int main () { int *i = (int*) malloc (sizeof(int) * 5); for (int n = 0; n < 5; n++) { i[n] = n;
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- Complete the combinations function to generate all combinations of the characters in the string s with length k recursivelyFind error #include<bits/stdc++.h> using namespace std; class Solution{ public: bool isPossible(vector<int>ank,int n,int mid) { int count=0; for(int i=0;i<rank.size()) { int val= (-1 + sqrt(1+(8*mid)/rank[i]))/2; count+=val; } return count>=n; } int findMinTime(int N, vector<int>&A, int L){ int low=*min_element(A.end()),high=1000000; int ans=high; while(low<=high) { int mid=low+(high)/2; if(isPossible(A,mid)) { ans=mid; high=mid; } else low=mid; } return ans; } }; int main() { int t; cin>>t; while(t--) { int l; cin >> l; vector<int>arr(l); for(int i = 0; i < l; i++){ cin >> arr[i]; } Solution ob; int ans = ob.findMin(n, *arr, l); cout…public class ArraySection { static void arraySection(int a[], int b[]) { int k = 0; int [] c = new int[a.length]; for(int i = 0; i < a.length; i++) { for(int j = 0; j < b.length; j++) if(a[i] == b[j]) c[k++] = a[i]; } for(int i = 0; i < k; i++) System.out.println(c[i]); System.out.println(); } public static void main(String[] args) { int a[] = { 1, 2, 3, 4, 5 }; int b[] = { 0, 2, 4,5 }; arraySection(a,b); }} Calculate the algorithm step number and algorithm time complexity of the above program?
- void func(vector<string>& names){ sort(names.begin(), names.end()); //(a) for(int i = names.size() - 1; i > 0; --i){. //(b) if(names[i] == names[i - 1]){ names[i] = move(names.back()); names.pop_back(); } } } What are the reasons or consequences for the programmer’s choice to loop backwards in the line marked (b), and under what circumstances would the algorithm behave differently if that line were replaced with: for (int i = 1; i < names.size(); ++i) ? What would change?public class arrayOutput ( public static void main (String [] args) { final int NUM ELEMENTS = 3; int[] userVals = new int [NUM_ELEMENTS]; int i; } Type the program's output userVals [0] = 2; userVals [1] = 6; userVals [2] = 8; for (i = userVals.length - 1; i >= 0; −−1) { System.out.println(userVals [1]); } C.C. ? ? ??Consider the Insertion Sort Function below:public static void insertionSort(int array[]){ int n = array.length; for (int j = 1; j < n; j++) { int key = array[j]; int i = j-1; while ( (i > -1) && ( array [i] > key ) ) { array [i+1] = array [i]; i--; } array[i+1] = key;} }Discuss when is its best case, and worst case performance.
- void ExtractMinimumWeightPath(int path[MAX], int weights[MAX][MAX], int pathWeights[MAX][MAX], int rowc, int colc){int rowsCount = rowc;int colsCount = colc;path[rowsCount];int col = 0;int i = 0;for(i=1;i<colsCount;i++){if(pathWeights[rowsCount-1][i]<pathWeights[rowsCount-1][col]){col = i;}} int row = rowsCount-1;do{path[row] = col + 1;if (col>0 && pathWeights[row-1][col-1] + weights[row][col] == pathWeights[row][col]){col = col - 1;} else if (col < colsCount-1 && pathWeights[row-1][col+1] + weights[row][col] == pathWeights[row][col]){col = col + 1;}row--;} while(row>0);path[0] = col + 1;} This is a code in C, I need to change the code as much as possible, but on condition to perform the same work as this. Changing the var names and the loops is enoughConsider the following C program: static int sum = 0; void fill (int* a, int l1) { for (int i = 0; i < 1; i += 1) { a[i] = i * i; sum = sum + a[i]; } int main (int argc, char** argv) { int length = 5; %3D int* array malloc (length * sizeof (int)); %3D fill(array, length); return sum; } Identify each of data spaces that will be exist during the lifetime of this program when it runs, and for each: (a) Specify when it would be allocated. (b) Specify in which region of the address space it would be placed. (c) Indicate how the space is created.DATA STRUCTURES - JAVA
- Background for Question 14-20: Below is a bubble sort program that sorts the elements in an array. static void bubbleSort(int] arr) { int n = arr.length; int temp = 0; for (int i = 0; i arr[]) { temp = arrj - 1]: arrj - 1] = arr[i]: arrj) = temp; } } Based on the program above, please draw a control flow graph for it. In your control flow graph, what are the test requirements for edge coverage? 2. 1. List test path(s) that achieves the edge coverage. 3. Provide test cases for each test path you list in the previous question. If it is not possible to find the test input for certain test path, describe the reason. 4. In your control flow graph, what are the test requirements for edge-pair coverage? 5. List test paths that achieve the edge-pair coverag. 6. Provide test cases for each test path you list in the previous question. If it is not possible to find the test input for certain test path, describe the reason.int sum, k, i, j; int x[4] [4]={1,2,3,4}, {5,6,7,8},{9,8,7,3},{2, 1,7,1}; sum=x[0] [0]; for (k=1; k<=3;k++) sum+=x[k] [k]; Give the value in sum after the statements are executed:Select the appropriate code that performs selection sort. a) int min; for(int j=0; j<arr.length-1; j++) { min = j; for(int k=j+1; k<=arr.length-1; k++) { if(arr[k] < arr[min]) min = k; } int temp = arr[min]; arr[min] = arr[j]; arr[j] = temp; } b) int min; for(int j=0; j<arr.length-1; j++) { min = j; for(int k=j+1; k<=arr.length; k++) { if(arr[k] < arr[min]) min = k; } int temp = arr[min]; arr[min] = arr[j]; arr[j] = temp; } c) int min; for(int j=0; j<arr.length-1; j++) { min = j; for(int k=j+1; k<=arr.length-1; k++) { if(arr[k] > arr[min]) min = k; } int temp = arr[min]; arr[min] = arr[j]; arr[j] = temp; } d) int min; for(int j=0; j<arr.length-1; j++) { min = j; for(int k=j+1; k<=arr.length; k++) { if(arr[k] > arr[min]) min = k; } int temp = arr[min]; arr[min] = arr[j]; arr[j] = temp; }