给定一个整数N ,任务是在1到N的范围内找到最大组的大小,如果两个数字的xor相同,则两个数字属于同一组。
例子:
Input: N = 13
Output: 2
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: 1
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.
方法:
为了解决上述问题,我们必须使用哈希图存储每个元素从1到N的数字的异或,并在重复时增加其频率。然后在哈希图中找到最大频率,该最大频率将是组的最大大小。最后,计数与最大组具有相同频率计数的所有组,然后返回计数。
下面是上述方法的实现:
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 digit
int 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 group
int find_count(int n)
{
// hash map for counting frquency
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 code
int 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 digit
static 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 group
static int find_count(int n)
{
// hash map for counting frquency
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 code
public 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 digit
def 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 group
def find_count(n):
# Hash map for counting frquency
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 N
N = 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 digit
static 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 group
static int find_count(int n)
{
// hash map for counting frquency
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 code
public static void Main(String[] args)
{
// initialise N
int N = 13;
Console.Write(find_count(N));
}
}
// This code is contributed by shikhasingrajput
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