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

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:

• 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 an vector of pair.
• Sort this vector in a way such that the pair with higher frequency appears first and among pairs having same frequency, those with smaller ASCII value comes 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; ` ` `  `        ``// Apppend 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; ` `} `

Output:

```611
```

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

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