给定整数N ，任务是查找具有最大大小的组数。从1到N的每个数字都根据其数字的总和进行分组。
例子：
Input: N = 13
Output: 4
Explanation:
There are 9 groups in total, they are grouped according to the sum of its digits of numbers from 1 to 13: [1, 10] [2, 11] [3, 12] [4, 13] [5] [6] [7] [8] [9].
Out of these, 4 groups have the largest size that is 2.
Input: n = 2
Output: 2
Explanation:
There are 2 groups in total. [1] [2] and both the groups have largest size 1.
方法：为解决上述问题，我们需要创建一个词典，其键代表从1到N的数字的唯一数字和。这些键的值将保持计数有多少个数字的总和等于其键。然后，我们将打印所有这些值中的最高值。
下面是上述方法的实现：
C++
// C++ implementation to Count the
// number of groups having the largest
// size where groups are according
// to the sum of its digits
#include
using namespace std;
// function to return sum of digits of i
int sumDigits(int n){
int sum = 0;
while(n)
{
sum += n%10;
n /= 10;
}
return sum;
}
// Create the dictionary of unique sum
map constDict(int n){
// dictionary that contain
// unique sum count
map d;
for(int i = 1; i < n + 1; ++i){
// calculate the sum of its digits
int sum1 = sumDigits(i);
if(d.find(sum1) == d.end())
d[sum1] = 1;
else
d[sum1] += 1;
}
return d;
}
// function to find the
// largest size of group
int countLargest(int n){
map d = constDict(n);
int size = 0;
// count of largest size group
int count = 0;
for(auto it = d.begin(); it != d.end(); ++it){
int k = it->first;
int val = it->second;
if(val > size){
size = val;
count = 1;
}
else if(val == size)
count += 1;
}
return count;
}
// Driver code
int main()
{
int n = 13;
int group = countLargest(n);
cout << group << endl;
return 0;
} Java
// Java implementation to Count the
// number of groups having the largest
// size where groups are according
// to the sum of its digits
import java.util.HashMap;
import java.util.Map;
class GFG{
// Function to return sum of digits of i
public static int sumDigits(int n)
{
int sum = 0;
while(n != 0)
{
sum += n % 10;
n /= 10;
}
return sum;
}
// Create the dictionary of unique sum
public static HashMap constDict(int n)
{
// dictionary that contain
// unique sum count
HashMap d = new HashMap<>();
for(int i = 1; i < n + 1; ++i)
{
// Calculate the sum of its digits
int sum1 = sumDigits(i);
if (!d.containsKey(sum1))
d.put(sum1, 1);
else
d.put(sum1, d.get(sum1) + 1);
}
return d;
}
// Function to find the
// largest size of group
public static int countLargest(int n)
{
HashMap d = constDict(n);
int size = 0;
// Count of largest size group
int count = 0;
for(Map.Entry it : d.entrySet())
{
int k = it.getKey();
int val = it.getValue();
if (val > size)
{
size = val;
count = 1;
}
else if (val == size)
count += 1;
}
return count;
}
// Driver code
public static void main(String[] args)
{
int n = 13;
int group = countLargest(n);
System.out.println(group);
}
}
// This code is contributed by divyeshrabadiya07 Python3
# Python3 implementation to Count the
# number of groups having the largest
# size where groups are according
# to the sum of its digits
# Create the dictionary of unique sum
def constDict(n):
# dictionary that contain
# unique sum count
d ={}
for i in range(1, n + 1):
# convert each number to string
s = str(i)
# make list of number digits
l = list(s)
# calculate the sum of its digits
sum1 = sum(map(int, l))
if sum1 not in d:
d[sum1] = 1
else:
d[sum1] += 1
return d
# function to find the
# largest size of group
def countLargest(n):
d = constDict(n)
size = 0
# count of largest size group
count = 0
for k, val in d.items():
if val > size:
size = val
count = 1
elif val == size:
count += 1
return count
# Driver Code
n = 13
group = countLargest(n)
print(group)
# This code is contributed by Sanjit_PrasadC#
// C# implementation to Count the
// number of groups having the largest
// size where groups are according
// to the sum of its digits
using System;
using System.Collections.Generic;
class GFG {
// Function to return sum of digits of i
static int sumDigits(int n)
{
int sum = 0;
while(n != 0)
{
sum += n % 10;
n /= 10;
}
return sum;
}
// Create the dictionary of unique sum
static Dictionary constDict(int n)
{
// dictionary that contain
// unique sum count
Dictionary d = new Dictionary();
for(int i = 1; i < n + 1; ++i)
{
// Calculate the sum of its digits
int sum1 = sumDigits(i);
if (!d.ContainsKey(sum1))
d.Add(sum1, 1);
else
d[sum1] += 1;
}
return d;
}
// Function to find the
// largest size of group
static int countLargest(int n)
{
Dictionary d = constDict(n);
int size = 0;
// Count of largest size group
int count = 0;
foreach(KeyValuePair it in d)
{
int k = it.Key;
int val = it.Value;
if (val > size)
{
size = val;
count = 1;
}
else if (val == size)
count += 1;
}
return count;
}
// Driver code
static void Main()
{
int n = 13;
int group = countLargest(n);
Console.WriteLine(group);
}
}
// This code is contributed by divyesh072019 输出：
4时间复杂度： O(N)