Problem Description

You need to create an array of positive integers with specific properties and find its minimum possible sum.

Given two positive integers n and target, you must construct a "beautiful" array that satisfies these three conditions:

1. The array has exactly n elements
2. All elements are distinct positive integers (no duplicates allowed)
3. No two different elements in the array can sum to target (if nums[i] + nums[j] == target for any i ≠ j, the array is not beautiful)

Your task is to find the minimum possible sum of all elements in such a beautiful array, and return this sum modulo 10^9 + 7.

For example, if n = 3 and target = 6, you cannot use both 2 and 4 in your array since 2 + 4 = 6. To minimize the sum, you'd want to use the smallest possible positive integers while avoiding pairs that sum to the target. One valid beautiful array could be [1, 2, 7] (where we skip 3, 4, 5 to avoid creating pairs that sum to 6), giving a sum of 10.

The key insight is that if you include a number x in your array, you cannot include target - x to maintain the beautiful property. This creates a constraint on which numbers you can select when trying to minimize the total sum.

Intuition

To minimize the sum, we want to use the smallest possible positive integers. Let's start by trying to pick 1, 2, 3, ... in order.

The key constraint is that if we pick a number x, we cannot pick target - x. This creates pairs that are "forbidden" together. For instance, if target = 10, then (1, 9), (2, 8), (3, 7), (4, 6) are forbidden pairs - we can only pick one number from each pair.

Notice that when x gets close to target/2, we're essentially choosing between two numbers that are roughly equal. For example, with target = 10, the pair (4, 6) has numbers close in value, while (1, 9) has numbers far apart. To minimize our sum, we should always pick the smaller number from each forbidden pair.

This leads us to a critical observation: we can safely pick all numbers from 1 up to ⌊target/2⌋ because for each of these numbers x, its forbidden partner target - x is larger than target/2, so we're always choosing the smaller option.

Let's call m = ⌊target/2⌋. If we need n numbers and n ≤ m, we can simply take 1, 2, 3, ..., n without any conflicts. The sum would be (1 + n) * n / 2.

But what if n > m? We can take all numbers from 1 to m (that's m numbers), but we still need n - m more numbers. Since we've exhausted all numbers less than target/2, we now need to pick numbers that don't have forbidden partners in our chosen set. The safe choice is to start from target itself (since no two numbers in our array can sum to create target, the number target has no forbidden partner in our array) and continue with target + 1, target + 2, and so on.

So when n > m, our array consists of:

• The numbers 1, 2, ..., m with sum (1 + m) * m / 2
• The numbers target, target + 1, ..., target + (n - m - 1) with sum (target + target + n - m - 1) * (n - m) / 2

This greedy approach guarantees we're always picking the smallest available numbers while respecting the constraint, thus minimizing the total sum.

Learn more about Greedy and Math patterns.

Solution Approach

The implementation follows a greedy mathematical approach based on our intuition about forbidden pairs.

First, we define the modulo value mod = 10^9 + 7 as required by the problem.

We calculate m = target // 2, which represents the threshold - all numbers from 1 to m can be safely included without creating any pairs that sum to target.

Case 1: When n <= m

If we need at most m numbers, we can simply take the first n positive integers: 1, 2, 3, ..., n. These are the smallest possible positive integers, and none of their pairs sum to target (since the maximum possible sum would be (m-1) + m < target).

The sum is calculated using the arithmetic series formula:

return ((1 + n) * n // 2) % mod

Case 2: When n > m

We need more than m numbers, so we:

1. Take all numbers from 1 to m (giving us m numbers)
2. Skip numbers from m + 1 to target - 1 (these would create forbidden pairs with our already chosen numbers)
3. Take numbers starting from target onwards: target, target + 1, ..., target + (n - m - 1)

The total sum consists of two parts:

• Sum of 1 to m: (1 + m) * m // 2
• Sum of n - m consecutive numbers starting from target: (target + target + n - m - 1) * (n - m) // 2

The second part uses the arithmetic series formula where:

• First term = target
• Last term = target + (n - m - 1)
• Number of terms = n - m

return ((1 + m) * m // 2 + (target + target + n - m - 1) * (n - m) // 2) % mod

The algorithm runs in O(1) time complexity since we're only performing mathematical calculations without any loops or recursion. The space complexity is also O(1) as we only use a constant amount of extra space.

Example Walkthrough

Let's walk through the solution with n = 5 and target = 8.

First, we calculate m = target // 2 = 8 // 2 = 4. This means we can safely use numbers 1 through 4 without creating any pairs that sum to 8.

Since n = 5 > m = 4, we fall into Case 2.

Step 1: Choose numbers from 1 to m We take: [1, 2, 3, 4]

• Note: We can't also include 7, 6, 5, or 4 (already chosen) as they would form forbidden pairs:
  • 1 + 7 = 8 (forbidden)
  • 2 + 6 = 8 (forbidden)
  • 3 + 5 = 8 (forbidden)
  • 4 + 4 = 8 (but we only have one 4)

Step 2: We still need n - m = 5 - 4 = 1 more number Starting from target = 8, we take the next available number: [8]

• Why 8 is safe: No other number in our array can sum with 8 to give 8 (that would require 0, which isn't a positive integer)

Final array: [1, 2, 3, 4, 8]

Calculating the sum:

• Sum of 1 to 4: (1 + 4) * 4 / 2 = 5 * 2 = 10
• Sum of numbers starting from 8 (just one number): 8
• Total sum: 10 + 8 = 18

Using the formula:

• Part 1: (1 + m) * m // 2 = (1 + 4) * 4 // 2 = 10
• Part 2: (target + target + n - m - 1) * (n - m) // 2 = (8 + 8 + 0) * 1 // 2 = 8
• Total: 10 + 8 = 18

The minimum possible sum is 18.

Solution Implementation

Time and Space Complexity

The time complexity is O(1) because the algorithm performs a fixed number of arithmetic operations regardless of the input size. The code consists of:

• A few variable assignments (mod, m)
• A single conditional check (if n <= m)
• A constant number of arithmetic operations (addition, multiplication, division, modulo)

All these operations execute in constant time, and there are no loops or recursive calls that depend on the input size.

The space complexity is O(1) because the algorithm uses only a fixed amount of extra space. The variables stored are:

• mod: a constant integer
• m: computed from target
• The intermediate results of arithmetic expressions

The space usage does not grow with the input size n or target, maintaining constant space complexity.

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

Common Pitfalls

1. Integer Overflow in Arithmetic Operations

The Pitfall: The most critical issue in this solution is potential integer overflow when calculating sums, especially for large values of n and target. In Python, while integers can grow arbitrarily large, performing modulo operations after large multiplications can still cause performance issues. More importantly, if this code were translated to languages like Java or C++, integer overflow would cause incorrect results.

For example, when calculating (target + target + remaining_count - 1) * remaining_count // 2, the intermediate multiplication could exceed the maximum integer value before the modulo operation is applied.

The Solution: Apply modulo operations more frequently during calculations to keep numbers manageable:

def minimumPossibleSum(self, n: int, target: int) -> int:
    MOD = 10**9 + 7
    max_before_target_half = target // 2
  
    if n <= max_before_target_half:
        # Apply modulo after calculation to prevent overflow
        sum_result = (n * (n + 1) // 2) % MOD
        return sum_result
  
    # Calculate first part with modulo
    sum_first_part = (max_before_target_half * (max_before_target_half + 1) // 2) % MOD
  
    # Calculate second part with modulo
    remaining_count = n - max_before_target_half
    first_term = target % MOD
    last_term = (target + remaining_count - 1) % MOD
    sum_second_part = ((first_term + last_term) % MOD * remaining_count % MOD * pow(2, MOD - 2, MOD)) % MOD
  
    return (sum_first_part + sum_second_part) % MOD

2. Off-by-One Error in Boundary Calculation

The Pitfall: A common mistake is incorrectly determining which numbers can be safely included. The code uses target // 2 as the boundary, but developers might accidentally use target / 2 (float division) or (target - 1) // 2, leading to incorrect results.

The Solution: Be explicit about the boundary logic and add comments:

# For any number x where x <= target//2, its pair (target - x) will be >= target//2 + 1
# This ensures no two numbers in range [1, target//2] sum to target
max_safe_value = target // 2  # Use integer division explicitly

3. Incorrect Arithmetic Series Formula Application

The Pitfall: When calculating the sum of consecutive integers starting from target, it's easy to miscalculate the last term or the count of terms. The formula requires:

• First term: target
• Last term: target + (n - max_before_target_half - 1)
• Number of terms: n - max_before_target_half

A common error is using target + n - max_before_target_half as the last term (forgetting to subtract 1).

The Solution: Break down the calculation with clear variable names:

remaining_count = n - max_before_target_half
first_value = target
last_value = target + remaining_count - 1  # -1 because we're counting from 0
sum_second_part = (first_value + last_value) * remaining_count // 2