# Generate a Number in Decreasing order of Frequencies of characters of a given String

• Difficulty Level : Medium
• Last Updated : 01 Oct, 2021

Given a string Str of length N, consisting of lowercase alphabets, the task is to generate a number in decreasing order of the frequency of characters in the given string. If two characters have the same frequency, the character with a smaller ASCII value appears first. Numbers assigned to characters {a, b, …., y, z} are {1, 2, …., 25, 26} respectively.
Note: For characters having values greater than 9 assigned to it, take its modulo 10.
Examples:

Input: N = 6, Str = “aaabbd”
Output: 124
Explanation:
Given characters and their respective frequencies are:

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• a = 3
• b = 2
• d = 1

Since the number needs to be generated in increasing order of their frequencies, the final generated number is 124.
Input: N = 6, Str = “akkzzz”
Output: 611
Explanation:
Given characters and their respective frequencies are:

• a = 1
• k = 2
• z = 3

For z, value to assigned = 26
Hence, the corresponding digit assigned = 26 % 10 = 6
For k, value to assigned = 11
Hence, the corresponding digit assigned = 11 % 10 = 1
Since the number needs to be generated in increasing order of their frequencies, the final generated number is 611

Approach:
Follow the steps below to solve the problem:

• Initialize a Map and store the frequencies of each character.
• Traverse the Map and insert all {Character, Frequency} pairs in a vector of pair.
• Sort this vector in a way such that the pair with higher frequency appears first and among pairs having the same frequency, those with smaller ASCII value come first.
• Traverse this vector and find the digit corresponding to each character.
• Print the final number generated.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement``// the above approach``#include ``using` `namespace` `std;` `// Custom comparator for sorting``bool` `comp(pair<``char``, ``int``>& p1,``          ``pair<``char``, ``int``>& p2)``{` `    ``// If frequency is same``    ``if` `(p1.second == p2.second)` `        ``// Character with lower ASCII``        ``// value appears first``        ``return` `p1.first < p2.first;` `    ``// Otherwise character with higher``    ``// frequency appears first``    ``return` `p1.second > p2.second;``}` `// Function to sort map accordingly``string sort(map<``char``, ``int``>& m)``{` `    ``// Declaring vector of pairs``    ``vector > a;` `    ``// Output string to store the result``    ``string out;` `    ``// Traversing map and pushing``    ``// pairs to vector``    ``for` `(``auto` `x : m) {` `        ``a.push_back({ x.first, x.second });``    ``}` `    ``// Using custom comparator``    ``sort(a.begin(), a.end(), comp);` `    ``// Traversing the Vector``    ``for` `(``auto` `x : a) {` `        ``// Get the possible digit``        ``// from assigned value``        ``int` `k = x.first - ``'a'` `+ 1;` `        ``// Ensures k does not exceed 9``        ``k = k % 10;` `        ``// Append each digit``        ``out = out + to_string(k);``    ``}` `    ``// Returning final result``    ``return` `out;``}` `// Function to generate and return``// the required number``string formString(string s)``{``    ``// Stores the frequencies``    ``map<``char``, ``int``> mp;` `    ``for` `(``int` `i = 0; i < s.length(); i++)``        ``mp[s[i]]++;` `    ``// Sort map in required order``    ``string res = sort(mp);` `    ``// Return the final result``    ``return` `res;``}` `// Driver Code``int` `main()``{``    ``int` `N = 4;` `    ``string Str = ``"akkzzz"``;` `    ``cout << formString(Str);` `    ``return` `0;``}`

## Python3

 `# Python3 Program to implement``# the above approach` `# Function to sort map``# accordingly``def` `sort(m):` `    ``# Declaring vector``    ``# of pairs``    ``a ``=` `{}` `    ``# Output string to``    ``# store the result``    ``out ``=` `""` `    ``# Traversing map and``    ``# pushing pairs to vector``    ``for` `x ``in` `m:``        ``a[x] ``=` `[]` `        ``a[x].append(m[x])` `    ``# Character with lower ASCII``    ``# value appears first``    ``a ``=` `dict``(``sorted``(a.items(),``                  ``key ``=` `lambda` `x : x[``0``]))` `    ``# Character with higher``    ``# frequency appears first``    ``a ``=` `dict``(``sorted``(a.items(),``                    ``reverse ``=` `True``,``                    ``key ``=` `lambda` `x : x[``1``]))` `    ``# Traversing the Vector``    ``for` `x ``in` `a:` `        ``# Get the possible digit``        ``# from assigned value``        ``k ``=` `ord``(x[``0``]) ``-` `ord``(``'a'``) ``+` `1` `        ``# Ensures k does``        ``# not exceed 9``        ``k ``=` `k ``%` `10` `        ``# Append each digit``        ``out ``=` `out ``+` `str``(k)` `    ``# Returning final result``    ``return` `out` `# Function to generate and return``# the required number``def` `formString(s):` `    ``# Stores the frequencies``    ``mp ``=` `{}``    ``for` `i ``in` `range``(``len``(s)):``        ``if` `s[i] ``in` `mp:``            ``mp[s[i]] ``+``=` `1``        ``else``:``            ``mp[s[i]] ``=` `1` `    ``# Sort map in``    ``# required order``    ``res ``=` `sort(mp)` `    ``# Return the``    ``# final result``    ``return` `res` `# Driver Code``N ``=` `4``Str` `=` `"akkzzz"``print``(formString(``Str``))` `# This code is contributed by avanitrachhadiya2155`
Output:
`611`

Time Complexity: O(NlogN)
Auxiliary Space: O(N)

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