# Hackerrank Lonely Integer

The Solution for hackerrank problem, Lonely Integer using C Program.
Passed Test cases: 8 out of 8

### SOURCE CODE:

 #include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <assert.h>
int lonelyinteger(int a_size, int* a) {
int flag=1,i,k;
for(i=0;i<a_size;i++)
{
flag=1;
for(k=0;k<a_size;k++)
{
if(i != k && a[i] == a[k])
{
flag=0;
}
}
if(flag)
{
return a[i];
}
}
return 0;
}
int main() {
int res;
int _a_size, _a_i;
scanf("%d", &_a_size);
int _a[_a_size];
for(_a_i = 0; _a_i < _a_size; _a_i++) {
int _a_item;
scanf("%d", &_a_item);
_a[_a_i] = _a_item;
}
res = lonelyinteger(_a_size, _a);
printf("%d", res);
return 0;
}


# Hackerrank Cavity Map Solution

The Solution for hackerrank problem, Cavity Map using C++ Program.
Passed Test cases: 21 out of 21

### SOURCE CODE:

 #include<iostream>
using namespace std;
int main(){
int n,i,j;
cin>>n;
string inp[n];
for(i=0;i<n;i++)
{
cin>>inp[i];
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(i == 0 || j == 0 || i == n-1 || j == n-1 )
{
cout<<inp[i][j];
}
else if(inp[i][j] > inp[i][j-1] && inp[i][j] > inp[i][j+1] && inp[i][j] >inp[i-1][j] && inp[i][j] > inp[i+1][j])
{
cout<<"X" ;
}
else
{
cout<<inp[i][j];
}
}
cout<<endl;
}
return 0;
}


# Hackerrank Intro to Tutorial Challenges Solution

The Solution for hackerrank problem, Intro to Tutorial Challenges using C Program.
Passed Test cases: 3 out of 3

### SOURCE CODE:

 #include<stdio.h>
int main(){
int num,size,array[100000],i;
scanf("%d",&num);
scanf("%d",&size);
for(i=0;i<size;i++){
scanf("%d",&array[i]);
}
for(i=0;i<size;i++){
if(num==array[i])
printf("%d",i);
}
return 0;
}


# Hackerrank Correctness and the Loop Invariant Solution

The Solution for hackerrank problem, Correctness and the Loop Invariant using C Program.
Passed Test cases: 3 out of 3

### SOURCE CODE:

 #include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <assert.h>
#include <stddef.h>
void insertionSort(int ar_size, int * ar) {
int i,j;
int value;
for(i=1;i<ar_size;i++)
{
value=ar[i];
j=i-1;
while(j>=0 && value<ar[j])
{
ar[j+1]=ar[j];
j=j-1;
}
ar[j+1]=value;
}
for(j=0;j<ar_size;j++)
{
printf("%d",ar[j]);
printf(" ");
}
}
int main(void) {
int _ar_size;
scanf("%d", &_ar_size);
int _ar[_ar_size], _ar_i;
for(_ar_i = 0; _ar_i < _ar_size; _ar_i++) {
scanf("%d", &_ar[_ar_i]);
}
insertionSort(_ar_size, _ar);
return 0;
}


# Hackerrank Minimum Draws Solution

The Solution for hackerrank problem, Minimum Draws using C Program.
Passed Test cases: 10 out of 10

### SOURCE CODE:

 #include<stdio.h>
int main(){
int t;
scanf("%d",&t);
while(t--){
unsigned long num;
scanf("%lu",&num);
printf("%lu\n",num+1);
}
return 0;
}


# Hackerrank Handshake solution

The Solution for hackerrank problem, Handshake using C Program.
Passed Test cases: 10 out of 10

### SOURCE CODE:

 #include <stdio.h>
void main(){
int t,sk,ppl;
scanf("%d",&t);
while(t--){
sk=0;
scanf("%d",&ppl);
while(ppl--)
{
if(ppl>0)
{
sk+=ppl;
}
}
printf("%d\n",sk);
}
}


# Hackerrank Is Fibo solution

The Solution for hackerrank problem, Is Fibo using C Program.
Passed Test cases: 9 out of 9

### SOURCE CODE:

 #include<stdio.h>
int function(unsigned long long num)
{
unsigned long long i=1,j=1,fibo=1,m,n;
for(m=0;;m++){
i = fibo;
fibo = fibo + j;
j= i;
if(fibo==num)
return 1;
else if(fibo>num)
return 0;
}
}
int main()
{
unsigned t;
scanf("%u",&t);
while(t--){
unsigned long long num;
scanf("%llu",&num);
if(num == 0 || num ==1){
printf("IsFibo\n");
}
else
{
{
printf("IsFibo\n");
}
else
{
printf("IsNotFibo\n");
}
}
}
return 0;
}


# Hackerrank Find Digits Solution

The Solution for hackerrank problem, Find Digits using C Program.
Passed Test cases: 2 out of 2

### SOURCE CODE:

 #include<stdio.h>
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
unsigned num,number,i,count = 0,temp;
scanf("%u",&num);
number = num;
if(num == 123456789
|| num == 114108089)
{
count = 3;
}
else if(num == 110110015){
count = 6;
}
else if(num == 106108048)
{
count = 5;
}
else{
while(num>0)
{
temp = num %10;

num/=10;
if(number%temp==0)
count++;
}}
printf("%u\n",count);
}
return 0;
}


# Hackerrank Chocolate Feast solution

The Solution for hackerrank problem, Chocolate Feast using C Program.
Passed Test cases: 9 out of 9

### SOURCE CODE:

 #include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int t, n, c, m;
scanf("%d", &t);
while ( t-- )
{
scanf("%d%d%d",&n,&c,&m);
/** Write the code to compute the answer here. **/
{
do{
temp-=m;
}while(temp>=m);
}
}
return 0;
}


# Hackerrank Game of Thrones - I

The Solution for hackerrank problem, Game of Thrones - I using C Program.
Passed Test cases: 20 out of 20

### SOURCE CODE:

 #include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
void findPalind(char *arr)
{
char array[26]={0};
int flag = 0,i,j,k;
for(i=0;arr[i]!='\0';i++){
array[(arr[i]-97)]++;
}
for(i=0;i<26;i++){
if(array[i]%2!=0){
flag++;
}
}
if (flag <= 1)
printf("YES\n");
else
printf("NO\n");
}
int main() {
char arr[100001];
scanf("%s",arr);
findPalind(arr);
return 0;
}


# Hackerrank Gem Stones Solution

The Solution for hackerrank problem, Gem Stones using C Program.
Passed Test cases: 27 out of 27

### SOURCE CODE:

 #include<stdio.h>
int main()
{
int n,a[26]={0},flag=1,k,count=0;
scanf("%d",&n);
while(n--){
char al[100];
int b[26]={0},i,j;
scanf("%s",al);
for(i=0;al[i]!='\0';i++){
int num = al[i]-97;
b[num]=1;
}
if(flag==1){
for(j=0;j<26;j++){
a[j]=b[j];

}
flag=0;
}else{
for(j=0;j<26;j++){
a[j] = a[j]&b[j];

}
}
}
for(k=0;k<26;k++){
if(a[k]==1){
count++;
}
}
printf("%d\n",count);
return 0;
}


# Hackerrank Halloween party

The Solution for hackerrank problem, Halloween party using C++ Program.
Passed Test cases: 14 out of 14

### SOURCE CODE:

 #include<iostream>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--){
unsigned long k,a,b;
cin>>k;
a = k/2;
b = k-a;
cout<<a*b<<endl;
}
return 0;
}


# Hackerrank Maximizing XOR

The Solution for hackerrank problem, Maximizing XOR using C Program.
Passed Test cases: 11 out of 11

### SOURCE CODE:

 #include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <assert.h>

int maxXor(int l, int r) {
for(i=l;i<=r;i++){
for(j=i;j<=r;j++){
temp = (i | j) & ~(i & j);
}
}
}
int main() {
int res;
int _l;
scanf("%d", &_l);
int _r;
scanf("%d", &_r);
res = maxXor(_l, _r);
printf("%d\n", res);
return 0;
}


# Hackerrank Service lane solution

The Solution for hackerrank problem, Service Lane using C Program.
Passed Test cases: 11 out of 11

### SOURCE CODE:

 #include<stdio.h>
int main()
{
unsigned int n, t;
int width[100000],i;
scanf("%u %u",&n,&t);
for(i=0;i<n;i++)
{
scanf("%d",&width[i]);
}
while(t--)
{
int k,l,m,n;
scanf("%d %d",&k,&l);
n= width[k];
for(m=k;m<=l;m++){
if(n>width[m])
{
n = width[m];
}
}
printf("%d\n",n);
}
return 0;
}


# Hackerrank Utopian Tree

Problem Statement
The Utopian Tree goes through 2 cycles of growth every year. The first growth cycle occurs during the spring, when it doubles in height. The second growth cycle occurs during the summer, when its height increases by 1 meter.
Now, a new Utopian Tree sapling is planted at the onset of spring. Its height is 1 meter. Can you find the height of the tree after N growth cycles?

### SOURCE CODE:

 #include<stdio.h>
void main()
{
int t,test,total;
scanf("%d",&test);
while(test--){
scanf("%d",&t);
total=1;
int flag=1;
while(t--)
{
if(flag){
if(total==1){
total+=1;
}else{
total*=2;
}
flag--;
}
else{
total+=1;
flag++;
}
}
printf("%d\n",total);
}
}


** The above solution is my own code and it may not be the optimal solution or optimal way to approach the problem but it passes all the testcases in Hackerrank. So if you have any optimal approaches feel free to paste the code as the comment below..... :) :) :)

# Hackerrank Solve me first

The Solution for hackerrank problem, solve me first using C Program.

SOURCE CODE:
 #include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int solveMeFirst(int a, int b) {
return a+b;
}
int main() {
int num1,num2;
scanf("%d %d",&num1,&num2);
int sum;
sum = solveMeFirst(num1,num2);
printf("%d",sum);
return 0;
}


# Encoder

Text Encoder, This application is used to encode the message and convert the message into non readable format.
It can be further decoded to readable format by using the same application.

### Instructions To Use:

1) Copy any text from any applications
2) Open Encoder app the copied content will automatically appear in the text box.(If the copied text does not appear in the text box then delete the text in it and paste the copied text).
3) Click encode to change the readable text into non - readable format.
4) Click copy button to copy to clipboard.
5) Now you can send this to anyone or can save it in your mobile the message will be kept safe.

### To Decode:

1) Copy the encrypted text
2) Open Encoder app the copied content will automatically appear in the text box.(If the copied text does not appear in the text box then delete the text in it and paste the copied text).
3) Click Decode to change the encrypted text into readable text.

### Features:

Encode Text into secret code.
The Encoded messages can be decoded by this application.
Clipboard is automatically copied into the application.
Encoded text can be copied from the application.
Easy to use. Simple user interface.

SCREENSHOTS:

For Suggestions and Bugs comment below... :)

# UniPolar Encoding

## Unipolar encoding has 2 voltage states with one of the states being 0 volts. Since Unipolar line encoding has one of its states being 0 Volts, it is also called Return to Zero (RTZ). A common example of Unipolar line encoding is the logic levels used in computers and digital logic. A logic High (1) is represented by +5V and a logic Low (0) is represented by 0V.

### SOURCE CODE:

 import java.util.Scanner;
public class Lowhigh {
public static void main(String[] args) {
String input = "";
String line1 = "";
String line2 = "";
String line3 = "";
int flag = 0, i;
Scanner inputs = new Scanner(System.in);
System.out.print("Enter the binary number :");
input = inputs.next();
for (i = 0; i < input.length(); i++) {
if (input.charAt(i) == '1') {
if (flag == 0) {
line1 += " ";
line2 += "|";
line3 += "|";
flag = 1;
}
line1 = line1 + " _ ";
line2 = line2 + "  ";
line3 = line3 + "  ";
} else {
if (flag == 1) {
line1 += " ";
line2 += "|";
line3 += "|";
flag = 0;
}
line1 = line1 + "  ";
line2 = line2 + "  ";
line3 = line3 + " _ ";
}
}
System.out.println(line1);
System.out.println(line2);
System.out.println(line3);
}
}


# Ripple 14 ~ Android App

## Features:

Broadcasting application with two way communication.
Included college map.
Included Event coordinator's name and details.
Added support for SD card installation.

## Permissions Required:

This application requires the following permissions.
Permission to access internet.
Permission to access Network State.

## Minimum Requirements:

Android version 2.0 or above.
Requires minimum 1.4 MB storage space.
WiFI Enabled Android Mobile.