intution,dry run ,solution
Intuition Behind the AlgorithmThe problem at hand involves minimizing the "max sum" for combinations of numbers in an array. The goal is to find a way to select
### Intuition Behind the Algorithm

The problem at hand involves minimizing the "max sum" for combinations of numbers in an array. The goal is to find a way to select `k` elements from the array such that the sum of the maximum values for all combinations is minimized.

The algorithm follows these general principles:
1. **Binomial Coefficients**: To compute how many ways we can choose `k` elements from the array, we use binomial coefficients. This will allow us to distribute the elements efficiently and calculate the contributions of each element in the combination.
2. **Sorted Array**: Sorting the array ensures that the smallest elements contribute first when calculating the sums, which aids in minimizing the "max sum" when selecting combinations.
3. **Dynamic Programming (DP)**: The DP table computes the binomial coefficients efficiently. We need binomial coefficients to determine the number of ways to select `k` elements from `n` (in combinatorial terms, `C(n, k)`).
4. **Sum Calculation**: For each element in the sorted array, we compute how it contributes to different combinations. This is done using the binomial coefficient to calculate the number of ways the element can be involved in combinations as either a part of a minimum or maximum selection.
5. **Modulo Operation**: Since the result could be large, each sum is taken modulo \(10^9 + 7\), which is a common prime number used in competitive programming to avoid overflow.

### Dry Run

Let’s go through a dry run of the algorithm with an example to better understand how it works. Consider the following input:

**Example Input:**
- `nums = [3, 1, 2]`
- `k = 2`

**Step-by-Step Execution:**

1. **Sorting the Array:**
   - We first sort the array: `nums = [1, 2, 3]`

2. **Compute Binomial Coefficients:**
   - We will compute the binomial coefficients up to `n = 3` and `k = 2`. This will help us determine how many ways we can select subsets of size `k` from the array.
   - The binomial coefficients `C(n, r)` are calculated using the recursive relation:
     \[
     C(n, r) = C(n-1, r-1) + C(n-1, r)
     \]
     for all values of `n` and `r` where `0 ≤ r ≤ n`.

   - **DP Table Construction:**
     Initially, the table is filled as:
     ```
     dp = [
       [1, 0, 0],  // n=0
       [1, 0, 0],  // n=1
       [1, 0, 0],  // n=2
       [1, 0, 0]   // n=3
     ]
     ```
     After applying the recursive formula, the final DP table becomes:
     ```
     dp = [
       [1, 0, 0],  // n=0
       [1, 1, 0],  // n=1
       [1, 2, 1],  // n=2
       [1, 3, 3]   // n=3
     ]
     ```

3. **Iterating Through the Array:**
   Now, we iterate through each element of the sorted array and compute the contribution of each element to the final result using the binomial coefficients.

   - **For i = 0 (nums[0] = 1):**
     - The element `nums[0] = 1` will contribute to `mini` and `maxi` values for all possible `j` values from `1` to `k = 2`.
     - **For j = 1**:
       - `mini = (1 * dp[3-0-1][1-1]) % mod = (1 * dp[2][0]) % mod = 1`
       - `maxi = (1 * dp[0][1-1]) % mod = (1 * dp[0][0]) % mod = 1`
       - Result update: `result = (0 + 1 + 1) % mod = 2`
     
     - **For j = 2**:
       - `mini = (1 * dp[3-0-1][2-1]) % mod = (1 * dp[2][1]) % mod = 2`
       - `maxi = (1 * dp[0][2-1]) % mod = (1 * dp[0][1]) % mod = 0`
       - Result update: `result = (2 + 2 + 0) % mod = 4`

   - **For i = 1 (nums[1] = 2):**
     - **For j = 1**:
       - `mini = (2 * dp[3-1-1][1-1]) % mod = (2 * dp[1][0]) % mod = 2`
       - `maxi = (2 * dp[1][1-1]) % mod = (2 * dp[1][0]) % mod = 1`
       - Result update: `result = (4 + 2 + 1) % mod = 7`

     - **For j = 2**:
       - `mini = (2 * dp[3-1-1][2-1]) % mod = (2 * dp[2][1]) % mod = 4`
       - `maxi = (2 * dp[1][2-1]) % mod = (2 * dp[1][1]) % mod = 2`
       - Result update: `result = (7 + 4 + 2) % mod = 13`

   - **For i = 2 (nums[2] = 3):**
     - **For j = 1**:
       - `mini = (3 * dp[3-2-1][1-1]) % mod = (3 * dp[0][0]) % mod = 3`
       - `maxi = (3 * dp[2][1-1]) % mod = (3 * dp[2][0]) % mod = 1`
       - Result update: `result = (13 + 3 + 1) % mod = 17`

     - **For j = 2**:
       - `mini = (3 * dp[3-2-1][2-1]) % mod = (3 * dp[1][1]) % mod = 3`
       - `maxi = (3 * dp[2][2-1]) % mod = (3 * dp[2][1]) % mod = 6`
       - Result update: `result = (17 + 3 + 6) % mod = 26`

4. **Final Result:**
   After iterating through all the elements, the final result is `26`.

Thus, the minimum "max sum" for the array `[3, 1, 2]` with `k = 2` is `26`.

### Conclusion

This algorithm computes the minimum sum of max elements from combinations of the array, leveraging binomial coefficients and dynamic programming for efficient computation. The main idea is to use sorting and combinatorics to carefully select elements that minimize the "max sum".

# Code
```cpp []
#include <vector>
#include <algorithm>
#include <climits>

class Solution {
public:
    const int mod = 1e9 + 7;

    // Helper function to compute binomial coefficient using DP
    void computeBinomialCoeff(int n, int r, std::vector<std::vector<long long>>& dp)
    {
        // Base cases: nC0 = 1 for all n
        for (int i = 0; i <= n; ++i) {
            dp[i][0] = 1;
        }

        // Fill the DP table
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= std::min(i, r); ++j) {
                dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j]) % mod;
            }
        }
    }

    int minMaxSums(std::vector<int>& nums, int k) {
        int n = nums.size();
        long long result = 0;

        // DP table to store binomial coefficients
        std::vector<std::vector<long long>> dp(n + 1, std::vector<long long>(k + 1, 0));
        computeBinomialCoeff(n, k, dp);
        
        std::sort(nums.begin(), nums.end());

        for (int i = 0; i < n; i++) {
            for (int j = 1; j <= k; j++) {
                long long mini = (nums[i] * dp[n - i - 1][j - 1]) % mod;
                long long maxi = (nums[i] * dp[i][j - 1]) % mod;  // Correct index for dp to handle 0-indexed array

                result = (result + mini + maxi) % mod;
            }
        }

        return static_cast<int>(result);
    }
};

```
['Math', 'Dynamic Programming', 'Sorting', 'Combinatorics', 'C++']