Problem Description

You need to create an array of n distinct positive integers where no two elements in the array sum up to k. This type of array is called a k-avoiding array.

The task is to find the minimum possible sum of all elements in such a k-avoiding array of length n.

For example, if n = 3 and k = 5, you need to pick 3 distinct positive integers such that no pair of them adds up to 5. One valid k-avoiding array could be [1, 2, 6] since:

• 1 + 2 = 3 (not 5)
• 1 + 6 = 7 (not 5)
• 2 + 6 = 8 (not 5)

The sum would be 1 + 2 + 6 = 9.

The solution uses a greedy approach: starting from i = 1, it tries to add the smallest possible positive integers to the array. When adding a number i, it marks k - i as forbidden (cannot be added later) to ensure no pair sums to k. The process continues until we have n numbers in our array.

The algorithm maintains a set vis to track which numbers cannot be added (because they would form a pair summing to k with an already-added number). For each position in the array, it finds the next smallest integer that isn't in the forbidden set, adds it to the running sum, and marks its complement with respect to k as forbidden.

Intuition

To minimize the sum of the array, we want to use the smallest possible positive integers. The natural strategy is to start with 1, 2, 3, and so on.

However, we face a constraint: no two numbers in our array can sum to k. This means if we include a number i in our array, we cannot include k - i because i + (k - i) = k.

Think of it as a pairing problem. For any positive integer i less than k, there exists a "partner" k - i such that they sum to k. We can choose one from each pair, but not both.

The key insight is that we should greedily pick the smallest available numbers to minimize the total sum. Starting from 1:

• If 1 is available, we take it and ban k - 1
• Move to 2: if it's not banned, take it and ban k - 2
• Continue this process, always taking the next smallest unbanned number

Why does this greedy approach work? Because we're always choosing the smaller number from each conflicting pair. For instance, if k = 10 and we need to choose between 3 and 7 (since 3 + 7 = 10), we pick 3 since it contributes less to the sum.

The algorithm tracks banned numbers in a set called vis (visited). When we add a number i to our result, we mark k - i as visited/banned. If we encounter a banned number while iterating, we skip it and move to the next integer. This ensures we build a valid k-avoiding array with the minimum possible sum.

Learn more about Greedy and Math patterns.

Solution Approach

The solution implements a greedy simulation approach to build the k-avoiding array with minimum sum.

Algorithm Steps:

1. Initialize variables:

   • s = 0: Running sum of the array elements
   • i = 1: Current positive integer candidate to add
   • vis = set(): Set to track forbidden numbers that cannot be added

2. Build the array iteratively:

   • Run a loop n times to add n elements to our array
   • For each iteration:
     • Skip forbidden numbers: While i is in the vis set, increment i to find the next available number
     • Mark the complement: Add k - i to vis to ensure we never add a number that would sum with i to make k
     • Accumulate the sum: Add i to our running sum s
     • Move to next candidate: Increment i for the next iteration

3. Return the result: After adding n elements, return the accumulated sum s

Why this works:

The algorithm ensures we always pick the smallest available positive integer at each step. By marking k - i as visited when we add i, we guarantee that no two numbers in our final array will sum to k.

For example, with n = 3 and k = 5:

• Iteration 1: i = 1, add 1 to sum, mark 5 - 1 = 4 as forbidden
• Iteration 2: i = 2, add 2 to sum, mark 5 - 2 = 3 as forbidden
• Iteration 3: i = 3 is forbidden, skip to i = 4 which is also forbidden, skip to i = 5, add 5 to sum

Final array: [1, 2, 5] with sum = 8

The time complexity is O(n + m) where m is the maximum number we might need to check (bounded by the numbers we skip). The space complexity is O(n) for storing the forbidden numbers in the set.

Example Walkthrough

Let's walk through the solution with n = 5 and k = 6.

We need to find 5 distinct positive integers where no two sum to 6, and we want the minimum possible sum.

Initial State:

• s = 0 (running sum)
• i = 1 (current candidate)
• vis = {} (forbidden numbers set)

Iteration 1:

• Current i = 1, not in vis, so we can use it
• Add 1 to sum: s = 0 + 1 = 1
• Mark k - i = 6 - 1 = 5 as forbidden: vis = {5}
• Move to next: i = 2

Iteration 2:

• Current i = 2, not in vis, so we can use it
• Add 2 to sum: s = 1 + 2 = 3
• Mark k - i = 6 - 2 = 4 as forbidden: vis = {5, 4}
• Move to next: i = 3

Iteration 3:

• Current i = 3, not in vis, so we can use it
• Add 3 to sum: s = 3 + 3 = 6
• Mark k - i = 6 - 3 = 3 as forbidden: vis = {5, 4, 3}
• Move to next: i = 4

Iteration 4:

• Current i = 4 is in vis, skip to i = 5
• Current i = 5 is in vis, skip to i = 6
• Current i = 6, not in vis, so we can use it
• Add 6 to sum: s = 6 + 6 = 12
• Mark k - i = 6 - 6 = 0 as forbidden: vis = {5, 4, 3, 0}
• Move to next: i = 7

Iteration 5:

• Current i = 7, not in vis, so we can use it
• Add 7 to sum: s = 12 + 7 = 19
• Mark k - i = 6 - 7 = -1 as forbidden: vis = {5, 4, 3, 0, -1}
• Move to next: i = 8

Result:

• Final array: [1, 2, 3, 6, 7]
• Final sum: 19

Let's verify no two elements sum to 6:

• 1 + 2 = 3 ✓
• 1 + 3 = 4 ✓
• 1 + 6 = 7 ✓
• 1 + 7 = 8 ✓
• 2 + 3 = 5 ✓
• 2 + 6 = 8 ✓
• 2 + 7 = 9 ✓
• 3 + 6 = 9 ✓
• 3 + 7 = 10 ✓
• 6 + 7 = 13 ✓

None sum to 6, confirming our k-avoiding array is valid with minimum sum 19.

Solution Implementation

Time and Space Complexity

The time complexity is O(n + k). In the worst case, we need to iterate n times to build the array. For each iteration, we may need to skip multiple values that are already in the vis set. The maximum number of values we might need to skip is bounded by the size of vis, which can grow up to O(k) in the worst case. This happens when we're avoiding pairs that sum to k, causing us to mark complementary values (through k - i) in the set. Therefore, the total iterations across all loops is O(n + k).

The space complexity is O(n + k). The vis set stores values to avoid, and in the worst case, it can contain up to O(n) entries (one for each element we add to our result). Additionally, since we're storing k - i for each selected value i, and these complementary values can range up to around k, the set size is bounded by O(n + k).

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

Common Pitfalls

Pitfall 1: Not Handling Edge Cases Where k - i ≤ 0

The Problem: When adding a number i to the array, the code marks k - i as forbidden. However, if i ≥ k, then k - i becomes zero or negative. Since we're only dealing with positive integers, adding negative numbers or zero to the forbidden set is unnecessary and could potentially cause confusion or inefficiency in certain implementations.

Example:

• If k = 5 and we're adding i = 6, we mark 5 - 6 = -1 as forbidden
• This doesn't affect correctness since -1 would never be considered anyway, but it clutters the forbidden set

Solution: Only add k - i to the forbidden set when it's a positive integer:

# Instead of unconditionally adding:
forbidden_numbers.add(k - current_number)

# Check if the complement is positive:
complement = k - current_number
if complement > 0:
    forbidden_numbers.add(complement)

Pitfall 2: Inefficient Memory Usage for Large k Values

The Problem: When k is very large compared to n, we might end up with many unnecessary entries in the forbidden set. For instance, if n = 3 and k = 1000000, we'd mark numbers like 999999, 999998, etc. as forbidden, even though we'd never reach them in our selection process.

Example:

• n = 5, k = 100
• We select [1, 2, 3, 4, 5] and mark [99, 98, 97, 96, 95] as forbidden
• These large numbers were never going to be selected anyway

Solution: Optimize by only marking numbers as forbidden if they're within the range we might actually select:

def minimumSum(self, n: int, k: int) -> int:
    total_sum = 0
    current_number = 1
    forbidden_numbers = set()
  
    for _ in range(n):
        while current_number in forbidden_numbers:
            current_number += 1
      
        # Only mark as forbidden if it could potentially be selected
        complement = k - current_number
        if 0 < complement <= k // 2:  # Optimization: only mark if in selectable range
            forbidden_numbers.add(complement)
      
        total_sum += current_number
        current_number += 1
  
    return total_sum

Pitfall 3: Mathematical Overflow for Large n

The Problem: For very large values of n, the sum could exceed integer limits in some programming languages. While Python handles arbitrary precision integers, this could be an issue in languages like C++ or Java.

Example:

• If n = 10^6, the minimum sum would be at least n * (n + 1) / 2, which could overflow

Solution: For languages with fixed integer sizes, use appropriate data types:

• In C++: Use long long instead of int
• In Java: Use long instead of int
• Consider modular arithmetic if the problem requires it