Depend permutations whose prefix & suffix AND are similar for every index

By
Sohaib
-
0
11
Adv1


Adv2

Given an array arr[] of measurement N, Return the variety of permutations of array arr[] which fulfill the situation arr[1] & arr[2] & . . . & arr[i] = arr[i+1] & arr[i+2] & . . . & arr[N] , for all i.

Be aware: Right here & denotes the bitwise AND operation.

Examples:

Enter: N = 3, arr[] = { 1, 1, 1 } 
Output: 6 
Clarification: Since all of the numbers are equal, no matter permutation we take, the sequence will observe the above situation. There are a complete of 6 permutations doable with index numbers from 1 to three : [1,  2,  3],  [1,  3,  2],  [2, 1, 3],  [2, 3, 1],  [3, 1,  2],  [3, 2, 1].

Enter: N = 4, arr[] = { 1, 3, 5, 1 }
Output: 4

Strategy: This downside will be solved primarily based on the next concept:

Take into account an arbitrary sequence b1, b2, . . ., bn. First, allow us to outline the arrays AND_pref and AND_suf of size N the place 

  • AND_prefi = b1 & b2 & . . . & bi and 
  • AND_sufi = bi & bi+1 & . . . & bn.

In keeping with the definition of the sequence: AND_pref1 = AND_suf2. Now AND_pref2 ≤ AND_pref1 = AND_suf2 ≤ AND_suf3. Additionally based on the definition, AND_pref2 = AND_suf3. Which means b1 = AND_pref2 = AND_suf3. 

Equally, for all i from 1 to n, we get AND_prefi = b1 and AND_sufi = b1.
Due to this fact for the sequence, b1 = bn and the bi should be a brilliant masks of  b1 for all i from 2 to n − 1.

Observe the steps beneath to unravel the issue:

  • Initialize a variable preAnd with ( 1 << 30 ) – 1.
  • Run a loop from i = 0 to n-1 and replace preAnd with ( preAnd & arr[i] ).   
  • Initialize a rely 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 retailer it within the reply variable.
  • Return the reply.

Beneath is the implementation of the above method:

C++

  

#embody <bits/stdc++.h>

utilizing namespace std;

#outline ll lengthy lengthy

  

ll mod = 1e9 + 7;

  

int countAndGood(int n, vector<int>& arr)

{

    

    int preAnd = (1 << 30) - 1;

  

    

    for (int i = 0; i < n; i++) {

        preAnd = (preAnd & arr[i]);

    }

  

    

    ll cnt = 0;

  

    

    

    for (int i = 0; i < n; i++) {

        if (preAnd == arr[i])

            cnt++;

    }

  

    

    ll ans = (cnt * (cnt - 1)) % mod;

  

    

    ll temp = 1;

    for (ll i = 2; i <= n - 2; i++) {

        temp = (temp * i) % mod;

    }

  

    

    ans = (ans * temp) % mod;

  

    

    return ans;

}

  

int most important()

{

    int N = 4;

    vector<int> arr = { 1, 3, 5, 1 };

  

    

    cout << countAndGood(N, arr);

  

    return 0;

}

Time Complexity: O(N) 
Auxiliary Area: O(1)

Associated Articles:

Adv3