def dont(wave): if len(wave) < 1: return [""] theg [] for its in range(len(wave)): for juh in dont(wave[:its] + wave[its+1:]): theg.append (wave[its] + juh) return theg If wave has length n, what is the worst-case time complexity of dont(wave)? O(T(n)) =

def dont(wave): if len(wave) < 1: return [""] theg [] for its in range(len(wave)): for juh in dont(wave[:its] + wave[its+1:]): theg.append (wave[its] + juh) return theg If wave has length n, what is the worst-case time complexity of dont(wave)? O(T(n)) =

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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solve in java Integer arrays originalArr and filterArr are read from input, each containing two elements. If an element in originalArr is not a multiple of the corresponding element in filterArr, replace the element in originalArr with the corresponding element in filterArr times 2. Ex: If the input is: 16 9 8 4 then the output is: 16 8 Note: Given integers A and B, A is a multiple of B only if A % B == 0. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 import java.util.Scanner; public class ReplaceNumbers { publicstaticvoidmain(String[] args) { Scannerscnr=newScanner(System.in); finalintNUM_VALS=2; int[] originalArr=newint[NUM_VALS]; int[] filterArr=newint[NUM_VALS]; inti; for (i=0; i<NUM_VALS; ++i) { originalArr[i] =scnr.nextInt(); } for (i=0; i<NUM_VALS; ++i) { filterArr[i] =scnr.nextInt(); } /* Your code goes here */ for (i = 0; i < originalArr.length; ++i) { System.out.print(originalArr[i] + " "); } System.out.println();…
Add comments in code #include <stdio.h>#include <stdlib.h>/* ADJACENCY MATRIX */int source,V,E,time,visited[20],G[20][20];void DFS(int i){ int j; visited[i]=1; printf(" %d->",i+1); for(j=0;j<V;j++) { if(G[i][j]==1&&visited[j]==0) DFS(j); }}int main(){ int i,j,v1,v2; printf("\t\t\tGraphs\n"); printf("Enter the no of edges:"); scanf("%d",&E); printf("Enter the no of vertices:"); scanf("%d",&V); for(i=0;i<V;i++) { for(j=0;j<V;j++) G[i][j]=0; } /* creating edges :P */ for(i=0;i<E;i++) { printf("Enter the edges (format: V1 V2) : "); scanf("%d%d",&v1,&v2); G[v1-1][v2-1]=1; } for(i=0;i<V;i++) { for(j=0;j<V;j++) printf(" %d ",G[i][j]); printf("\n"); } printf("Enter the source: "); scanf("%d",&source); DFS(source-1); return 0;}
in Python Minima in permutations. Write a program that takes an integer n from the command line, generates a random permutation, prints the permutation, and prints the number of left-to-right minima in the permutation (the number of times an element is the smallest seen so far). Then write a program that takes integers m and n from the command line, generates m random permutations of length n, and prints the average number of left-to-right minima in the permutations generated
Add comments in code to explain #include <stdio.h>#include <stdlib.h>/* ADJACENCY MATRIX */int source,V,E,time,visited[20],G[20][20];void DFS(int i){ int j; visited[i]=1; printf(" %d->",i+1); for(j=0;j<V;j++) { if(G[i][j]==1&&visited[j]==0) DFS(j); }}int main(){ int i,j,v1,v2; printf("\t\t\tGraphs\n"); printf("Enter the no of edges:"); scanf("%d",&E); printf("Enter the no of vertices:"); scanf("%d",&V); for(i=0;i<V;i++) { for(j=0;j<V;j++) G[i][j]=0; } /* creating edges :P */ for(i=0;i<E;i++) { printf("Enter the edges (format: V1 V2) : "); scanf("%d%d",&v1,&v2); G[v1-1][v2-1]=1; } for(i=0;i<V;i++) { for(j=0;j<V;j++) printf(" %d ",G[i][j]); printf("\n"); } printf("Enter the source: "); scanf("%d",&source); DFS(source-1);…
The split method in the String class returns an array of strings consisting of the substrings split by the delimiters. However, the delimiters are not returned. Implement the following new method that returns an array of strings consisting of the substrings split by the matching delimiters, including the matching delimiters. public static String[] split(String s, String regex) For example, split("ab#12#453", "#") returns ab, #, 12, #, and 453 in an array of String and split("a?b?gf#e", "[?#]") returns a, ?, b, ?, gf, #, and e in an array of String.
Implement Flatten Arrays.Given an array that may contain nested arrays,produce a single resultant array."""from collections.abc import Iterable # return listdef flatten(input_arr, output_arr=None): if output_arr is None: output_arr = [] for ele in input_arr: if not isinstance(ele, str) and isinstance(ele, Iterable): flatten(ele, output_arr) #tail-recursion else: output_arr.append(ele) #produce the result return output_arr # returns iteratordef flatten_iter(iterable): """ Takes as input multi dimensional iterable and returns generator which produces one dimensional output. """ for element in iterable: if not isinstance(element, str) and.
Suppose that intArray is an array of integers, and length specifies the number of elements in intArray. Also, suppose that low and high are two integers such that 0 <= low < length, 0 <= high < length, and low < high. That is, low and high are two indices in intArray. Write a recursive definition that reverses the elements in intArray between low and high.
Given the following function, what happens if a[] contains just one element that doesn't match val? int binarySearch(int a[], int first, int last, int val){ if (first > last) return -1; int middle = (first + last) / 2; if (a[middle] == val) return middle; if (a[middle] < val) return binarySearch(a, middle+1, last, val); else return binarySearch(a, first, middle-1, val);} Group of answer choices binarySearch never calls itself again and terminates (recursion never happens) binarySearch calls itself once then terminates (recursion happens once) binarySearch calls itself twice then terminates (recursion happens twice) binarySearch calls itself 3 times then terminates (recursion happens 3 times)
FOR JAVA Write a Java method that takes two 2 dimensional int arrays (a and b) and a 2 dimensional boolean array (c) where all three arrays have the same size. Your method should return an array (result) such that, if a partic- ular element in c is true, then the corresponding (same indexed) element in result must be the multiplication of the the corresponding elements of a and b. If the element in c is false, then the the corresponding element in result must be the negative of multiplication of the the corresponding element in a and b. For example ifa = {{1,2,3},{4,5}},b = {{1,2,1},{0,2}} andc = {{true,false,true},{false,true}}, then the result should be result={{1, -4, 3}{0, 10}}.
Given an array of strings, return all groups of strings that are anagrams. The groups must be created in order of their appearance in the original array. Look at the sample case for clarification. Note: The final output will be in lexicographic order. Example 1: Input:N = 5words[] = {act,god,cat,dog,tac}Output:act cat tac god dogExplanation:There are 2 groups ofanagrams "god", "dog" make group 1."act", "cat", "tac" make group 2.
mplement a Java program that applies the Newton-Raphson's method xn+1 = xn – f(xn) / f '(xn) to search the roots for this polynomial function ax6 – bx5 + cx4 – dx3+ ex2 – fx + g = 0. Fill out a, b, c, d, e, f, and g using the first 7 digits of your ID, respectively. For example, if ID is 4759284, the polynomial function would be 4x6 – 7x5 + 5x4 – 9x3+ 2x2 – 8x + 4 = 0. The program terminates when the difference between the new solution and the previous one is smaller than 0.00001 within 2000 iterations. Otherwise, it shows Not Found as the final solution.
6. In Java Write a static method transpose() that receives as parameter, a 2D square array numArof integers, and swaps the row and column of each element. For example:6 7 8 0 6 3 1 23 2 4 5 → 7 2 5 01 5 8 2 8 4 8 92 0 9 3 0 5 2 3Hint: Once you find the largest-sum column, you will need to swap its entries with the first column,element-by-element.