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Count of triplets whose XOR has even set bits in range [L, R] for Q queries

  • Last Updated : 08 Jul, 2021

Given an array arr[] of size N, and an array Q[] consisting of M queries of type {L, R}, the task is to print the count of triplets whose XOR has an even number of set bits in the range [L, R], for each of the M queries.

Examples:

Input: arr[] = {1, 2, 3, 4, 5}, N = 5, Q[] = {{1, 4}, {3, 4}}, M = 2
Output: 3
              0
Explanation:
Perform the query as following:

  1. Query(1, 4): Print 3. The triplets whose xor has an even number of set bits are {1, 2, 3}, {1, 3, 4} and {2, 3, 4}.
  2. Query(3, 4): Print 0. There are no triplets that satisfy the condition.

Input: arr[] = {3, 3, 3}, N = 2, Q[] = {{1, 3}}, M = 1
Output: 1

Approach: The given problem can be solved based on the following observations: 



  1. The XOR of two numbers X and Y i.e X^Y can have an even number of set bits only if:
    1. Both X and Y have an even number of set bits.
    2. Both X and Y have an odd number of set bits.
  2. Thus, there are only cases, where the XOR of three numbers has an even number of set bits, which are:
    1. All three numbers have an even number of set bits.
    2. Exactly two of them have an odd number of set bits and the other has an even number of set bits.
  3. Therefore, the problem can be solved with the concept of prefix arrays by applying the above observations.

Follow the steps below to solve the problem:

  • Initialize a prefix array say preo[] of size N+1, where preo[i] stores the number of elements up to index i, that have an odd number of set bits.
  • Create another prefix array say pree of size N+1, where pree[i] stores the number of elements up to index i, that have an even number of set bits.
  • Traverse the array arr[], and do the following:
    • Use the __builtin_popcount() function to calculate the number of set bits in the current element.
    • If the number of set bits is odd, increment preo[i]
    • Otherwise, increment pree[i]
  • Traverse the array Q[] and perform the following steps:
    • Calculate the number of elements that have an odd number of set bits and store it in a variable say odd, using the preo[] array.
    • Calculate the number of elements that have an even number of set bits and store it in a variable say even, using the pree [] array.
    • Initialize a variable ans to 0 to store the count of triplets.
    • If even is not less than 3, add evenC3 to ans.
    • If even is not less than 1 and odd is not less than 2, add (oddC2)*(evenC1) to ans.
    • Finally, print the count of triplets obtained in ans for the current query {L, R}.

Below is an implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Utility function to calculate NCR
int NCR(int N, int R)
{
    if (R > N - R)
        R = N - R;
    int ans = 1;
    for (int i = 1; i <= R; i++) {
        ans *= N - R + i;
        ans /= i;
    }
    return ans;
}
// Function to answer queries about
// the number of triplets  in range
// whose XOR has an even number of
// set bits
void solveXorTriplets(int arr[], int N,
                      vector<pair<int, int> > Q, int M)
{
    // Prefix arrays to store
    // number that have odd
    // and even setbits
    int preo[N + 1] = { 0 };
    int pree[N + 1] = { 0 };
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
        // Update
        preo[i + 1] += preo[i];
        pree[i + 1] += pree[i];
 
        // Find bits in current
        // element
        int setbits = __builtin_popcount(arr[i]);
 
        // If number of setbits
        // is odd
        if (setbits % 2)
            preo[i + 1]++;
        // Otherwise
        else
            pree[i + 1]++;
    }
    // Traverse the query Q[][]
    for (int i = 0; i < M; i++) {
        // Stores Left boundary
        int L = Q[i].first;
        // Stores Right boundary
        int R = Q[i].second;
 
        // Stores number of elements
        // that have odd set bits in
        // the given range
        int odd = preo[R] - preo[L - 1];
 
        // Store number of elements
        // that have even set bits
        // in given range
        int even = pree[R] - pree[L - 1];
 
        // Store the count
        // of the triplets
        int ans = 0;
 
        // If even is greater
        // than ore equal to 3
        if (even >= 3)
            ans += NCR(even, 3);
 
        // If odd is greater than
        // or equal to and even is
        // greater than or equal to
        // 1
        if (odd >= 2 && even >= 1)
            ans += NCR(odd, 2) * NCR(even, 1);
 
        // Print the answer
        // for current query
        cout << ans << endl;
    }
}
// Driver Code
int main()
{
 
    int arr[] = { 1, 2, 3, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    vector<pair<int, int> > Q = { { 1, 4 }, { 3, 4 } };
    int M = Q.size();
 
    solveXorTriplets(arr, N, Q, M);
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
static class pair
{
    int first, second;
     
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }  
   
// Utility function to calculate NCR
static int NCR(int N, int R)
{
    if (R > N - R)
        R = N - R;
         
    int ans = 1;
     
    for(int i = 1; i <= R; i++)
    {
        ans *= N - R + i;
        ans /= i;
    }
    return ans;
}
 
// Function to answer queries about
// the number of triplets  in range
// whose XOR has an even number of
// set bits
static void solveXorTriplets(int arr[], int N,
                             Vector<pair> Q, int M)
{
     
    // Prefix arrays to store
    // number that have odd
    // and even setbits
    int preo[] = new int[N + 1];
    int pree[] = new int[N + 1];
 
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
         
        // Update
        preo[i + 1] += preo[i];
        pree[i + 1] += pree[i];
 
        // Find bits in current
        // element
        int setbits = Integer.bitCount(arr[i]);
 
        // If number of setbits
        // is odd
        if (setbits % 2 != 0)
            preo[i + 1]++;
             
        // Otherwise
        else
            pree[i + 1]++;
    }
     
    // Traverse the query Q[][]
    for(int i = 0; i < M; i++)
    {
         
        // Stores Left boundary
        int L = Q.elementAt(i).first;
         
        // Stores Right boundary
        int R =  Q.elementAt(i).second;
 
        // Stores number of elements
        // that have odd set bits in
        // the given range
        int odd = preo[R] - preo[L - 1];
 
        // Store number of elements
        // that have even set bits
        // in given range
        int even = pree[R] - pree[L - 1];
 
        // Store the count
        // of the triplets
        int ans = 0;
 
        // If even is greater
        // than ore equal to 3
        if (even >= 3)
            ans += NCR(even, 3);
 
        // If odd is greater than
        // or equal to and even is
        // greater than or equal to
        // 1
        if (odd >= 2 && even >= 1)
            ans += NCR(odd, 2) * NCR(even, 1);
 
        // Print the answer
        // for current query
        System.out.println(ans);
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int N = arr.length;
 
    Vector<pair> Q = new Vector<>();
      Q.add(new pair(1, 4));
      Q.add(new pair(3, 4));
   
    int M = Q.size();
 
    solveXorTriplets(arr, N, Q, M);
}
}
 
// This code is contributed by Dharanendra L V.

Python3




# Python3 program for the above approach
 
 
# Utility function to calculate NCR
def NCR(N, R):
    if (R > N - R):
        R = N - R
    ans = 1
    for i in range(1,R+1):
        ans *= N - R + i
        ans //= i
    return ans
 
# Function to answer queries about
# the number of triplets  in range
# whose XOR has an even number of
# set bits
def solveXorTriplets(arr, N, Q, M):
    # Prefix arrays to store
    # number that have odd
    # and even setbits
    preo = [0]*(N + 1)
    pree = [0]*(N + 1)
 
    # Traverse the array arr[]
    for i in range(N):
        preo[i + 1] += preo[i]
        pree[i + 1] += pree[i]
 
        # Find bits in current
        # element
        setbits =bin(arr[i]).count('1')
 
        # If number of setbits
        # is odd
        if (setbits % 2):
            preo[i + 1] +=1
        # Otherwise
        else:
            pree[i + 1]+=1
 
    # Traverse the query Q[][]
    for i in range(M):
       
        # Stores Left boundary
        L = Q[i][0]
         
        # Stores Right boundary
        R = Q[i][1]
 
        # Stores number of elements
        # that have odd set bits in
        # the given range
        odd = preo[R] - preo[L - 1]
 
        # Store number of elements
        # that have even set bits
        # in given range
        even = pree[R] - pree[L - 1]
 
        # Store the count
        # of the triplets
        ans = 0
 
        # If even is greater
        # than ore equal to 3
        if (even >= 3):
            ans += NCR(even, 3)
 
        # If odd is greater than
        # or equal to and even is
        # greater than or equal to
        # 1
        if (odd >= 2 and even >= 1):
            ans += NCR(odd, 2) * NCR(even, 1)
 
        # Prthe answer
        # for current query
        print (ans)
         
# Driver Code
if __name__ == '__main__':
 
    arr = [1, 2, 3, 4, 5]
    N = len(arr)
 
    Q = [ [ 1, 4 ], [ 3, 4 ] ]
    M = len(Q)
 
    solveXorTriplets(arr, N, Q, M)
 
# This code is contributed by mohit kumar 29.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class pair{
 
    int first, second;
     
    pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }  
     
    public static int BitCount(int n)
    {
        var count = 0;
        while (n != 0)
        {
            count++;
            n &= (n - 1); //walking through all the bits which are set to one
        }
     
        return count;
    }
       
    // Utility function to calculate NCR
    static int NCR(int N, int R)
    {
        if (R > N - R)
            R = N - R;
             
        int ans = 1;
         
        for(int i = 1; i <= R; i++)
        {
            ans *= N - R + i;
            ans /= i;
        }
        return ans;
    }
     
    // Function to answer queries about
    // the number of triplets  in range
    // whose XOR has an even number of
    // set bits
    static void solveXorTriplets(int[] arr, int N,
                                 List<pair> Q, int M)
    {
         
        // Prefix arrays to store
        // number that have odd
        // and even setbits
        int[] preo = new int[N + 1];
        int[] pree = new int[N + 1];
     
        // Traverse the array arr[]
        for(int i = 0; i < N; i++)
        {
             
            // Update
            preo[i + 1] += preo[i];
            pree[i + 1] += pree[i];
     
            // Find bits in current
            // element
            int setbits = BitCount(arr[i]);
     
            // If number of setbits
            // is odd
            if (setbits % 2 != 0)
                preo[i + 1]++;
                 
            // Otherwise
            else
                pree[i + 1]++;
        }
         
        // Traverse the query Q[][]
        for(int i = 0; i < M; i++)
        {
             
            // Stores Left boundary
            int L = Q[i].first;
             
            // Stores Right boundary
            int R =  Q[i].second;
     
            // Stores number of elements
            // that have odd set bits in
            // the given range
            int odd = preo[R] - preo[L - 1];
     
            // Store number of elements
            // that have even set bits
            // in given range
            int even = pree[R] - pree[L - 1];
     
            // Store the count
            // of the triplets
            int ans = 0;
     
            // If even is greater
            // than ore equal to 3
            if (even >= 3)
                ans += NCR(even, 3);
     
            // If odd is greater than
            // or equal to and even is
            // greater than or equal to
            // 1
            if (odd >= 2 && even >= 1)
                ans += NCR(odd, 2) * NCR(even, 1);
     
            // Print the answer
            // for current query
            Console.WriteLine(ans);
        }
    }
     
    // Driver Code
    public static void Main()
    {
        int[] arr = { 1, 2, 3, 4, 5 };
        int N = arr.Length;
     
        List<pair> Q = new List<pair>();
          Q.Add(new pair(1, 4));
          Q.Add(new pair(3, 4));
       
        int M = Q.Count;
     
        solveXorTriplets(arr, N, Q, M);
    }
}
 
// This code is contributed by ShubhamSingh10

C++




#include <iostream>
using namespace std;
 
int main() {
 
    cout<<"GFG!";
    return 0;
}

Javascript




<script>
 
// JavaScript program for the above approach
 
function bitCount1(n) {
  return n.toString(2).match(/1/g).length
}
 
// Utility function to calculate NCR
function NCR(N, R)
{
    if (R > N - R)
        R = N - R;
    var ans = 1;
    for (var i = 1; i <= R; i++) {
        ans *= N - R + i;
        ans /= i;
    }
    return ans;
}
// Function to answer queries about
// the number of triplets  in range
// whose XOR has an even number of
// set bits
function solveXorTriplets(arr, N, Q, M)
{
    // Prefix arrays to store
    // number that have odd
    // and even setbits
    var preo = new Array(N+1).fill(0);
    var pree = new Array(N+1).fill(0);
 
    // Traverse the array arr[]
    for (var i = 0; i < N; i++) {
        // Update
        preo[i + 1] += preo[i];
        pree[i + 1] += pree[i];
 
        // Find bits in current
        // element
        var setbits = bitCount1(arr[i]);
 
        // If number of setbits
        // is odd
        if (setbits % 2)
            preo[i + 1]++;
        // Otherwise
        else
            pree[i + 1]++;
    }
    // Traverse the query Q[][]
    for (var i = 0; i < M; i++) {
        // Stores Left boundary
        var L = Q[i][0];
        // Stores Right boundary
        var R = Q[i][1];
 
        // Stores number of elements
        // that have odd set bits in
        // the given range
        var odd = preo[R] - preo[L - 1];
 
        // Store number of elements
        // that have even set bits
        // in given range
        var even = pree[R] - pree[L - 1];
 
        // Store the count
        // of the triplets
        var ans = 0;
 
        // If even is greater
        // than ore equal to 3
        if (even >= 3)
            ans += NCR(even, 3);
 
        // If odd is greater than
        // or equal to and even is
        // greater than or equal to
        // 1
        if (odd >= 2 && even >= 1)
            ans += NCR(odd, 2) * NCR(even, 1);
 
        // Print the answer
        // for current query
        document.write(ans + "<br>");
    }
}
// Driver Code
var arr = [1, 2, 3, 4, 5 ];
var N = arr.length;
 
var Q = [ [ 1, 4 ], [ 3, 4 ] ];
var M = Q.length;
 
solveXorTriplets(arr, N, Q, M);
 
// This code is contributed by Shubhamsingh10
 
</script>
Output
3
0

Time Complexity: O(N+M)
Auxiliary Space: O(N)

 

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