# Sansa and XOR

### Problem Statement :

```Sansa has an array. She wants to find the value obtained by XOR-ing the contiguous subarrays, followed by XOR-ing the values thus obtained. Determine this value.

Example

Subarray	Operation	Result
3		None		3
4		None		4
5		None		5
3,4		3 XOR 4		7
4,5		4 XOR 5		1
3,4,5		3 XOR 4 XOR 5	2
Now we take the resultant values and XOR them together:

. Return .

Function Description

Complete the sansaXor function in the editor below.

sansaXor has the following parameter(s):

int arr[n]: an array of integers
Returns

int: the result of calculations

Input Format

The first line contains an integer , the number of the test cases.

Each of the next  pairs of lines is as follows:
- The first line of each test case contains an integer , the number of elements in .
- The second line of each test case contains  space-separated integers .```

### Solution :

```                            ```Solution in C :

In  C :

#include <stdio.h>
#include <stdlib.h>
#define MOD 1000000007
#define s(k) scanf("%d",&k);
#define sll(k) scanf("%lld",&k);
#define p(k) printf("%d\n",k);
#define pll(k) printf("%lld\n",k);
#define f(i,N) for(i=0;i<N;i++)
#define f1(i,N) for(i=0;i<=N;i++)
#define f2(i,N) for(i=1;i<=N;i++)
typedef long long ll;
int main(){
int t,n,k,sum,i;
s(t)
while(t--){
s(n)
if(!(n%2)){
printf("0\n");
f(i,n)
s(k)
} else {
sum=0;
f(i,n){
s(k)
if(!(i%2))
sum^=k;
}
p(sum);
}
}
return 0;
}```
```

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

In  C++  :

#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <limits>
#include <cstring>
#include <string>
using namespace std;

#define pairii pair<int, int>
#define llong long long
#define pb push_back
#define sortall(x) sort((x).begin(), (x).end())
#define INFI  numeric_limits<int>::max()
#define INFLL numeric_limits<llong>::max()
#define INFD  numeric_limits<double>::max()
#define FOR(i,s,n) for (int (i) = (s); (i) < (n); (i)++)
#define FORZ(i,n) FOR((i),0,(n))

void solve() {
int n;
scanf("%d",&n);
int res = 0;
FORZ(i,n) {
int x;
scanf("%d",&x);
if ((i+1)%2 && (n-i)%2) {
res ^= x;
}
}
printf("%d\n",res);
}

int main() {
#ifdef DEBUG
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
int t;
scanf("%d",&t);
FORZ(i,t) solve();
return 0;
}```
```

```                        ```Solution in Java :

In  Java :

import java.io.IOException;
import java.util.HashMap;
import java.util.Map;

public class Solution {
public static void main(String[] args) throws IOException {
int num = Integer.parseInt(line);

for (int i = 0; i < num; i++) {
sansaXor(ns, Integer.valueOf(count));
}

}

public static void sansaXor(String[] strs, int count){
Map<Integer,Integer> map = new HashMap<Integer,Integer>();
long result = 0;
for(int i = 0; i < count; i++){
int numCount = (i + 1)*(count - i);
int tmp = Integer.valueOf(strs[i]);
if(map.containsKey(tmp)){
map.put(tmp, numCount+map.get(tmp));
}else{
map.put(tmp, numCount);
}
}

for(int k: map.keySet()){
int value = map.get(k);
if(value%2!=0){
result = result^k;
}
}

System.out.println(result);
}
}```
```

```                        ```Solution in Python :

In  Python3 :

for _ in range(int(input())):
N = int(input())
A = tuple(map(int,input().split()))
X = 0
for i, x in enumerate(A):
if ((i+1) * (N-i)) % 2 == 1:
X ^= x
print(X)```
```

## 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.

## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

## Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

## Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t

## 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