Given an array arr[] of size N, Return the number of permutations of array arr[] which satisfy the condition arr[1] & arr[2] & . . . & arr[i] = arr[i+1] & arr[i+2] & . . . & arr[N], for all i.
Note: Here & denotes the bitwise AND operation.
Examples:
Input: N = 3, arr[] = { 1, 1, 1 }
Output: 6
Explanation: Since all the numbers are equal, whatever permutation we take, the sequence will follow the above condition. There are a total of 6 permutations possible with index numbers from 1 to 3 : [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1].Input: N = 4, arr[] = { 1, 3, 5, 1 }
Output: 4
Approach: This problem can be solved based on the following idea:
Consider an arbitrary sequence b1, b2, . . ., bn. First, let us define the arrays AND_pref and AND_suf of length N where
- AND_prefi = b1 & b2 & . . . & bi and
- AND_sufi = bi & bi+1 & . . . & bn.
According to the definition of the sequence: AND_pref1 = AND_suf2. Now AND_pref2 ? AND_pref1 = AND_suf2 ? AND_suf3. Also according to the definition, AND_pref2 = AND_suf3. This means that b1 = AND_pref2 = AND_suf3.
Similarly, for all i from 1 to n, we get AND_prefi = b1 and AND_sufi = b1.
Therefore for the sequence, b1 = bn and the bi must be a super mask of b1 for all i from 2 to n ? 1.
Follow the steps below to solve the problem:
- Initialize a variable preAnd with ( 1 << 30 ) – 1.
- Run a loop from i = 0 to n-1 and update preAnd with ( preAnd & arr[i] ).
- Initialize a count variable (say cnt) with 0.
- Run a loop from i = 0 to n – 1
- If preAnd = arr[i], then Increment cnt by 1.
- Compute (cnt * ( cnt – 1 ) * (n – 2) !) % (1e9 + 7) and store it in the answer variable.
- Return the answer.
Below is the implementation of the above approach:
// C++ code to implement the above approach #include <bits/stdc++.h> using namespace std;
#define ll long long // Given mod number . ll mod = 1e9 + 7; // Function performing calculation int countAndGood( int n, vector< int >& arr)
{ // Initializing preAnd .
int preAnd = (1 << 30) - 1;
// Precomputing the And of the array arr
for ( int i = 0; i < n; i++) {
preAnd = (preAnd & arr[i]);
}
// Initializing cnt with 0
ll cnt = 0;
// Counting the total number in arr which
// are equal to preAnd
for ( int i = 0; i < n; i++) {
if (preAnd == arr[i])
cnt++;
}
// Finding (cnt)P(cnt-2)
ll ans = (cnt * (cnt - 1)) % mod;
// Finding (n-2)!
ll temp = 1;
for (ll i = 2; i <= n - 2; i++) {
temp = (temp * i) % mod;
}
// Multiplying temp and ans
ans = (ans * temp) % mod;
// Returning ans variable
return ans;
} // Driver code int main()
{ int N = 4;
vector< int > arr = { 1, 3, 5, 1 };
// Function call
cout << countAndGood(N, arr);
return 0;
} |
// Java code to implement the above approach import java.util.*;
class GFG {
// Given mod number .
static final int mod = ( int )(1e9 + 7 );
// Function performing calculation
static int countAndGood( int n, List<Integer> arr)
{
// Initializing preAnd .
int preAnd = ( 1 << 30 ) - 1 ;
// Precomputing the And of the array arr
for ( int i = 0 ; i < n; i++) {
preAnd = (preAnd & arr.get(i));
}
// Initializing cnt with 0
long cnt = 0 ;
// Counting the total number in arr which
// are equal to preAnd
for ( int i = 0 ; i < n; i++) {
if (preAnd == arr.get(i))
cnt++;
}
// Finding (cnt)P(cnt-2)
long ans = (cnt * (cnt - 1 )) % mod;
// Finding (n-2)!
long temp = 1 ;
for ( long i = 2 ; i <= n - 2 ; i++) {
temp = (temp * i) % mod;
}
// Multiplying temp and ans
ans = (ans * temp) % mod;
// Returning ans variable
return ( int )ans;
}
// Driver Code
public static void main(String[] args)
{
int N = 4 ;
List<Integer> arr = new ArrayList<>();
arr.add( 1 );
arr.add( 3 );
arr.add( 5 );
arr.add( 1 );
// Function Call
System.out.println(countAndGood(N, arr));
}
} // This Code is Contributed by Prasad Kandekar(prasad264) |
# Python code to implement the above approach # Given mod number . MOD = int ( 1e9 + 7 )
# Function performing calculation def countAndGood(n, arr):
# Initializing preAnd .
preAnd = ( 1 << 30 ) - 1
# Precomputing the And of the array arr
for i in range (n):
preAnd & = arr[i]
# Initializing cnt with 0
cnt = 0
# Counting the total number in arr which
# are equal to preAnd
for i in range (n):
if preAnd = = arr[i]:
cnt + = 1
# Finding (cnt)P(cnt-2)
ans = (cnt * (cnt - 1 )) % MOD
# Finding (n-2)!
temp = 1
for i in range ( 2 , n - 2 + 1 ):
temp = (temp * i) % MOD
# Multiplying temp and ans
ans = (ans * temp) % MOD
# Returning ans variable
return ans
# Driver Code N = 4
arr = [ 1 , 3 , 5 , 1 ]
# Function Call print (countAndGood(N, arr))
# This Code is Contributed by Prasad Kandekar(prasad264) |
// C# code to implement the above approach using System;
using System.Collections;
class Gfg
{ // Given mod number .
static long mod = 1000000007;
// Function performing calculation
static long countAndGood( int n, int [] arr)
{
// Initializing preAnd .
int preAnd = (1 << 30) - 1;
// Precomputing the And of the array arr
for ( int i = 0; i < n; i++) {
preAnd = (preAnd & arr[i]);
}
// Initializing cnt with 0
long cnt = 0;
// Counting the total number in arr which
// are equal to preAnd
for ( int i = 0; i < n; i++) {
if (preAnd == arr[i])
cnt++;
}
// Finding (cnt)P(cnt-2)
long ans = (cnt * (cnt - 1)) % mod;
// Finding (n-2)!
long temp = 1;
for ( long i = 2; i <= n - 2; i++) {
temp = (temp * i) % mod;
}
// Multiplying temp and ans
ans = (ans * temp) % mod;
// Returning ans variable
return ans;
}
// Driver code
static void Main( string [] args)
{
int N = 4;
int [] arr = { 1, 3, 5, 1 };
// Function call
Console.Write(countAndGood(N, arr));
}
} // this code is contributed by poojaagarwal2. |
// Javascript code to implement the above approach // Given mod number . const MOD = 1e9 + 7; // Function performing calculation function countAndGood(n, arr) {
// Initializing preAnd .
var preAnd = (1 << 30) - 1;
// Precomputing the And of the array arr
for ( var i = 0; i < n; i++) {
preAnd &= arr[i];
}
// Initializing cnt with 0
var cnt = 0;
// Counting the total number in arr which
// are equal to preAnd
for ( var i = 0; i < n; i++) {
if (preAnd === arr[i]) {
cnt++;
}
}
// Finding (cnt)P(cnt-2)
var ans = (cnt * (cnt - 1)) % MOD;
// Finding (n-2)!
var temp = 1;
for ( var i = 2; i <= n - 2; i++) {
temp = (temp * i) % MOD;
}
// Multiplying temp and ans
ans = (ans * temp) % MOD;
// Returning ans variable
return ans;
} // Driver Code var N = 4;
var arr = [1, 3, 5, 1];
// Function Call console.log(countAndGood(N, arr)); // This Code is Contributed by Prasad Kandekar(prasad264) |
4
Time Complexity: O(N)
Auxiliary Space: O(1)
Related Articles: