Calculating sum of digits using Recursion

Input: 253

Output: 10 
(we need to add the individual digits in the number) 

Source code:

import java.util.*;

import java.io.*;

public class Main

{

public static void main(String[] args) 

{

    Scanner s=new Scanner(System.in);

    System.out.println("Enter number");

    Long n =s.nextLong();

System.out.println("Answer: "+sumOfDigits(n));

}

public static Long sumOfDigits(Long n)

{

    if(n==0)

        return 0L;

        

    return (n%10)+sumOfDigits(n/10);

}

}

Output:

Enter number

1155588

Answer: 33

Palindrome Check using Recursion

 Problem: 

You are given a String 'S'. You need to recursively find the if the string is Palindrome or not.

Test case 1: 

Input: abba

Output: True

Test case 2: 

Input: abbcbba

Output: True

Test case 3: 

Input: abbad

Output: False

Source code:

import java.util.*;

import java.io.*;

public class Main

{

    static int k=1;

public static void main(String[] args) 

{

    Scanner s=new Scanner(System.in);

    System.out.println("Enter String");

    String str =s.next();

System.out.println("Answer: "+isPalindrome(str,0,str.length()-1));

}

public static boolean isPalindrome(String str,int start,int end)

{

    if(start>=end)

        return true;

        

    return ((str.charAt(start)==str.charAt(end)) && isPalindrome(str,start+1,end-1));

}

}

 

Union of Two arrays.

Given two different sorted arrays. We need to print union of two arrays. ( Elements which are common in both the arrays)

arr1={1,3,4,5,7}

arr2={2,3,5,6}

Output: {1,2,3,4,5,6,7}

Source code:

public class Main
{
public static void main(String[] args)
{
int arr1[]={1,3,4,5,7};
int arr2[]={2,3,5,6};
union(arr1,arr2);
}
public static void union(int arr1[],int arr2[])
{
int i=0,j=0;

while(i<arr1.length&&j<arr2.length)
{
if(arr1[i]<arr2[j])
{
System.out.print(arr1[i]+" ");
i++;
}
else if(arr1[i]>arr2[j])
{
System.out.print(arr2[j]+" ");

j++;
}
else if(arr1[i]==arr2[j])
{
System.out.print(arr1[i]+" ");

i++;
j++;
}
}

while(i<arr1.length)
System.out.print(arr1[i++]+" ");

while(j<arr2.length)
System.out.print(arr2[j++]+" ");

}
}

Initializing a string with repeated character of specific length

Source code:

import java.util.*;

import java.io.*;

public class Main

{

public static void main(String[] args) 

{

   char c='0';

            int n=10;

  char ch[]=new char[n];

  Arrays.fill(ch,c);

  String s=new String(ch);

  System.out.println(s);   

}

}

output:

0000000000

Minimum flips required to get given binary String.

Given a string S, the task is to find minimum flips required to convert an initial binary string consisting of only zeroes to S where every flip of a character flips all succeeding characters as well.

Input: S = “01011”
Output: 3
Explanation:
Initial String – “00000”
Flip the 2nd bit – “01111”
Flip the 3rd bit – “01000”
Flip the 4th bit – “01011”
Total Flips = 3
Input: S = “01001”
Output: 3
Explanation:
Initial String – “00000”
Flip the 2nd bit – “01111”
Flip the 3rd bit – “01000”
Flip the 5th bit – “01001”
Total Flips = 3

Source Code:

public class FlipDigitsOfStringToGetTarget {

public static void main(String[] args) {

String target="01011";
int n=target.length();
System.out.print("Minimum flips required: "+minFlips(target,n));

}
public static int minFlips(String target,int n)
{
int count=0;
char curr='1';

for(int i=0;i<n;i++)
{
char ch=target.charAt(i);

if(ch==curr)
{
++count;
curr=(char)(48+(curr%2));
}
}
return count;
}

}

Anagram Check of 2 strings

Given a program to check whether two strings are anagram or not. 

An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

 Example 1:

SILENT

LISTEN

Example 2:

aab

aba

In the above case both strings have same number of characters(characters used exactly once).

Solution:

Approach 1:
Time Complexity: O(n2)

Source code:

public class Main {
public static void main(String[] args)
{
String s="aaab",s2="abba";
boolean isAnagram=false,visited[]=new boolean[s2.length()];
if(s.length()==s2.length())
{
for(int i=0;i<s.length();i++)
{
char ch=s.charAt(i);
isAnagram=false;
for(int j=0;j<s2.length();j++)
{
if(s2.charAt(j)==ch&&visited[j]==false)
{
isAnagram=true;
visited[j]=true;
break;
}
}
if(!isAnagram)
break;
}
}
if(isAnagram)
System.out.println("Anagram");
else
System.out.println("Not an Anagram");
}
}

Approach 2:

Taking two 256 arrays and making a note of count at its respective indices.
Eg: if 'a' is encountered then index at 97 is increased.

public class Main
{
public static void main(String[] args)
{
String a="aab";
String b="aba";
boolean isAnagram=true;
int[] al=new int[256];
int[] bl=new int[256];

for(char c:a.toCharArray()){
int index=(int)c;
al[index]++;
}
for(char c:b.toCharArray()){
int index=(int) c;
bl[index]++;
}
for(int i=0;i<256;i++){
if(al[i]!=bl[i]){
isAnagram=false;
break;
}
}
if(isAnagram)
System.out.println("Anagram");
else
System.out.println("Not an anagram");
}
}

Approach 3:

Taking one integer array and when character encountered in first string the increment and after iterating second string decrement the same index. At last check whether all the elements in array are zeros.

public class Main
{
public static void main(String[] args)
{
String a="aab";
String b="aba";
boolean isAnagram=true;
int[] al=new int[256];

for(char c:a.toCharArray()){
int index=(int)c;
al[index]++;
}
for(char c:b.toCharArray()){
int index=(int) c;
al[index]--;
}
for(int i=0;i<256;i++){
if(al[i]!=0)
{
isAnagram=false;
break;
}
}
if(isAnagram)
System.out.println("Anagram");
else
System.out.println("Not an anagram");
}
}

Maximum in sub-arrays of length K:

Given an array of integers and a number k, where 1 <= k <= length of the array, compute the maximum values of each subarray of length k.

For example, given array = [10, 5, 2, 7, 8, 7] and k = 3, we should get: [10, 7, 8, 8], since:

10 because {10,5,2} (10 is maximum in sub array)

7   because {5,2,7} (7 is maximum in sub array)

8   because {2,7,8} (8 is maximum in sub array)

8   because {7,8,7} (8 is maximum in sub array)

Imput :

first line of  input contains integer N,

second line of input contains array of integers of length N;

third line of input contains integer K,the length of sub array

Output:

otuput consists of an array of maximum numbers of each subarray.

Program Code:

#include <stdio.h>

int main()

{

    int k,n,a[100],ind=0;

    int i,j,max[100],h,m;

    scanf("%d",&n);

    for(i=0;i<n;i++)

        scanf("%d",&a[i]);

    scanf("%d",&k);

    h=k;

    for(i=0;i<=n-k;i++)

    {

        

        m=a[i];

for(j=i+1;j<h;j++)

        {

if(a[j]>m)

            {

m=a[j];

            }

        }

        max[i]=m;

        h++;

    }

    

    for(i=0;i<=n-k;i++)

    {

printf("%d ",max[i]);

    }

    return 0;

}

Pattern 3 44 555 6666 6666 555 44 3

Input:
4

Output:
3
44
555
6666
6666
555
44
3

Source code:
#include<stdio.h>
int main()
{
    int n,i,j,k,s=3;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        for(k=1;k<=i;k++)
        {
            printf("%d",s);
        }
        s++;
        printf("\n");
    }
    s--;
    for(i=n;i>=1;i--)
    {
        for(k=1;k<=i;k++)
        {
            printf("%d",s);
        }
        s--;
        printf("\n");
    }
    return 0;
}

                                               SWITCHING 100 BULBS

There are 100 light bulbs lined up in a row in a long room. Each bulb has its own switch and is currently switched off. The room has an entry door and an exit door. There are 100 people lined up outside the entry door. Each bulb is numbered consecutively from 1 to 100. So is each person.

Person No. 1 enters the room, switches on every bulb, and exits. Person No. 2 enters and flips the switch on every second bulb (turning off bulbs 2, 4, 6...). Person No. 3 enters and flips the switch on every third bulb (changing the state on bulbs 3, 6, 9...). This continues until all 100 people have passed through the room.

Find the state of nth switch(i.e.ON/OFF)?
Input:
Nth integer to find state
Output :
State of the Nth switch.
Assume 0 as OFF and 1 as ON.

Sample:
Input:
64
Output:
1

Program:
#include<stdio.h>
int a[150]={0};
int n=100;
int main()
{
    int i,j,k,swit;
    scanf("%d",&swit);
    for(i=1;i<=n;i++)
    {
        for(k=1;k<=n;k++)
        {
            if(k%i==0)
            {
                if(a[k]==0)
                    a[k]=1;
                else if(a[k]==1)
                    a[k]=0;
            }
        }
    }
    for(i=1;i<=n;i++)
    {
        if(i==swit)
        {
            printf("%d",a[i]);
        }
    }
    return 0;
}

1. To Print Butterfly Pattern using loops

Solution:

#include <stdio.h>

int main()

{

    int i,j,n;

    scanf("%d",&n);

    for(i=1;i<=n;i++)

    {

        for(j=1;j<=2*n;j++)

        {

            if(i<j)

            printf(" ");

            else

            printf("&");

            if(i<=2*n-j)

            printf(" ");

            else

            printf("&");

        }

        printf("\n");

    }

        for(i=1;i<=n;i++)

    {

        for(j=1;j<=2*n;j++)

        {

            if(i>(n-j+1))

            printf(" ");

            else

            printf("&");

            if((i+n)>j)

            printf(" ");

            else

            printf("*");

        }

        printf("\n");

    }

    return 0;

}