Given an array **arr[]** of size **N**, and an integer** K** representing a digit, the task is to print the given array in increasing order according to the increasing frequency of the digit **K** in the array elements.

**Examples:**

Input:arr[] = {15, 66, 26, 91}, K = 6Output:15 91 26 66Explanation:

The frequency of digit 6 in array elements is {0, 2, 1, 0}. The elements in increasing order of frequency of the digit 6 in {15, 91, 26, 66}.

Input:arr[] = {20, 21, 0}, K = 0Output:21 20 0Explanation:

The frequency of digit 0 in array elements is {1, 0, 1}. The elements in increasing order of frequency of the digit 0 is {21, 20, 0}.

**Approach:** The idea is to count the occurrences of the digit **K** for each element of the array and sort the array according to it. Follow the steps below to solve the problem:

- Initialize a multimap, say
**mp**, to store occurrences of**K**in each element in sorted order. - Traverse the given array
**arr[]**using the variable**i**and perform the following:- Store the occurrences of digit
**K**in**arr[i]**in a variable**cnt**. - Insert a pair of
**cnt**and**arr[i]**in**mp**.

- Store the occurrences of digit
- Print the array elements in
**mp**to get the required sorted order.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `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<` `int` `, ` `int` `> 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<` `int` `, ` `int` `>(` ` ` `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.` |

**Output:**

15 91 26 66

**Time Complexity:** O(N*log_{10}N)**Auxiliary Space:** O(N)

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