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📜  根据数组元素中数字K递增的频率对数组进行排序

📅  最后修改于: 2021-04-17 18:24:40             🧑  作者: Mango

给定大小为N的数组arr []和代表数字的整数K ，任务是根据数字K在数组元素中的递增频率，以升序打印给定数组。

例子：

方法：想法是对数组中每个元素的数字K的出现进行计数，并根据该数组对数组进行排序。请按照以下步骤解决问题：

  • 初始化一个多重映射(例如mp) ，以按排序顺序在每个元素中存储K的出现。
  • 使用变量i遍历给定的数组arr []并执行以下操作：
    • 将数字K的出现存储在变量cnt中的arr [i]中。
    • 在mp中插入一对cnt和arr [i] 。
  • 在mp中打印数组元素以获取所需的排序顺序。

下面是上述方法的实现：

C++
// C++ program for the above approach
#include 
using namespace std;
 
// Function to count the occurrences
// of digit K in an element
int countOccurrences(int num, int K)
{
    // If num and K are both 0
    if (K == 0 && num == 0)
        return 1;
 
    // Initialize count as 0
    int count = 0;
 
    // Count for occurrences of digit K
    while (num > 0) {
        if (num % 10 == K)
            count++;
        num /= 10;
    }
 
    // Return the count
    return count;
}
 
// Function to print the given array
// in increasing order of the digit
// K in the array elements
void sortOccurrences(int arr[],
                     int N, int K)
{
    // Stores the occurrences of K
    // in each element
    multimap mp;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Count the frequency
        // of K in arr[i]
        int count = countOccurrences(
            arr[i], K);
 
        // Insert elements in mp
        mp.insert(pair(
            count, arr[i]));
    }
 
    // Print the elements in the map, mp
    for (auto& itr : mp) {
        cout << itr.second << " ";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 15, 66, 26, 91 };
    int K = 6;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    sortOccurrences(arr, N, K);
 
    return 0;
}

Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  // Function to count the occurrences
  // of digit K in an element
  static int countOccurrences(int num, int K)
  {
    // If num and K are both 0
    if (K == 0 && num == 0)
      return 1;
 
    // Initialize count as 0
    int count = 0;
 
    // Count for occurrences of digit K
    while (num > 0) {
      if (num % 10 == K)
        count++;
      num /= 10;
    }
 
    // Return the count
    return count;
  }
 
  // Pair class
  static class Pair {
 
    int idx;
    int freq;
 
    Pair(int idx, int freq)
    {
      this.idx = idx;
      this.freq = freq;
    }
  }
 
  // Function to print the given array
  // in increasing order of the digit
  // K in the array elements
  static void sortOccurrences(int arr[], int N, int K)
  {
 
    // Stores the Pair with index
    // and occurences of K of each element
    Pair mp[] = new Pair[N];
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
      // Count the frequency
      // of K in arr[i]
      int count = countOccurrences(arr[i], K);
 
      // Insert Pair in mp
      mp[i] = new Pair(i, count);
    }
 
    // sort the mp in increasing order of freq
    // if freq equal then according to index
    Arrays.sort(mp, (p1, p2) -> {
      if (p1.freq == p2.freq)
        return p1.idx - p2.idx;
      return p1.freq - p2.freq;
    });
 
    // Print the elements in the map, mp
    for (Pair p : mp) {
      System.out.print(arr[p.idx] + " ");
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    int arr[] = { 15, 66, 26, 91 };
    int K = 6;
    int N = arr.length;
 
    // Function Call
    sortOccurrences(arr, N, K);
  }
}
 
// This code is contributed by Kingash.

Python3
# Python program for the above approach
 
# Function to count the occurrences
# of digit K in an element
def countOccurrences( num, K):
   
    # If num and K are both 0
    if (K == 0 and num == 0):
        return 1
       
    # Initialize count as 0
    count = 0
     
    # Count for occurrences of digit K
    while (num > 0):
        if (num % 10 == K):
            count += 1
        num //= 10
         
    # Return the count
    return count
 
# Function to print the given array
# in increasing order of the digit
# K in the array elements
def sortOccurrences(arr, N, K):
   
    # Stores the occurrences of K
    # in each element
    mp = []
     
    # Traverse the array
    for i in range(N):
       
        # Count the frequency
        # of K in arr[i]
        count = countOccurrences(arr[i], K)
         
        # Insert elements in mp
        mp.append([count, arr[i]])
 
    # Print the elements in the map, mp
    mp = sorted(mp)
     
    for itr in mp:
        print(itr[1], end = ' ')
 
# Driver Code
arr = [ 15, 66, 26, 91 ]
K = 6
N = len(arr)
 
# Function Call
sortOccurrences(arr, N, K);
 
# This code is contributed by rohitsingh07052.

输出： 
15 91 26 66

时间复杂度： O(N * log 10 N)
辅助空间： O(N)