以很大的数字打印所有K位重复数字

给定一个非常大的数字N (字符串和数字K) ，任务是打印频率大于1的所有K位重复数字。
例子：

Input: str = “123412345123456”, K = 4 
Output: 
1234 – 3 
2345 – 2 
Explanation: 
The 4-digit numbers having frequency greater than 1 is 1234 and 2345.

Input: N = 1432543214325432, K = 5 
Output: 
14325 – 2 
32543 – 2 
43254 – 2 
Explanation: 
The 5-digit numbers having frequency greater than 1 is 14325, 32543, and 43254.

方法：由于数字是以字符串形式给出的，因此，其想法是将大小为K的所有子字符串及其频率存储在映射中。现在，在迭代Map时，仅打印那些频率大于1的子字符串及其出现的次数。

下面是上述方法的实现：

C++

// C++ program for the above approach 
#include  
using namespace std; 
  
// Function to print all K digit 
// repeating numbers 
void print_Kdigit(string S, int K) 
{ 
  
    // Map to store the substrings 
    // with their frequencies 
    map m; 
  
    // Iterate over every substring 
    // and store their frequencies 
    // in the map 
    for (int i = 0; i < S.length() - K; 
        i++) { 
        string a = S.substr(i, K); 
  
        // Increment the count of 
        // substrings in map 
        m[a]++; 
    } 
  
    // Iterate over all the substrings 
    // present in the map 
    for (auto x : m) { 
  
        // Condition to check if the 
        // frequency of the substring 
        // present in the map 
        // is greater than 1 
        if (x.second > 1) { 
            cout << x.first << " - "
                << x.second << "\n"; 
        } 
    } 
} 
  
// Driver Code 
int main() 
{ 
    // Given Number in form of string 
    string str = "123412345123456"; 
  
    // Given K 
    int K = 4; 
  
    // Function Call 
    print_Kdigit(str, K); 
}

Java

// Java program for the above approach 
import java.util.*; 
  
class GFG{ 
  
// Function to print all K digit 
// repeating numbers 
static void print_Kdigit(String S, int K) 
{ 
      
    // Map to store the substrings 
    // with their frequencies 
    Map m = new HashMap<>(); 
  
    // Iterate over every substring 
    // and store their frequencies 
    // in the map 
    for(int i = 0; i < S.length() - K; i++) 
    { 
        String a = S.substring(i, i + K); 
  
        // Increment the count of 
        // substrings in map 
        m.put(a, m.getOrDefault(a, 0) + 1); 
          
    } 
  
    // Iterate over all the substrings 
    // present in the map 
    for(Map.Entry x : m.entrySet()) 
    { 
          
        // Condition to check if the 
        // frequency of the substring 
        // present in the map 
        // is greater than 1 
        if (x.getValue() > 1) 
        { 
            System.out.println(x.getKey() + " - " + 
                            x.getValue()); 
        } 
    } 
} 
  
// Driver code 
public static void main(String[] args) 
{ 
      
    // Given Number in form of string 
    String str = "123412345123456"; 
      
    // Given K 
    int K = 4; 
      
    // Function call 
    print_Kdigit(str, K); 
} 
} 
  
// This code is contributed by offbeat

Python3

# Python3 program of the above approach 
def print_Kdigit(S, K): 
      
    # Dictionary to store the substrings 
    # with their frequencies 
    m = {} 
  
    # Iterate over every substring 
    # and store their frequencies 
    # in the dictionary 
    for i in range(len(S) - K): 
        a = S[i:i + K] 
          
        # Initialize the count of 
        # substrings in dictionary with 0 
        m[a] = 0
  
    for i in range(len(S) - K): 
        a = S[i:i + K] 
          
        # Increment the count of 
        # substrings in dictionary 
        m[a] += 1
  
    # Iterate over all the substrings 
    # present in the dictionary 
    for key, value in m.items(): 
        if value > 1: 
            print(key, "-", value) 
  
# Driver Code 
str = "123412345123456"
  
# Given K 
K = 4
  
# Function Call 
print_Kdigit(str, K) 
  
# This code is contributed by Vishal Maurya

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
  
class GFG{
  
// Function to print all K digit
// repeating numbers
static void print_Kdigit(string S, int K)
{
      
    // Map to store the substrings
    // with their frequencies
    Dictionary m = new Dictionary();
  
    // Iterate over every substring
    // and store their frequencies
    // in the map
    for(int i = 0; i < S.Length - K; i++) 
    {
        string a = S.Substring(i, K);
  
        // Increment the count of
        // substrings in map
        m[a] = m.GetValueOrDefault(a, 0) + 1;
    }
  
    // Iterate over all the substrings
    // present in the map
    foreach(KeyValuePair x in m)
    {
          
        // Condition to check if the
        // frequency of the substring
        // present in the map
        // is greater than 1
        if (x.Value > 1)
        {
            Console.Write(x.Key + " - " + 
                          x.Value + "\n"); 
        }
    }
}
  
// Driver code
public static void Main(string[] args)
{
      
    // Given number in form of string
    string str = "123412345123456";
      
    // Given K
    int K = 4;
      
    // Function call
    print_Kdigit(str, K);
}
}
  
// This code is contributed by rutvik_56

输出：

1234 - 3
2345 - 2

时间复杂度： O(N * K)