According to counting sort, which line is not correct?void solve(){ int n; cin >> n; vector<int> a(n); for (int i = 0; i < n; i++) cin >> a[i]; int maxi = *max_element(a.begin(), a.end()); int mini = *min_element(a.begin(), a.end()); vector<int> f(maxi - mini + 2);//Statement A for (int i = 0; i < n; i++) { f[(a[i] - mini)++];//Statement B } for (int i = 0; i < f.size(); i++) { for (int j = 0; j < f[i]; j++) { cout << (i + mini) << " ";// Statement C } }}

Sorting ProgramIn answering each questions, consider the program below:#include <iostream>#include <algorithm>using namespace std;// Function to perform Bubble Sort on an arrayvoid bubbleSort(int arr[], int n) { for (int i = 0; i < n - 1; i++) { for (int j = 0; j < n - 1 - i; j++) { // Swap if the element found is greater than the next element if (arr[j] > arr[j + 1]) { swap(arr[j], arr[j + 1]); } } }}int main() { // Given array of 10 numbers int numbers[] = {4, 2, 7, 1, 9, 5, 8, 3, 10, 6}; int n = sizeof(numbers) / sizeof(numbers[0]); // Call the bubbleSort function to sort the array bubbleSort(numbers, n); // Display the sorted array cout << "Sorted array: "; for (int i = 0; i < n; i++) { cout << numbers[i] << " "; } return 0;}---------------------------------------------------------------------------Definition of Terms in the given questions below:In the context of the Bubble Sort algorithm, a "pass" refers to a complete iteration through the array. During each pass, the algorithm compares adjacent elements and swaps them if they are in the wrong order. After each pass, the largest unsorted element is guaranteed to be in its correct position.The number of passes needed to sort an array of size n is generally n - 1. This is because, after the first pass, the largest element is in its correct place, and after the second pass, the second largest element is in its correct place, and so on. After n - 1 passes, the smallest element is also in its correct place, and the array is fully sorted.For example, if you have an array of 5 elements, you would typically need 4 passes to sort it. Each pass contributes to placing one more element in its correct position within the sorted portion of the array.

What is the worst case time complexity of the following code :/* * V is sorted * V.size() = N * The function is initially called as searchNumOccurrence(V, k, 0, N-1) */int searchNumOccurrence(vector<int> &V, int k, int start, int end) { if (start > end) return 0; int mid = (start + end) / 2; if (V[mid] < k) return searchNumOccurrence(V, k, mid + 1, end); if (V[mid] > k) return searchNumOccurrence(V, k, start, mid - 1); return searchNumOccurrence(V, k, start, mid - 1) + 1 + searchNumOccurrence(V, k, mid + 1, end);}NOTE : This question involves recursion which will be explained later in topic Backtracking. So, if you are not able to approach this question now, you can give it a try later.

std::vector<int> temp; for (int num : arr) { if (num <= x) { temp.push_back(num); } } if (temp.size() < 2) { return -1; } std::sort(temp.begin(), temp.end()); int minDiff = temp[1] - temp[0]; for (int i = 2; i < temp.size(); i++) { minDiff = std::min(minDiff, temp[i] - temp[i - 1]); } return minDiff;}

Which of the following returns the correct minimum of an array of num integers called values? There can be multiple answers. Question 9Answer a. int Min( int *values, int num){ int i; int min=0; for (i=0; i<num; i++) if (values[i] < min) min = values[i]; return min; } b. int Min( int *values, int num){ int i; int mini=num-1; for (i=0; i<num-1; i++) if (values[i] < values[mini]) mini = i; return values[mini]; } c. int Min( int *values, int num){ int i; int mini=0; for (i=1; i<num; i++) if (values[i] < values[mini]) mini = i; return values[mini]; } d. int Min( int *values, int num){ int i; int min= values[0]; for (i=1; i<num; i++) if (values[i] < min) min = values[i]; return min; } e. int Min( int *values, int num){ int i; int mini=0; for (i=0; i<num; i++) if (values[i] < values[mini]) mini = values[i]; return values[mini]; } f. int Min( int *values, int num){ int i; int mini=num; for (i=0; i<num; i++) if (values[i] < values[mini]) mini = i; return values[mini]; }

Which header file is required to use the sort() and stable_sort() functions in C++?Options: Pick one correct answer from below<iostream><string><vector><algorithm>