# Permuting Two Arrays

### Problem Statement :

```There are two -element arrays of integers,  and . Permute them into some  and  such that the relation  holds for all  where .

There will be  queries consisting of , , and . For each query, return YES if some permutation ,  satisfying the relation exists. Otherwise, return NO.

A valid  is  and :  and . Return YES.

Function Description

Complete the twoArrays function in the editor below. It should return a string, either YES or NO.

twoArrays has the following parameter(s):

int k: an integer
int A[n]: an array of integers
int B[n]: an array of integers
Returns
- string: either YES or NO

Input Format

The first line contains an integer , the number of queries.

The next  sets of  lines are as follows:

The first line contains two space-separated integers  and , the size of both arrays  and , and the relation variable.
The second line contains  space-separated integers .
The third line contains  space-separated integers .```

### Solution :

```                            ```Solution in C :

In  C  :

#include<stdio.h>
void sort(long long int x[],int first,int last){
long long int pivot,j,temp,i;
if(first<last){
pivot=first;
i=first;
j=last;

while(i<j){
while(x[i]<=x[pivot]&&i<last)
i++;
while(x[j]>x[pivot])
j--;
if(i<j){
temp=x[i];
x[i]=x[j];
x[j]=temp;
}
}

temp=x[pivot];
x[pivot]=x[j];
x[j]=temp;
sort(x,first,j-1);
sort(x,j+1,last);

}
}

main()
{
long long int a[1000],b[1000],i,j,k,flag,n,t;
scanf("%lld",&t);
while(t--)
{
scanf("%lld%lld",&n,&k);
for(i=0;i<n;++i)
scanf("%lld",&a[i]);
for(i=0;i<n;++i)
scanf("%lld",&b[i]);
sort(a,0,n-1);
sort(b,0,n-1);
flag=1;
for(i=0;i<n;++i)
{
if((a[i]+b[n-1-i])<k)
{
flag=0;
break;
}
}
if(flag==1)
printf("YES\n");
else
printf("NO\n");
}
}```
```

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

In  C++  :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int A[1005], B[1005];
int main() {
int T;
scanf("%d", &T);
while(T--){
int N, K;
scanf("%d%d", &N, &K);
for(int i=0; i < N; ++i)
scanf("%d", A+i);
for(int i=0; i < N; ++i)
scanf("%d", B+i);
sort(A, A+N);
sort(B, B+N);
bool ok=1;
for(int i=0; i < N; ++i)
if(A[i]+B[N-1-i] < K)
ok=0;
if(ok)
printf("YES\n");
else
printf("NO\n");
}

return 0;
}```
```

```                        ```Solution in Java :

In Java :

import java.io.*;
import java.math.BigInteger;
import java.util.*;

import static java.util.Arrays.*;

public class Solution {
private static final int mod = (int)1e9+7;

final Random random = new Random(1);
final IOFast io = new IOFast();

public void run() throws IOException {
int T = io.nextInt();
LOOP: while(T-- != 0) {
int n = io.nextInt();
int k = io.nextInt();
int[] A = new int[n];
int[] B = new int[n];
for(int i = 0; i < A.length; i++) {
A[i] = io.nextInt();
}
for(int i = 0; i < A.length; i++) {
B[i] = io.nextInt();
}
sort(A);
sort(B);
for(int i = 0; i < A.length; i++) {
if(A[i] + B[n-1-i] < k) {
io.out.println("NO");
continue LOOP;
}
}
io.out.println("YES");
}
}

void main() throws IOException {
//		IOFast.setFileIO("rle-size.in", "rle-size.out");
try {
run();
}
catch (EndOfFileRuntimeException e) { }
io.out.flush();
}

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

static class EndOfFileRuntimeException extends RuntimeException {
private static final long serialVersionUID = -8565341110209207657L; }

static
public class IOFast {
private PrintWriter out = new PrintWriter(System.out);

void setFileIO(String ins, String outs) throws IOException {
out = new PrintWriter(new FileWriter(outs));
}

//		private static final int BUFFER_SIZE = 50 * 200000;
private static final char[] buffer = new char[1024 * 8];
private static final char[] str = new char[500000*8*2];
private static boolean[] isDigit = new boolean[256];
private static boolean[] isSpace = new boolean[256];
private static boolean[] isLineSep = new boolean[256];

static {
for(int i = 0; i < 10; i++) { isDigit['0' + i] = true; }
isDigit['-'] = true;
isSpace[' '] = isSpace['\r'] = isSpace['\n'] = isSpace['\t'] = true;
isLineSep['\r'] = isLineSep['\n'] = true;
}

public int read() throws IOException {
pos = 0;
if(readLen <= 0) { throw new EndOfFileRuntimeException(); }
}
return buffer[pos++];
}

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

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

public char nextChar() throws IOException {
while(true) {
if(!isSpace[c]) { return (char)c; }
}
}

int reads(char[] cs, int len, boolean[] accept) throws IOException {
try {
while(true) {
if(accept[c]) { break; }
str[len++] = (char)c;
}
}
catch(EndOfFileRuntimeException e) { ; }

return len;
}

public char[] nextLine() throws IOException {
int len = 0;
str[len++] = nextChar();

try {
if(str[len-1] == '\r') { len--; read(); }
}
catch(EndOfFileRuntimeException e) { ; }

return Arrays.copyOf(str, len);
}

public String nextString() throws IOException {
return new String(next());
}

public char[] next() throws IOException {
int len = 0;
str[len++] = nextChar();
return Arrays.copyOf(str, len);
}

public double nextDouble() throws IOException {
return Double.parseDouble(nextString());
}

}

}```
```

```                        ```Solution in Python :

In  Python3 :

def isGood(listA, listB, k):
n = len(listA)
listA.sort()
listB.sort(reverse=True)
for i in range(n):
if listA[i]+listB[i] < k:
return False
return True

T = int(input().strip())
for i in range(T):
[n, k] = [int(x) for x in input().strip().split()]
listA = [int(x) for x in input().strip().split()]
listB = [int(x) for x in input().strip().split()]
if isGood(listA, listB, k):
print("YES")
else:
print("NO")```
```

## Binary Search Tree : Lowest Common Ancestor

You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b

## Swap Nodes [Algo]

A binary tree is a tree which is characterized by one of the following properties: It can be empty (null). It contains a root node only. It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees. In-order traversal is performed as Traverse the left subtree. Visit root. Traverse the right subtree. For this in-order traversal, start from

## Kitty's Calculations on a Tree

Kitty has a tree, T , consisting of n nodes where each node is uniquely labeled from 1 to n . Her friend Alex gave her q sets, where each set contains k distinct nodes. Kitty needs to calculate the following expression on each set: where: { u ,v } denotes an unordered pair of nodes belonging to the set. dist(u , v) denotes the number of edges on the unique (shortest) path between nodes a

## Is This a Binary Search Tree?

For the purposes of this challenge, we define a binary tree to be a binary search tree with the following ordering requirements: The data value of every node in a node's left subtree is less than the data value of that node. The data value of every node in a node's right subtree is greater than the data value of that node. Given the root node of a binary tree, can you determine if it's also a

## Square-Ten Tree

The square-ten tree decomposition of an array is defined as follows: The lowest () level of the square-ten tree consists of single array elements in their natural order. The level (starting from ) of the square-ten tree consists of subsequent array subsegments of length in their natural order. Thus, the level contains subsegments of length , the level contains subsegments of length , the

## Balanced Forest

Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to a