Answered: I Evaluate the time complexity of…

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I Evaluate the time complexity of following code: int a-0; for (i-0; i< N; i++){ aati;) for (i- 0; i< N; it+) { a-a+i;)

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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Recursion Practice Welcome back! In this lab, we will be reviewing recursion by practicing with some basic recursion problems. Objectives Increase familiarity with recursive logic by working through several recursive problems. Taking into consideration a few corner cases through analyzing the test cases. Using a regex expression that will remove punctuation. Getting Started This lab includes the following .java file: L4/└── Recursion.java└── Main.java**Main.java is a read-only file used for testing. It is not included in the starter jar.Here is the starter jar if you would like to code in a different environment: L4.jar. Please complete ALL functions. Make sure to read the description for each function carefully. Do not include any for or while loops in your methods. These can all be completed in a purely recursive style, so do it recursively! In the spirit of incremental development, implement each method one at a time, look at the test cases and take into consideration what is…
Provide answer in C language please You are tasked with creating a program that takes in an array of integers and sorts them in ascending order using a custom sorting algorithm. The custom sorting algorithm should be based on the following rules: The sorting algorithm must use a divide-and-conquer approach. The sorting algorithm must be recursive. The sorting algorithm must have an average case time complexity of O(n log n). The sorting algorithm must have a worst case time complexity of O(n^2). The sorting algorithm must be stable, meaning that the relative order of equal elements should be preserved.
Explain this c code in details. #include <stdio.h> int main() { int max_weight_truck, n; scanf("%d", &max_weight_truck); scanf("%d", &n); int weight_item[n]; int priority[n]; int p_sort[n]; int w_sort[n]; int a[n]; for(int i=0; i<n; i++) scanf("%d", &weight_item[i]); for(int i=0; i<n; i++) scanf("%d", &priority[i]); //sorting priority array for(int i=0; i<n; i++) p_sort[i]= priority[i]; int temp=0; for (int i = 0; i < n; i++) { for (int j = i+1; j < n; j++) { if(p_sort[i] < p_sort[j]) { temp = p_sort[i]; p_sort[i] = p_sort[j]; p_sort[j] = temp; } } } for(int i=0; i<n; i++) a[i]= 0; for(int i=0; i<n; i++) { int index; for(int j=0; i<n; j++) { if(p_sort[i]== priority[j] && a[j]==0) { index= j; break; } } w_sort[i]= weight_item[index]; a[index]=1; } int s=0, i=0,j=0,truck=5, l=0; int o_item[truck], unused_space[truck]; // initializing arrays for(i=0; i<truck; i++) { o_item[i]= 0; unused_space[i]= 0; } i=0; while(truck>0)…
суmath 1/ 2 90% SEM 1 (2020 -21) Calculate the running time of the following algorithm/program and find the big-oh order of complexity. a) void funPrint({ int k-6; k++; int s=k/3; cout>n; int a[n][n]; int b[n][n]; for (int i-0; i>a[i][j]; cin>>b[i][j]:} int c[n][n]; for (int i 0; i
#include<iostream>#include<math.h>using namespace std;class TotalResistance{double series_res,parallel_res,sp_res;public:TotalResistance(){series_res=parallel_res=sp_res=0;}void seriesResistance(double resistance[],int n);void parallelResistance(double resistance[],int n);void spResistance(double resistance[],int n);};void TotalResistance::seriesResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];cout<<"Total Resistance in series is: "<<series_res<<endl;}void TotalResistance::parallelResistance(double resistance[],int n){double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in parallel is: "<<parallel_res<<endl;}void TotalResistance::spResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in series…
11 - The code segment below has time complexity? for (int i=0; i
lint power (int base, int exponent); Define the recursive function power() that when invoked returns baseonent Assume that exponent is an integer greater than or equal to 1. Hint: The recursion step would use the relationship basenent = base * baseonert- and the terminating condition occurs when exponent is equal to 1 because base' = base or when exponent is equal to o because base = 1.
int X[900]; int Y[600]; int sum, sum1, sum2, sum3; //parallelism : dividing outer loop in three parts //i = 1 to 300 for(i=1;i<=300;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } //i = 301 to 600 for(i=301;i<=600;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } //i = 601 to 900 for(i=601;i<=900;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } sum = sum1 + sum2 + sum3;} another way to solve the question that send in the pic
Find 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…
#include<bits/stdc++.h> using namespace std; void bubbleSort(int arr[], int n) { for (int i = 0; i < n - 1; i++) { for (int j = 0; j < n - i - 1; j++) { if (arr[j] > arr[j + 1]) { int temp = arr[j]; arr[j] = arr[j + 1]; arr[j + 1] = temp; } } } } int binarySearch(int arr[], int l, int r, int x, int& comp) { comp++; if (r >= l) { int mid = l + (r - l) / 2; if (arr[mid] == x) { return mid; } if (arr[mid] > x) { return binarySearch(arr, l, mid - 1, x, comp); } return binarySearch(arr, mid + 1, r, x, comp); } return -1; } int main() { int Num[8192]; srand(time(NULL)); for (int i = 0; i < 8192; i++) { Num[i] = rand() % 10001; } clock_t starting_time = clock(); bubbleSort(Num, 8192); clock_t ending_time = clock(); clock_t result =…
main.cpp:4:17: error: 'string' has not been declared 4 | void KMPSearch(string pat, string txt) main.cpp:4:29: error: string' has not been declared 4 | void KMPSearch(string pat, string txt) Aununun main.cpp:31:25: error: string' has not been declared 31 | void computelPSArray(string pat, int M, int[] lps) main.cpp:31:50: error: expected or . before lps' 31 | void computeLlPSArray(string pat, int M, int[] lps) main.cpp:55:8: error: expected :' before 'static' 55 | public static void Main() main.cpp:61:2: error: expected ;' after class definition 61 | } | main.cpp: In member function 'void GFG::KMPSearch(int, int)': main.cpp:6:15: error: request for member Length' in 'pat', which is of non-class type 'int' 6 | int M = pat.Length; main.cpp:7:15: error: request for member Length' in txt’, which is of non-class type 'int' int N = txt.Length; main.cpp:8:6: warning: structured bindings only available with -std3c++17' or (-std=gnu++17' 8 | int[] lps new int[M];
C++ Coding Assignment: Redo the 8 queens 1-dimensional array program with backtracking by REMOVING ALL GOTO's, but implementing the same algorithm. GIVEN C++ CODE - WITH GOTO's: #include <iostream>using namespace std; int main() { int q[8], c=0; q[0] = 0; // place a queen in row 0 of column 0 int counter = 0; // to count solutions // c: column as q[c]: row in column c next_col: ++c; // or c++; // next column if(c==8) goto print; q[c]=-1; // 0 next_row: ++q[c]; // or q[c]++; // next row if(q[c]==8) goto backtrack; for(int i=0; i<c; ++i) if(q[i]==q[c] || (c-i)== abs(q[c]-q[i])) // || = row test OR diagonal tests goto next_row; goto next_col; backtrack: --c; // or c-- if(c==-1) // once we backtrack to column -1, we have all the solutions. return 0; // ends the program goto next_row; print: counter++; // new solutions cout << counter << ":" << endl; for(int…