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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 <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;
 
        // 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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