# Two Strings

### Problem Statement :

```Given two strings, determine if they share a common substring . A substring may be as small as one character.

For example, the words "a", "and", "art" share the common substring a. The words "be" and "cat" do not share a substring.

Function Description

Complete the function twoStrings in the editor below. It should return a string, either YES or NO based on whether the strings share a common substring.

twoStrings has the following parameter(s):

s1, s2: two strings to analyze .

Input Format

The first line contains a single integer p, the number of test cases.

The following p pairs of lines are as follows:

1. The first line contains string  s1
2. The second line contains string s2.

Constraints

s1 and s2 consist of characters in the range ascii[a-z].
1 <= p <= 10
1 <= | s1 |, | s2 | <= 10 ^5

Output Format

For each pair of strings, return YES or NO.

.```

### Solution :

```                            ```Solution in C :

In C :

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

int main() {
int k;
scanf("%d",&k);
while(k>0)
{
char str[110000],btr[110000];
int cn1[500],cn2[500],i,j,t,a,b;
for(i=0;i<500;i++)
{
cn1[i]=0;
cn2[i]=0;
}
scanf("%s",str);
scanf("%s",btr);
a=strlen(str);
b=strlen(btr);
for(i=0;i<a;i++)
{
cn1[str[i]]++;
}
for(i=0;i<b;i++)
{
cn2[btr[i]]++;
}
t=0;
for(i=0;i<500;i++)
{
if(cn1[i]*cn2[i]>0)
{
t=1;
break;
}
}
if(t)
{
printf("YES\n");
}
else
{
printf("NO\n");
}
k--;
}
return 0;
}```
```

```                        ```Solution in C++ :

In C++ :

#include <bits/stdc++.h>
using namespace std;
int main()
{
int a; cin >> a;
for (int g=0;g<a; g++)
{
string b,c; cin >> b >> c; map <char,int> k;
for (int y=0;y<b.length(); y++) k[b[y]]=1; int counter=0;
for (int y=0;y<c.length(); y++)
{
if (k[c[y]]) counter=1;
}
if (counter) cout << "YES" << '\n';
else cout << "NO" << '\n';
}return 0;
}```
```

```                        ```Solution in Java :

In Java :

import java.io.*;
import java.util.*;

public class TwoStrings {

PrintWriter out;
StringTokenizer st;
boolean eof;

void solve() throws IOException {
int t = nextInt();
outer: while (t-- > 0) {
char[] a = nextToken().toCharArray();
char[] b = nextToken().toCharArray();
boolean[] has = new boolean[26];
for (char c : a) {
has[c - 'a'] = true;
}
for (char c : b) {
if (has[c - 'a']) {
out.println("YES");
continue outer;
}
}
out.println("NO");
}
}

TwoStrings() throws IOException {
out = new PrintWriter(System.out);
solve();
out.close();
}

public static void main(String[] args) throws IOException {
new TwoStrings();
}

String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
} catch (Exception e) {
eof = true;
return null;
}
}
return st.nextToken();
}

String nextString() {
try {
} catch (IOException e) {
eof = true;
return null;
}
}

int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}

long nextLong() throws IOException {
return Long.parseLong(nextToken());
}

double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
}```
```

```                        ```Solution in Python :

In Python3 :

a = int(input())
for i in range(a):
i = (input())
j = (input())
setx = set([a for a in i])
sety = set([y for y in j])
if setx.intersection(sety) == set():
print("NO")
else:
print("YES")```
```

## Reverse a doubly linked list

This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.

## Tree: Preorder Traversal

Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's

## Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your \$inOrder* func

## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.