# Yet Another Minimax Problem

### Problem Statement :

```You are given n non-negative integers, a0,a1,...,an-1. We define the score for some permutation () of length  to be the maximum of api ^ apI+1 for 0 <= i < n-1.

Find the permutation with the minimum possible score and print its score.

Note: ^ is the exclusive-OR (XOR) operator.

Input Format

The first line contains single integer, n, denoting the number of integers.
The second line contains n space-separated integers, a0,a1,...,an-1, describing the respective integers.

Constraints
2 <= n <= 3000
0 <= ai <= 10^9

Output Format

Print a single integer denoting the minimum possible score.```

### Solution :

```                            ```Solution in C :

In C++ :

#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
#include <cmath>

using namespace std;

int n,i,j,a[3005],p[3005],siz[3005];

struct edge{
int x,y,cost;
bool operator < (const edge& T) const{
return cost < T.cost;
}
};

vector <edge> b;

int find(int s){
if(p[s] == s)
return s;
return p[s] = find(p[s]);
}

int main(){
scanf("%d", &n);
for(i = 0; i < n; i++)
scanf("%d", &a[i]);
for(i = 0; i < n; i++){
p[i] = i;
siz[i] = 1;
}
for(i = 0; i < n; i++){
for(j = i + 1; j < n; j++){
b.push_back({i ,j , a[i] ^ a[j]});
}
}
sort(b.begin(), b.end());
for(i = 0; i < b.size(); i++){
int x = find(b[i].x);
int y = find(b[i].y);
if(x != y){
p[x] = y;
siz[y] += siz[x];
}
if(siz[y] == n)
return 0 * printf("%d\n", b[i].cost);
}
}

In Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] a = new int[n];

for (int i = 0; i < n; i++) {
a[i] = sc.nextInt();
}
Arrays.sort(a);

if (a[0]==a[n-1]) {
System.out.println(0);
return;
}

int min = 0;
int max = 0;

for (int i = 1; i < 31; i++) {
if (a[0] >= (1<<i))
min++;
}

for (int i = 1; i < 31; i++) {
if (a[n-1] >= (1<<i))
max++;
}

while (min==max&&min!=0) {
for (int i = 0; i < n; i++) {
a[i] -= 1<<min;
}
min = 0;
max = 0;
for (int i = 1; i < 31; i++) {
if (a[0] >= (1<<i))
min++;
}
for (int i = 1; i < 31; i++) {
if (a[n-1] >= (1<<i))
max++;
}
}

if (max==0) {
System.out.println(0);
return;
}

ArrayList<Integer> l = new ArrayList<Integer>();
ArrayList<Integer> h = new ArrayList<Integer>();

for (int i = 0; i < n; i++) {
if (a[i] < (1<<max))
else
}

int result = Integer.MAX_VALUE;
for (int i : l) {
for (int j : h) {
if ((i^j)<result)
result = i^j;
if (result == (1<<max))
break;
}
if (result == (1<<max))
break;
}
System.out.println(result);
}
}

In C :

#include <stdio.h>
#define N 5000

int main() {
int n, i, j, m, t, p2, a[N];
scanf("%d", &n);
for(i=0; i<n; i++) scanf("%d", &a[i]);
m=0;
for(i=0; i<n; i++) for(j=i+1; j<n; j++) if((a[i]^a[j])>m) m=a[i]^a[j];
if(m==0) {
printf("0\n");
return 0;
}
for(p2=t=m; t; t>>=1) p2|=t;
p2=(p2>>1)+1;
for(i=0; i<n; i++) for(j=i+1; j<n; j++) if((a[i]^a[j])&p2)
if((a[i]^a[j])<m) m=a[i]^a[j];
printf("%d\n", m);
return 0;
}

In Python3 :

def solve(As) :
a = max(As)
b = -1
while a > 0 :
a >>= 1
b += 1

if b < 0 :
return 0

l = 1<<b
while all(a & l for a in As) :
l >>= 1

if l == 0 :
return 0

Al = [a for a in As if not a & l]
Au = [a for a in As if a & l]

best = l << 1
for al in Al :
for au in Au :
if al ^ au < best :
best = al ^ au

return best

n = int(input())
As = set(map(int,input().split()))
print(solve(As))```
```

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