In Ocaml Write a function l3_of_3l : ’a list * ’b list * ’c list -> (’a * ’b * ’c) list = that transforms a triple of lists into lists of triples. If lengths don’t match then ignore leftover elements. examples l3_of_3l ([1;2;3] ,[1.;2.] ,[ ’ a ’; ’b ’; ’c ’; ’d ’]) ;; - : ( int * float * char ) list = [(1 , 1. , ’a ’) ; (2 , 2. ,’b ’) ] l3_of_3l ([1.;2.] ,[ ’ a ’; ’b ’; ’c ’; ’d ’] ,[1;2;3]) ;; - : ( float * char * int ) list = [(1. , ’a ’ , 1) ; (2. , ’b’ , 2) ]
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In Ocaml Write a function l3_of_3l : ’a list * ’b list * ’c list -> (’a * ’b * ’c) list = that transforms a triple of lists into lists of triples. If lengths don’t match then ignore leftover elements.
examples
l3_of_3l ([1;2;3] ,[1.;2.] ,[ ’ a ’; ’b ’; ’c ’; ’d ’]) ;;
- : ( int * float * char ) list = [(1 , 1. , ’a ’) ; (2 , 2. ,’b ’) ]
l3_of_3l ([1.;2.] ,[ ’ a ’; ’b ’; ’c ’; ’d ’] ,[1;2;3]) ;;
- : ( float * char * int ) list = [(1. , ’a ’ , 1) ; (2. , ’b’ , 2) ]
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- Please program in C 8.18 LAB: Simple linked list Given an IntNode struct and the operating functions for a linked list, complete the following functions to extend the functionality of the linked list: IntNode* IntNode_GetNth(IntNode* firstNode, int n)- Return a pointer to the nth node of the list starting at firstNode. void IntNode_PrintList(IntNode* firstNode) - Call IntNode_PrintNodeData() to output values of the list starting at firstNode. Do not add extra space characters in between values. int IntNode_SumList(IntNode* firstNode) - Return the sum of the values of all nodes starting at firstNode. Note: The code for IntNode_Create() provided here differs from the code shown in the book. The given main() performs various actions to test IntNode_GetNth(), IntNode_PrintList(), and IntNode_SumList(). main() reads 5 integers from a user: The number of nodes to be added to a new list The value of the first node of the list An increment between the values of two consecutive…Write a function set_elements() that assigns 1 to the last and fifth elements of a list parameter. Ex: If the input is: One man's trash is another man's treasure then the output is: ['One', "man's", 'trash', 'is', 1, "man's", 1] ''' Your code goes here ''' vals_to_update = input().split() set_elements(vals_to_update) print(vals_to_update)Consider the following function access_element_by_index that returns the iterator to the element at the index i in the list l. typedef std::list<int> int_list; int_list::iterator access_element_by_index(size_t I, int_list &) { assert( ? ); … } Assume the first element of the list is at the index 0, the second at the index 1, and so on. If the function is required to return an iterator that can be dereferenced, determine the precondition of the function, and write an assertion to validate it. (You don’t need to implement the functions).
- Suppose a node of a linked list is defined as follows in your program: typedef struct{ int data; struct Node* link; } Node; Write the definition of a function, printOdd(), which displays all the odd numbers of a linked list. What will be the time complexity of printOdd()? Express the time complexity in terms of Big-O notation.Using C languge, implement programmer defined-data types with linked lists. A set of integers may be implemented using a linked list.Implement the following functions given the definition:typedef struct node* nodeptr;typedef struct node{int data;nodeptr next;}Node;typedef Node* Set;Set initialize();- simply initialize to NULLvoid display(Set s);- display on the screen all valid elements of the listSet add(Set s, elem);- simply store elem in the listint contains(Set s, int elem);- search the array elements for the value elemSet getUnion(Set result, Set s1, Set s2);- store in the set result the set resulting from the union of s1 and s2- x is an element of s1 union s2 if x is an element of s1 or x is an element of s2Set intersection(Set result, Set s1, Set s2);- store in the set result the set resulting from the intersection of s1 and s2- x is an element of s1 intersection s2 if x is an element of s1 and x is an element of s2Set difference(Set result, Set s1, Set s2);- store in the set…Write a function that gets a linked list of ints, and reverses it. For example - on input 1 -> 2 -> 3 -> 4, after the function finishes the execution, the list becomes 4 -> 3 -> 2 -> 1 - If the list has one element, then it doesn’t change - If the list is empty, then it doesn’t change You may use the data fields in the struct and the functions provided in LL.h and LL.c. // reverses a linked list void LL_reverse(LL_t* list); Test for the Function: void test_q3() { LL_t* lst = LLcreate(); LL_add_to_tail(lst, 1); LL_add_to_tail(lst, 3); LL_add_to_tail(lst, 8); LL_add_to_tail(lst, 4); LL_add_to_tail(lst, 3); LL_reverse(lst); intcorrect[] = {3,4,8,3,1}; inti; node_t* n = lst->head; for(i=0;i<5;i++) { if (n==NULL) { printf("Q3 ERROR: node %d==NULL unexpected\n", i); return; } if (n->data != correct[i]) { printf("Q3 ERROR: node%d->data==%d, expected %d\n", i, n->data, correct[i]); return; } n = n->next; } if (n==NULL) printf("Q3 ok\n"); } Support…
- Write a function that gets a linked list of ints, and reverses it. For example - on input 1 -> 2 -> 3 -> 4, after the function finishes the execution, the list becomes 4 -> 3 -> 2 -> 1 - If the list has one element, then it doesn’t change - If the list is empty, then it doesn’t change You may use the data fields in the struct and the functions provided in LL.h and LL.c. // reverses a linked list void LL_reverse(LL_t* list); Test for the Function; void test_q3() { LL_t* lst = LLcreate(); LL_add_to_tail(lst, 1); LL_add_to_tail(lst, 3); LL_add_to_tail(lst, 8); LL_add_to_tail(lst, 4); LL_add_to_tail(lst, 3); LL_reverse(lst); intcorrect[] = {3,4,8,3,1}; inti; node_t* n = lst->head; for(i=0;i<5;i++) { if (n==NULL) { printf("Q3 ERROR: node %d==NULL unexpected\n", i); return; } if (n->data != correct[i]) { printf("Q3 ERROR: node%d->data==%d, expected %d\n", i, n->data, correct[i]); return; } n = n->next; } if (n==NULL) printf("Q3 ok\n"); } Support File…Write a function cmid to count elements of an integer list between 5 and 10. One function only.cmid [12,3,6,4,5,1] → 2In Ocaml Map functions left Write a function map_fun_left : (’a -> ’a) list -> ’a list -> ’a list = that is given a list of functions and a list of elements. For each element apply all functions from left to it and keep the result. Return the results as a list. examplesmap_fun_left [((+) 1) ;( * ) 2; fun x - >x -10] [1;2;3;4];;- : int list = [ -6; -4; -2; 0]map_fun_left [( fun x - > int_of_char x | > (+) 32 | >char_of_int ) ;( fun x - > char_of_int (( int_of_char x )+1) ) ] [ ’A ’; ’B ’; ’C ’; ’D ’];;- : char list = [ ’b ’; ’c ’; ’d ’; ’e ’]
- C++ Create a generic function average({elements}) that returns the average of all the given elements. For example, average({x, y, z}), returns (x + y + z)/3. The argument is an initializer list that is assumed not to be empty. For the sake of the exercise, don’t use other STL algorithms in the implementation of the function. Write a test driverprogram by using C language In the struct structure given below, the data of a student is stored in a singly linear linked list. Write a function named "secondHighest" that takes the starting address of the list as a parameter and returns the student number with the second highest grade? typedef struct node { int OgrNo; int not; struct node *next; };Ocaml List of triples or a triple of lists? Write a function l3_of_3l : ’a list * ’b list * ’c list -> (’a * ’b * ’c) list = that transforms a triple of lists into lists of triples. If lengths don’t match then ignore leftover elements. Do not copypaste the symbol ’ in the examples as it will not work. examples l3_of_3l ([1;2;3],[1.;2.],[’a’;’b’;’c’;’d’]);; - : (int * float * char) list = [(1, 1., ’a’); (2, 2., ’b’)] l3_of_3l ([1.;2.],[’a’;’b’;’c’;’d’],[1;2;3]);; - : (float * char * int) list = [(1., ’a’, 1); (2., ’b’, 2)]