Just programming

Time Complexity for Binary Search will be Log(n).  But the main Condition is, the inputted elements in the array must be sorted (ascending order). import java.util.*; class BSearch {   public static void main(String[] args)   {   Scanner s=new Scanner(System.in);   int n=s.nextInt();   int a[]=new int[n];     for(int i=0;i<n;i++)  a[i]=s.nextInt();     System.out.println("enter the element to be find");   int f=s.nextInt();     int k=bs(a,0,n-1,f);   if(k==-1)  System.out.println("the element not present");   else       System.out.println("the element is at index  "+(k+1));   }       static int bs(int a[],int st,int en,int f)   {  if(en>=st)  {   int mid=(st+en)/2;   if(a[mid]==f)  return mid;   else if(a[mid]<f)  return bs(a,mid+1,en,f);   else ...

import java.util.*; class Intnode {   public int data;   public Intnode next;     Intnode()   {   this.next=null;   } } public class Linklist {   public static void main(String[] args)   {    Scanner s = new Scanner(System.in);    Intnode start= new Intnode();    Intnode temp=start;       System.out.println("Please enter data in the node");     int k=s.nextInt(); start.data=k; while(true) {  System.out.println("do u want to make more such node ? press y/n");   char a= s.next().charAt(0);     if(a=='n')    break;   Intnode temp2= new Intnode();   System.out.println("Please enter data in the node");        k=s.nextInt();   temp2.data=k;   temp.next=temp2; temp=temp.next; } temp =start; while(temp!=null) {  System.out.print(temp.d...

import java.util.*; class qsort {  public static void main(String args[])  { Scanner s=new Scanner(System.in); int n=s.nextInt(); int a[]=new int [n]; for(int i=0;i<n;i++) a[i]=s.nextInt(); quick_sort(a,0,n-1); for(int i=0;i<n;i++) System.out.print(a[i]+" ");  }   static void quick_sort(int a[],int st,int en)   {  if(st<en)  {        int k=pivot_ele(a,st,en);   quick_sort(a,st,k);   quick_sort(a,k+1,en);  }   }   static int pivot_ele(int a[],int st,int en)   {  int i,j,pivot; pivot=st; i=st+1; j=en; while(i<=j) { while(a[i]<a[pivot] && i<en) i++; while(a[j]>a[pivot]) j--; if(i<=j) { int t=a[i]; a[i]=a[j]; a[j]=t; i++; j--; } }  ///while close  int t=a[j]; a[j]=a[st]; a[...

import java.util.*; class msort {  public static void main(String args[])  { Scanner s=new Scanner(System.in); int n=s.nextInt();         int a[]=new int [n]; for(int i=0;i<n;i++) a[i]=s.nextInt(); merge_sort(a,0,n-1); for(int i=0;i<n;i++) System.out.print(a[i]+" ");  }   static void merge_sort(int a[],int st,int en)   {  if(st<en)  {        int k=(st+en)/2;   merge_sort(a,st,k);   merge_sort(a,k+1,en);   merge(a,st,k,en);  }   }       static void merge(int array[],int lowerIndex,int middle,int higherIndex) { int tempMergArr[]=new int[higherIndex-lowerIndex+1]; int k=0;        for (int i = lowerIndex; i <= higherIndex; i++)             tempMergArr[k++] = array[i];         int i = lowerIndex; int r=i...

Similarity between Merge_sort  and  Quick_sort In quick sort , the main thing is to choose the PIVOT  element. In Merge sort , the main thing is merging part.

import java.util.*; class LCM {  public static void main(String args[])  {    Scanner s=new Scanner(System.in);    int k1 = s.nextInt();    int k2= s.nextInt();      int gcd=hcf(k1,k2);    System.out.println(" hcf of these two number is "+gcd);      int lcm=(k1*k2)/gcd;    System.out.println(" lcm of these two number is "+lcm);      int xor=k1^k2;    System.out.println(" xor of these two number is " +xor);      int addd=Add(k1,k2);    System.out.println("addition method without using arthemetic operators is "+addd);      }  static int hcf(int a,int b)  { int r=a%(b); int diviser=a; int divident=b; int quotient; while(r>0) {   quotient=divident/diviser; r=divident%diviser; divident=diviser; diviser=r; } return divident;  }  /*  Sum ...

import java.util.*; import java.io.Console; import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.IOException; class Input {  public static void main(String []args) throws IOException  {   Scanner a = new Scanner(System.in); // creating object of scanner class   String s=a.nextLine();              // way to take input through Scanner(nextLine for String)   System.out.println(s);   int i=a.nextInt();                  // way to take input through Scanner(nextInt for int)   System.out.println(i);   String str=System.console().readLine();  // way to take input from console   System.out.println(str);   BufferedReader in = new BufferedReader( new InputStreamReader(System.in)); // creating object of BufferedReader   String str1 = in.readLine();         System.out.println(str1); ...

def main ( ) : while True: w= input ( ) if w== 42 : break else : print w main ( ) -----------------------------------------------------------------------------------------============================================================================================= I. Yet another A + B codeforces def main():             a=int(input())             b=int(input())             c=int(input())                         if a+b == c:                 print("YES")             elif a+a == c:                 print("YES")             elif b+b == c: ...

<?php fscanf(STDIN, "%d\n", $test); while($test--) { $ooo=array(); for($i=0;$i<26;$i++) $ooo[$i]=0; fscanf(STDIN, "%s\n", $raa); $qwe=0; $p=strlen($raa); for($i=0;$i<$p;$i++) { $w=ord($raa[$i]); $qwe= $w-97; $ooo[$qwe]=$ooo[$qwe]+1; } $f=0; $c=0; for($i=0;$i<26;$i++) { if($ooo[$i]%2!=0) { $f=1; $c++; } } if($f==0) { echo"Yes"; } else if($f==1 && strlen($raa)%2==1 && $c==1) { echo "Yes"; } else { echo "No"; } echo "<br>"; }