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 <bits/stdc++.h> 
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<pair<char, int> > 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:

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

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C++

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