Program to find sum of differences between max and min elements from randomly selected k balls from n balls in Python

Suppose we have n balls which are numbered by an array nums, whose size is n and nums[i] represents the number of ball i. Now we have another value k. In each turn we pick k balls from n different balls and find the difference of maximum and minimum values of k balls and store the difference in a table. Then put these k balls again into that pot and pick again until we have selected all possible selections. Finally find the sum of all differences from the table. If the answer is too large, then return result mod 10^9+7.

So, if the input is like n = 4, k = 3, nums = [5, 7, 9, 11], then the output will be 20 because combinations are −

• [5,7,9], difference 9-5 = 4
• [5,7,11], difference 11-5 = 6
• [5,9,11], difference 11-5 = 6
• [7,9,11], difference 11-7 = 4

so 4+6+6+4 = 20.

Algorithm Approach

The key insight is to calculate the contribution of each element to the total sum. For a sorted array, we can determine how many times each element appears as maximum or minimum in all possible combinations.

To solve this, we will follow these steps −

• m := 10^9 + 7
• inv := a new list with elements [0, 1] for modular inverse calculation
• for i in range 2 to n, do
  • insert (m - floor of (m / i) * inv[m mod i] mod m) at the end of inv
• comb_count := 1 (tracks combination count)
• res := 0
• for pick in range k - 1 to n - 1, do
  • res := res +(nums[pick] - nums[n - 1 - pick]) * comb_count mod m
  • res := res mod m
  • comb_count := comb_count *(pick + 1) mod m * inv[pick + 2 - k] mod m
• return res

Example

Let us see the following implementation to get better understanding −

def solve(n, k, nums):
    m = 10**9 + 7
    
    # Calculate modular inverses
    inv = [0, 1]
    for i in range(2, n + 1):
        inv.append(m - m // i * inv[m % i] % m)
    
    comb_count = 1
    res = 0
    for pick in range(k - 1, n):
        res += (nums[pick] - nums[n - 1 - pick]) * comb_count % m
        res %= m
        comb_count = comb_count * (pick + 1) % m * inv[pick + 2 - k] % m
    
    return res

n = 4
k = 3
nums = [5, 7, 9, 11]
print(solve(n, k, nums))

20

How It Works

The algorithm leverages the fact that in a sorted array, for any combination of k elements, the maximum and minimum are fixed. By calculating the number of times each element contributes as maximum or minimum across all combinations, we can compute the sum efficiently without generating all combinations explicitly.

The modular inverse calculation helps handle large numbers and ensures the result stays within bounds while maintaining mathematical correctness.

Conclusion

This solution efficiently calculates the sum of differences between maximum and minimum elements from all possible k-combinations using combinatorial mathematics. The time complexity is O(n) with proper modular arithmetic handling.