In the Time- complexity column, write how many times would every line be executed. Thereafter, below the cable, write the worst-case complexity of the whole code. Code Time-complexity 1 int i, j,k = 0 1 2 for (i = n/2; i < n; i + +) { for (j = 2; i < n; j = j* 2) { k = k + n/2; }} 2 3 4 4 Worst case complexity of the code is

In the Time- complexity column, write how many times would every line be executed. Thereafter, below the cable, write the worst-case complexity of the whole code. Code Time-complexity 1 int i, j,k = 0 1 2 for (i = n/2; i < n; i + +) { for (j = 2; i < n; j = j* 2) { k = k + n/2; }} 2 3 4 4 Worst case complexity of the code is

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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Consider the following loop, which is the same loop as in the Free Form part: int low, high, A[100]; //assume array A has been initialized //Precondition: 0 A[best]) best=sub; ++sub; } //Postcondition: A[i]
sum = 0; for (int i = 1; i< n; i = sum++ || 2*i)
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in c++, use parallel arrays , and srand(time(NULL)) to randomize the two cards dealt at the start Create a Blackjack (21) game. Your version of the game will imagine only a SINGLE suit of cards, so 13 unique cards, {2,3,4,5,6,7,8,9,10,J,Q,K,A}. Upon starting, you will be given two cards from the set, non-repeating. Your program MUST then tell you the odds of receiving a beneficial card (that would put your value at 21 or less), and the odds of receiving a detrimental card (that would put your value over 21). Recall that the J, Q, and K cards are worth ‘10’ points, the A card can be worth either ‘1’ or ‘11’ points, and the other cards are worth their numerical values.
A/ find code optimization method to this code and then find type to this code 1- For(i=1;i
int functionC (int n) { int i, j, sumC = 0; for (i=n; i > 0; i=i-5) for (j=1; j 0) { if (functionC(n) % 2 == 0) { for (i=m; i > 0; i=i/3) sumE++; } else (10) Asymptotic runtime of functionE { for (i=m; i > 0; i=i-3) sumE--; } n--; } return sumE;
make flowchart for this one plz #include<stdio.h>int sortingbook(int array[], int N){ int count = 0, i , j , k , minimum , index , flag = 0; for(i=0;i<N;i++) { minimum = array[i]; //ith element as the minimum index = i; //setting index value as i for( j = i; j < N; j++) { if(array[j] < minimum) { minimum = array[j]; //calculating minimum index = j; //assigning the minimum position into index } } for(k = index; k>i ; k--) //shifting one position to the right from index to i array[k] = array[k-1]; if(array[i]!=minimum) count++; //if the value of min chganged from previous assumption array[i] = minimum; } return count;}int main(){ int array[20],i,N,c; printf("\nEnter number of books(N) : "); scanf("%d",&N); //number of books printf("\nEntercurrent sequence of %d books : \n=> ",N); for(i=0;i<N;i++) scanf("%d",&array[i]); //reading sqeuence c = sortingbook(array,N); printf("\nSorted sequence is :\n=>…
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int getMax(int arr[], int n) { intmx=arr[0]; for (inti=1; i<n; i++) if (arr[i] >mx) mx=arr[i]; returnmx; } Can u give me the code for this one as well....this is the first function and countsort is the second
Java Programming : A party has been organised on cruise. The party is organised for a limited time (T). The number of guests entering (E[i]) and leaving (L[i]) the party at every hour is represented as elements of the array. The task is to find the maximum number of guests present on the cruise at any given instance within T hours. Example 1: Input : 5 > Value of T [7,0,5,1,3] -> E[], Element of E[0] to E[N-1], where input each element is separated by new line [1,2,1,3,4] -> L[], Element of L[0] to L[N-1], while input each element is separate by new line. Output : 8 -> Maximum number of guests on cruise at an instance. Explanation: 1st hour: Entry 7 Exit: 1 No. of guests on ship : 6 2nd hour : Entry Exit : 2 No. of guests on ship: 6-2=4 Hour 3: Entry: 5 Exit: 1 No. of guests on ship : 4+5-1-8 Hour 4: Entry 1 Exit 3 No. of guests on ship: 8+1-3=6 Hour 5: Entry 3 Exit: 4 No. of guests on ship: 6+3-4-5 Hence, the maximum number of guests within 5 hours is 8.
In c++