to increase or decrease the sum - we only care about the absolute difference d = abs(sum(nums) - goal). We can always choose to add either limit or -limit based on which direction we need to go. The minimum number of elements needed is simply the ceiling of d/limit, which can be computed as (d + limit - 1) // limit using integer arithmetic.

Learn more about Greedy patterns.

Solution Approach

The implementation is straightforward once we understand the greedy approach:

1. Calculate the current sum: First, we compute the sum of all elements in the array nums. This gives us our starting point.

2. Find the difference: We calculate the absolute difference between the current sum and the goal: d = abs(sum(nums) - goal). This tells us how much we need to add (in absolute terms) to reach the goal.

3. Calculate minimum elements needed: To minimize the number of elements, we want each added element to have the maximum possible absolute value, which is limit. The number of such elements needed is ⌈d/limit⌉.

   To compute the ceiling division without using floating-point arithmetic, we use the formula: (d + limit - 1) // limit. This is a common pattern for ceiling division with integers.

Why does (d + limit - 1) // limit give us the ceiling?

• If d is divisible by limit, then (d + limit - 1) // limit = d // limit
• If d has a remainder when divided by limit, adding (limit - 1) ensures we round up to the next integer

Important Note on Data Types: As mentioned in the reference, with array elements in the range [-10^6, 10^6] and up to 10^5 elements, the sum could be as large as 10^11 in absolute value. This exceeds the range of 32-bit integers, so 64-bit integers should be used in languages where this matters (Python handles this automatically).

The entire solution condenses to just three lines: calculate sum, find absolute difference, and return the ceiling division result.

Example Walkthrough

Let's walk through a concrete example to see how this solution works.

Example: nums = [2, -3, 1], limit = 4, goal = 10

Step 1: Calculate the current sum

• Sum of array: 2 + (-3) + 1 = 0

Step 2: Find the difference

• We need to go from 0 to 10
• Difference: d = |0 - 10| = 10

Step 3: Determine minimum elements needed

• We need to cover a gap of 10 using elements with maximum absolute value of 4
• Since we need to increase the sum (from 0 to 10), we'll add positive elements
• Using the greedy approach, we add elements with value 4 (the maximum allowed)
• How many 4s do we need? ⌈10/4⌉ = ⌈2.5⌉ = 3
• Using integer arithmetic: (10 + 4 - 1) // 4 = 13 // 4 = 3

Verification:

• We add three elements: 4, 4, and 2
• New array: [2, -3, 1, 4, 4, 2]
• New sum: 2 + (-3) + 1 + 4 + 4 + 2 = 10 ✓

Another Example: nums = [5, 2], limit = 3, goal = -2

Step 1: Current sum = 5 + 2 = 7

Step 2: Difference = |7 - (-2)| = 9

Step 3: Minimum elements = (9 + 3 - 1) // 3 = 11 // 3 = 3

• Since we need to decrease from 7 to -2, we add three -3s
• New sum: 7 + (-3) + (-3) + (-3) = -2 ✓

The key insight is that we always use elements with absolute value equal to limit to minimize the count, only using a smaller value for the last element if needed.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. This is because the sum(nums) operation needs to iterate through all elements in the array once to calculate the total sum. The remaining operations - subtraction, absolute value calculation, and integer division - are all O(1) constant time operations.

The space complexity is O(1) as the algorithm only uses a constant amount of extra space. The variable d stores a single integer value, and no additional data structures that scale with the input size are created. The sum() function computes the sum iteratively without creating additional space proportional to the input.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow in Languages with Fixed Integer Types

The Pitfall: In languages like Java, C++, or C#, using 32-bit integers (int) can cause overflow when calculating the sum of the array. With up to 10^5 elements each having values up to ±10^6, the sum can reach ±10^11, which exceeds the 32-bit integer range (approximately ±2.1 × 10^9).

Example of the Problem:

// Java - WRONG approach
public int minElements(int[] nums, int limit, int goal) {
    int sum = 0;  // Using 32-bit int - can overflow!
    for (int num : nums) {
        sum += num;
    }
    int difference = Math.abs(sum - goal);  // Can also overflow!
    return (difference + limit - 1) / limit;
}

The Solution: Use 64-bit integers (long in Java/C#, long long in C++) for all sum and difference calculations:

// Java - CORRECT approach
public int minElements(int[] nums, int limit, int goal) {
    long sum = 0;  // Using 64-bit long
    for (int num : nums) {
        sum += num;
    }
    long difference = Math.abs(sum - (long)goal);  // Cast goal to long
    return (int)((difference + limit - 1) / limit);
}

2. Incorrect Ceiling Division Formula

The Pitfall: Attempting to use regular division and then rounding up, which can lead to floating-point precision issues or incorrect implementations:

# WRONG approaches:
result = int(difference / limit) + (1 if difference % limit != 0 else 0)  # Verbose
result = int(difference / limit) + int(difference % limit > 0)  # Still verbose
result = math.ceil(difference / limit)  # Works but requires import

The Solution: Use the integer arithmetic formula (difference + limit - 1) // limit which handles all cases correctly without floating-point operations or conditional logic.

3. Forgetting the Absolute Value

The Pitfall: Not taking the absolute value of the difference can lead to negative results when the current sum exceeds the goal:

# WRONG - missing abs()
difference = sum(nums) - goal  # Could be negative!
return (difference + limit - 1) // limit  # Wrong result for negative difference

The Solution: Always use abs(sum(nums) - goal) to ensure you're working with the magnitude of the difference, regardless of whether you need to add positive or negative values to reach the goal.