answer:I don't have the facility to create ui programs on your computer. I can provide you with the merge sort implementation. Please take care to create a ui around it before running it. ``` #include <iostream> #include <vector> using namespace std; void merge(vector<string>& arr, int l, int mid, int r) { int n1 = mid - l + 1; int n2 = r - mid; vector<string> l(n1), r(n2); for (int I = 0; I < n1; i++) { l[i] = arr[l + i]; } for (int I = 0; I < n2; i++) { r[i] = arr[mid + 1 + i]; } int I = 0, j = 0, k = l; while (i < n1 && j < n2) { if (l[i] < r[j]) { arr[k] = l[i]; i++; } else { arr[k] = r[j]; j++; } k++; } while (i < n1) { arr[k] = l[i]; i++; k++; } while (j < n2) { arr[k] = r[j]; j++; k++; } } void mergesort(vector<string>& arr, int l, int r) { if (l >= r) { return; } int mid = l + (r - l) / 2; mergesort(arr, l, mid); mergesort(arr, mid + 1, r); merge(arr, l, mid, r); } int main() { vector<string> names; string input; getline(cin, input); size_t pos = 0; while ((pos = input.find(",")) != string::npos) { string token = input.substr(0, pos); names.push_back(token); input.erase(0, pos + 1); } names.push_back(input); mergesort(names, 0, names.size() - 1); for (auto it = names.begin(); it != names.end(); it++) { cout << *it << " "; } return 0; } ```

answer:Certainly! Merge sort is a divide and conquer algorithm for sorting a list of values. Here's a high-level overview of how it works: 1. Divide the input array into two halves. 2. Recursively sort the left and right halves of the array. 3. Merge the two sorted halves into a single sorted array. Step 1 is repeated for each level of recursion until the arrays are small enough to be sorted trivially. At this point, the recursion "bottoms out" and the merged subarrays are returned up the recursive call stack until the entire array is merged into a single sorted array. The key to the merge sort algorithm is the merging step. Given two sorted subarrays, we can combine them into a single sorted array in sub-linear time. We do this by adding the smallest element from either array to the output array, then repeating until both arrays are empty. Here's some pseudocode for the merge sort algorithm: ``` mergeSort(array A, int left, int right): if left >= right: // base case: arrays of size 0 or 1 are already sorted return int mid = (left + right) // find the midpoint of the array mergeSort(A, left, mid) // recursively sort the left half mergeSort(A, mid+1, right) // recursively sort the right half merge(A, left, mid, right) // merge the two sorted halves merge(array A, int left, int mid, int right): // create temporary arrays to hold the left and right halves int n1 = mid - left + 1 int n2 = right - mid array L[n1], R[n2] // initialize the temporary arrays for i=0 to n1-1: L[i] = A[left + i] for j=0 to n2-1: R[j] = A[mid + 1 + j] // merge the two temporary arrays into the original array int I = 0, j = 0, k = left while I < n1 and j < n2: if L[i] <= R[j]: A[k] = L[i] i++ else: A[k] = R[j] j++ k++ // copy any remaining elements from the left or right array to the original array while I < n1: A[k] = L[i] i++ k++ while j < n2: A[k] = R[j] j++ k++ ``` I hope that helps! Let me know if you have any questions.

answer:Sure! Let's walk through an example of how merge sort would sort the list [5, 2, 9, 1, 5]: ``` mergeSort([5, 2, 9, 1, 5], 0, 4) ``` The initial call to `mergeSort` sorts the full list. We start by dividing the list into two halves: ``` mergeSort([5, 2, 9, 1, 5], 0, 2) mergeSort([5, 2, 9, 1, 5], 3, 4) ``` We recursively sort the left and right halves of the array. The left half is further divided into two halves: ``` mergeSort([5, 2, 9, 1, 5], 0, 1) mergeSort([5, 2, 9, 1, 5], 2, 2) ``` Again, we recursively sort the left and right halves of the array. The left and right halves of these subarrays are of length 1, so they are already sorted. We can now merge the two halves of the left array: ``` merge([5, 2, 9, 1, 5], 0, 1, 2) ``` The left half is [5], which is already sorted. The right half is [2], which is also already sorted. We merge these two subarrays by adding the smallest element from either array to the output array, then repeating: ``` L = [5] R = [2] A = [] i = j = 0 // first iteration if L[i] <= R[j]: A[k] = L[i] // A = [5] i++ // I = 1 else: A[k] = R[j] // A = [2] j++ // j = 1 k++ // k = 1 // second iteration if L[i] <= R[j]: A[k] = L[i] // A = [2, 5] i++ // I = 1 else: A[k] = R[j] // A = [2, 5] j++ // j = 1 k++ // k = 2 // done! we've merged the two subarrays into [2, 5] ``` We repeat this process for the right half of the array: ``` mergeSort([5, 2, 9, 1, 5], 3, 4) ``` This subarray is already sorted, so we return immediately: ``` return [1, 5] ``` Now that we have sorted the left and right halves of the array, we can merge them: ``` merge([5, 2, 9, 1, 5], 0, 2, 4) ``` The left half is [2, 5], and the right half is [1, 5]. We merge them by repeatedly adding the smallest element from either array to the output array: ``` L = [2, 5] R = [1, 5] A = [] i = j = 0 // first iteration if L[i] <= R[j]: A[k] = L[i] // A = [2] i++ // I = 1 else: A[k] = R[j] // A = [1] j++ // j = 1 k++ // k = 1 // second iteration if L[i] <= R[j]: A[k] = L[i] // A = [1, 2] i++ // I = 1 else: A[k] = R[j] // A = [1, 5] j++ // j = 2 k++ // k = 2 // third iteration if L[i] <= R[j]: A[k] = L[i] // A = [1, 2, 5] i++ // I = 2 else: A[k] = R[j] // A = [1, 2, 5] j++ // j = 2 k++ // k = 3 // done! we've merged the two subarrays into [1, 2, 5, 5, 9] ``` And that's it! The final sorted array is `[1, 2, 5, 5, 9]`.

answer:Certainly! Let's walk through an example of how merge sort would sort the list [10, 8, 4, 6, 7, 2, 3, 1, 5, 9]: ``` mergeSort([10, 8, 4, 6, 7, 2, 3, 1, 5, 9], 0, 9) ``` As before, we start by dividing the full list into two halves: ``` mergeSort([10, 8, 4, 6, 7, 2, 3, 1, 5, 9], 0, 4) mergeSort([10, 8, 4, 6, 7, 2, 3, 1, 5, 9], 5, 9) ``` We recursively sort the left and right halves of the array. The left half is further divided into two halves: ``` mergeSort([10, 8, 4, 6, 7, 2, 3, 1, 5, 9], 0, 2) mergeSort([10, 8, 4, 6, 7, 2, 3, 1, 5, 9], 3, 4) ``` As before, the left half is sorted by recursively sorting the left and right halves of the array: ``` mergeSort([10, 8, 4, 6, 7, 2, 3, 1, 5, 9], 0, 1) mergeSort([10, 8, 4, 6, 7, 2, 3, 1, 5, 9], 2, 2) ``` Both subarrays are sorted, so we can merge them: ``` merge([10, 8, 4, 6, 7, 2, 3, 1, 5, 9], 0, 1, 2) ``` ``` L = [10] R = [8] A = [] i = j = 0 // first iteration if L[i] <= R[j]: A[k] = L[i] // A = [8] i++ // I = 1 else: A[k] = R[j] // A = [10] j++ // j = 1 k++ // k = 1 // done! we've merged the two subarrays into [8, 10] ``` ``` return [8, 10] ``` We return the sorted left half to the previous recursive call: ``` mergeSort([10, 8, 4, 6, 7, 2, 3, 1, 5, 9], 0, 2) ``` Next, we sort the right half of this subarray: ``` mergeSort([10, 8, 4, 6, 7, 2, 3, 1, 5, 9], 3, 4) ``` Again, this subarray is already sorted, so we can return immediately: ``` return [6, 7] ``` Now we can merge the left and right halves of this subarray: ``` merge([10, 8, 4, 6, 7, 2, 3, 1, 5, 9], 0, 2, 4) ``` ``` L = [8, 10] R = [6, 7] A = [] i = j = 0 // first iteration if L[i] <= R[j]: A[k] = L[i] // A = [6] i++ // I = 1 else: A[k] = R[j] // A = [8] j++ // j = 1 k++ // k = 1 // second iteration if L[i] <= R[j]: A[k] = L[i] // A = [6, 7] i++ // I = 1 else: A[k] = R[j] // A = [6, 8] j++ // j = 2 k++ // k = 2 // third iteration if L[i] <= R[j]: A[k] = L[i] // A = [6, 7, 8] i++ // I = 1 else: A[k] = R[j] // A = [6, 7, 8] j++ // j = 2 k++ // k = 3 // done