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Find the size of largest group where groups are according to the xor of digits

• Last Updated : 06 Jul, 2021

Given an integer N and the task is to find the size of the largest group in a range 1 to N, where two numbers belong to the same group if xor of its digits is the same.

Examples:

Input: N = 13
Output:
Explanation:
There are 10 groups in total, they are grouped according to the xor of its digits of numbers from 1 to 13: [11] [1, 10] [2, 13] [3, 12] [4] [5] [6] [7] [8] [9].
Out of these, 3 groups have the largest size that is 2.

Input: N = 2
Output:
Explanation:
There are 2 groups in total, they are grouped according to the xor of its digits of numbers from 1 to 2: [1] [2].
Out of these, both groups have the largest size that is 1.

Approach:
To solve the problem mentioned above we have to store the xor of digit of every element from 1 to N using a hash map and increment its frequency if it repeats. Then find the maximum frequency within the hash map, which would be the largest size of the group. And finally, count all the groups who have the same frequency count as the largest group and return the count.

Below is the implementation of above the approach:

C++14

 // c++ implementation to Find the// size of largest group, where groups// are according to the xor of its digits.#include using namespace std; // Function to find out xor of digitint digit_xor(int x){    int xorr = 0;     // calculate xor digitwise    while (x) {        xorr ^= x % 10;        x = x / 10;    }     // return xor    return xorr;} // Function to find the// size of largest groupint find_count(int n){    // hash map for counting frequency    map mpp;     for (int i = 1; i <= n; i++) {         // counting freq of each element        mpp[digit_xor(i)] += 1;    }     int maxm = 0;    for (auto x : mpp) {        // find the maximum        if (x.second > maxm)             maxm = x.second;    }     return maxm;} // Driver codeint main(){    // initialise N    int N = 13;     cout << find_count(N);     return 0;}

Java

 // Java implementation to Find the// size of largest group, where groups// are according to the xor of its digits.import java.util.*;class GFG{ // Function to find out xor of digitstatic int digit_xor(int x){    int xorr = 0;     // calculate xor digitwise    while (x > 0)    {        xorr ^= x % 10;        x = x / 10;    }     // return xor    return xorr;} // Function to find the// size of largest groupstatic int find_count(int n){    // hash map for counting frequency    HashMap mpp = new HashMap();     for (int i = 1; i <= n; i++)    {        // counting freq of each element        if(mpp.containsKey(digit_xor(i)))            mpp.put(digit_xor(i),                    mpp.get(digit_xor(i)) + 1);        else            mpp.put(digit_xor(i), 1);    }     int maxm = 0;    for (Map.Entry x : mpp.entrySet())    {        // find the maximum        if (x.getValue() > maxm)             maxm = x.getValue();    }    return maxm;} // Driver codepublic static void main(String[] args){    // initialise N    int N = 13;     System.out.print(find_count(N));}} // This code is contributed by shikhasingrajput

Python3

 # Python3 implementation to find the# size of largest group, where groups# are according to the xor of its digits. # Function to find out xor of digitdef digit_xor(x):     xorr = 0     # Calculate xor digitwise    while (x != 0):        xorr ^= x % 10        x = x // 10     # Return xor    return xorr # Function to find the# size of largest groupdef find_count(n):     # Hash map for counting frequency    mpp = {}     for i in range(1, n + 1):         # Counting freq of each element        if digit_xor(i) in mpp:            mpp[digit_xor(i)] += 1        else:            mpp[digit_xor(i)] = 1     maxm = 0         for x in mpp:                 # Find the maximum        if (mpp[x] > maxm):            maxm = mpp[x]     return maxm # Driver code # Initialise NN = 13 print(find_count(N)) # This code is contributed by divyeshrabadiya07

C#

 // C# implementation to Find the// size of largest group, where groups// are according to the xor of its digits.using System;using System.Collections.Generic;class GFG{ // Function to find out xor of digitstatic int digit_xor(int x){    int xorr = 0;     // calculate xor digitwise    while (x > 0)    {        xorr ^= x % 10;        x = x / 10;    }     // return xor    return xorr;} // Function to find the// size of largest groupstatic int find_count(int n){    // hash map for counting frequency    Dictionary mpp = new Dictionary();     for (int i = 1; i <= n; i++)    {        // counting freq of each element        if(mpp.ContainsKey(digit_xor(i)))            mpp[digit_xor(i)] =                mpp[digit_xor(i)] + 1;        else            mpp.Add(digit_xor(i), 1);    }     int maxm = 0;    foreach (KeyValuePair x in mpp)    {        // find the maximum        if (x.Value > maxm)             maxm = x.Value;    }    return maxm;} // Driver codepublic static void Main(String[] args){    // initialise N    int N = 13;     Console.Write(find_count(N));}}  // This code is contributed by shikhasingrajput

Javascript


Output:
2

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