alternatingSum: Given a list of integers, compule the alternating sum of its elements. 3 (bc 1 10 (b/c e # [1, 2, 3, 4, 5] -> # [@ , 1© , 20] -> : [9, 9, 9, 9] -> o (b/c 9 - 9 + 9 - 9 = 0) def alternatingSum(nums): 2 + 3 - 4 + 5 = 3) 10 + 20 = O) %3D
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- 2 = 0 ) { numList [ i ] i } } } [1, 2, 3, 4, 5] [0, 2, 4, 6, 8] [2, 4, 6, 8, 10] [1, 3, 5, 7, 9]В.width; } { t = B.type; w = { T.type = C.type; T.width = C.width; } T → B C В > int { B.type = integer; B.width 4; } В — foat { B.type = float; B.width = 8; } { C.type = t; C.width = w; } C - [ num] C1 = array(num. value, C1.type); { C.type C.width = num. value x C1. width; }#ifndef lab5ExF_h #define lab5ExF_h typedef struct point { char label[10]; double x ; // x coordinate for point in a Cartesian coordinate system double y; // y coordinate for point in a Cartesian coordinate system double z; // z coordinate for point in a Cartesian coordinate system }Point; void reverse (Point *a, int n); /* REQUIRES: Elements a[0] ... a[n-2], a[n-1] exists. * PROMISES: places the existing Point objects in array a, in reverse order. * The new a[0] value is the old a[n-1] value, the new a[1] is the * old a[n-2], etc. */ int search(const Point* struct_array, const char* target, int n); /* REQUIRES: Elements struct-array[0] ... struct_array[n-2], struct_array[n-1] * exists. target points to string to be searched for. * PROMISES: returns the index of the element in the array that contains an * instance of point with a matching label. Otherwise, if there is * no point in the array that its label matches the target-label, * it should return -1. * If there are more than…
- #include <stdio.h> int arrC[10] = {0}; int bSearch(int arr[], int l, int h, int key); int *joinArray(int arrA[], int arrB[]) { int j = 0; if ((arrB[0] + arrB[4]) % 5 == 0) { arrB[0] = 0; arrB[4] = 0; } for (int i = 0; i < 5; i++) { arrC[j++] = arrA[i]; if (arrB[i] == 0 || (bSearch(arrA, 0, 5, arrB[i]) != -1)) { continue; } else arrC[j++] = arrB[i]; } for (int i = 0; i < j; i++) { int temp; for (int k = i + 1; k < j; k++) { if (arrC[i] > arrC[k]) { temp = arrC[i]; arrC[i] = arrC[k]; arrC[k] = temp; } } } for (int i = 0; i < j; i++) { printf("%d ", arrC[i]); } return arrC; } int bSearch(int arr[], int l, int h, int key) { if (h >= l) { int mid = l + (h - l) / 2; if…#ifndef LLCP_INT_H#define LLCP_INT_H #include <iostream> struct Node{ int data; Node *link;}; bool DelOddCopEven(Node* headPtr);int FindListLength(Node* headPtr);bool IsSortedUp(Node* headPtr);void InsertAsHead(Node*& headPtr, int value);void InsertAsTail(Node*& headPtr, int value);void InsertSortedUp(Node*& headPtr, int value);bool DelFirstTargetNode(Node*& headPtr, int target);bool DelNodeBefore1stMatch(Node*& headPtr, int target);void ShowAll(std::ostream& outs, Node* headPtr);void FindMinMax(Node* headPtr, int& minValue, int& maxValue);double FindAverage(Node* headPtr);void ListClear(Node*& headPtr, int noMsg = 0); // prototype of DelOddCopEven of Assignment 5 Part 1 #endif // definition of DelOddCopEven of Assignment 5 Part 1//Algorithm should: /*NOT destroy any of the originally even-valued node. This means that the originally even-valued nodes should be retained as part of the resulting list. Destroy…#ifndef LLCP_INT_H#define LLCP_INT_H #include <iostream> struct Node{ int data; Node *link;};void DelOddCopEven(Node*& headPtr);int FindListLength(Node* headPtr);bool IsSortedUp(Node* headPtr);void InsertAsHead(Node*& headPtr, int value);void InsertAsTail(Node*& headPtr, int value);void InsertSortedUp(Node*& headPtr, int value);bool DelFirstTargetNode(Node*& headPtr, int target);bool DelNodeBefore1stMatch(Node*& headPtr, int target);void ShowAll(std::ostream& outs, Node* headPtr);void FindMinMax(Node* headPtr, int& minValue, int& maxValue);double FindAverage(Node* headPtr);void ListClear(Node*& headPtr, int noMsg = 0); // prototype of DelOddCopEven of Assignment 5 Part 1 #endif
- x=(-10:-1:-15;-2,3]; How many elements are generated in x 12 O 5 O 10 8 No elements; errorSKELETON CODE IS PROVIDED ALONG WITH C AND H FILES. #include <stdio.h> #include <stdlib.h> #include <string.h> #include <stdbool.h> #include "node.h" #include "stack_functions.h" #define NUM_VERTICES 10 /** This function takes a pointer to the adjacency matrix of a Graph and the size of this matrix as arguments and prints the matrix */ void print_graph(int * graph, int size); /** This function takes a pointer to the adjacency matrix of a Graph, the size of this matrix, the source and dest node numbers along with the weight or cost of the edge and fills the adjacency matrix accordingly. */ void add_edge(int * graph, int size, int src, int dst, int cost); /** This function takes a pointer to the adjacency matrix of a graph, the size of this matrix, source and destination vertex numbers as inputs and prints out the path from the source vertex to the destination vertex. It also prints the total cost of this…T/F Suffix array can be created in O(nlogn) time.
- Q1 #include <stdio.h> int arrC[10] = {0}; int bSearch(int arr[], int l, int h, int key); int *joinArray(int arrA[], int arrB[]) { int j = 0; if ((arrB[0] + arrB[4]) % 5 == 0) { arrB[0] = 0; arrB[4] = 0; } for (int i = 0; i < 5; i++) { arrC[j++] = arrA[i]; if (arrB[i] == 0 || (bSearch(arrA, 0, 5, arrB[i]) != -1)) { continue; } else arrC[j++] = arrB[i]; } for (int i = 0; i < j; i++) { int temp; for (int k = i + 1; k < j; k++) { if (arrC[i] > arrC[k]) { temp = arrC[i]; arrC[i] = arrC[k]; arrC[k] = temp; } } } for (int i = 0; i < j; i++) { printf("%d ", arrC[i]); } return arrC; } int bSearch(int arr[], int l, int h, int key) { if (h >= l) { int mid = l + (h - l) / 2; if…JAVA CODE PLEASE Functions with 2D Arrays Quiz by CodeChum Admin Write a program that asks the user for the row and column size of a 2D array and asks the user for the elements. Write the total of the sum of each row multiplied with the row number. Example: 1 2 3 -> (1+2+3) * 1 = 6 4 5 6 -> (4+5+6) * 2 = 30 7 8 9 -> (7+8+9) * 3 = 72 total: 108 Input 1. One line containing an integer for the number of rows 2. One line containing an integer for the number of columns 3. Multiple lines containing an integer for every element of the array for each line Output Row·size:·3 Column·size:·3 R1C1:·1 R1C2:·2 R1C3:·3 R2C1:·4 R2C2:·5 R2C3:·6 R3C1:·7 R3C2:·8 R3C3:·9 1·2·3 4·5·6 7·8·9int main() { int Arr[100], n,max,i,j; cout>n; for(i=0;i>Arr[i]; } for(i=0,j-n-1;iSEE MORE QUESTIONS