# Manipulative Numbers

### Problem Statement :

```Suppose that  is a list of  numbers  and  is a permutation of these numbers, we say B is K-Manipulative if and only if:

is not less than , where  represents the XOR operator.

You are given . Find the largest  such that there exists a K-manipulative permutation .

Input:

The first line is an integer . The second line contains  space separated integers - .

Output:
The largest possible , or  if there is no solution.```

### Solution :

```                            ```Solution in C :

In  C :

#include<stdio.h>
#include<string.h>
int compare (const void * a, const void * b)
{
return ( *(int*)a - *(int*)b );
}
int isMajority(int a[], int size, int cand)
{
int i, count = 0;
for (i = 0; i < size; i++)
if(a[i] == cand)
count++;
if (count > size/2)
return 1;
else
return 0;
}

int findCandidate(int a[], int size)
{
int maj_index = 0, count = 1;
int i;
for(i = 1; i < size; i++)
{
if(a[maj_index] == a[i])
count++;
else
count--;
if(count == 0)
{
maj_index = i;
count = 1;
}
}
return a[maj_index];
}

int printMajority(int a[], int size)
{
int cand = findCandidate(a, size);
if(isMajority(a, size, cand))
return -1;
return 0;
}

/*void print(int *A,int n)
{
int i;
for(i=0;i<n;i++)
printf("%d ",A[i]);
printf("\n");
}*/
int main()
{
int n,A[1000],i,j,B[1000],flag=-1;
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&A[i]);
//print(A,n);
qsort (A,n,sizeof(int),compare);
//print(A,n);
for(i=31;i>=1;i--)
{
memcpy(&B,A,n* sizeof(int));
//print(B,n);
for(j=0;j<n;j++)
B[j]=B[j]>>i;
//print(B,n);
if(printMajority(B,n)==0)
{
printf("%d\n",i);
flag=0;
break;
}
}
if(flag==-1)
printf("-1\n");
return 0;
}```
```

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

In  C ++  :

#include <iostream>
#include <algorithm>
using namespace std;

int highbit(int x)
{
int k = -1;
while (x != 0)
{
x >>= 1;
k++;
}
return k;
}

int main()
{
int a[100], b[100], n;
cin >> n;
for (int i = 0; i < n; i++)
cin >> a[i];
int *p = a, *q = b;
int m = n / 2;
sort(p, p + n);
int k = highbit(p[m]);
while (k >= 0)
{
int x = 1 << k;
for (int i = 0; i < n; i++)
q[i] = p[i] ^ x;
sort(q, q + n);
int s = highbit(q[m]);
if (s == k) break;
swap(p, q);
k = s;
}
cout << k << endl;
return 0;
}```
```

```                        ```Solution in Java :

In  Java :

import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;

public class Solution {
static InputStream is;
static PrintWriter out;
static String INPUT = "";

static void solve()
{
int n = ni();
int[] a = new int[n];
int[] b = new int[n];
for(int i = 0;i < n;i++)a[i] = ni();

for(int k = 30;k >= 0;k--){
for(int i = 0;i < n;i++)b[i] = a[i] >>> k;
if(iskmul(b)){
out.println(k);
return;
}
}
out.println(-1);
}

static boolean iskmul(int[] a)
{
int n = a.length;
Arrays.sort(a);
int ct = 0;
int max = 1;
for(int i = 0;i < n;i++){
if(i == 0 || a[i] != a[i-1]){
ct = 1;
}else{
ct++;
max = Math.max(max, ct);
}
}
return max <= n/2;
}

public static void main(String[] args) throws Exception
{
long S = System.currentTimeMillis();
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);

solve();
out.flush();
long G = System.currentTimeMillis();
tr(G-S+"ms");
}

static boolean eof()
{
try {
is.mark(1000);
int b;
while((b = is.read()) != -1 && !(b >= 33 && b <= 126));
is.reset();
return b == -1;
} catch (IOException e) {
return true;
}
}

static int ni()
{
try {
int num = 0;
boolean minus = false;
while((num = is.read()) != -1 && !((num >= '0' && num <= '9') || num == '-'));
if(num == '-'){
num = 0;
minus = true;
}else{
num -= '0';
}

while(true){
int b = is.read();
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
}
} catch (IOException e) {
}
return -1;
}

static void tr(Object... o) { if(INPUT.length() != 0)System.out.println(Arrays.deepToString(o)); }
}```
```

```                        ```Solution in Python :

In  Python3 :

from collections import Counter
input()
a = list(map(int, input().split()))
for i in range(32):
if Counter(x >> i for x in a).most_common(1)[0][1] <= len(a) // 2:
else:
break
```

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