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 them, 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: 

1. Initialize a Map and store the frequencies of each character.
2. Traverse the Map and insert all {Character, Frequency} pairs in a vector of pair.
3. 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.
4. Traverse this vector and find the digit corresponding to each character.
5. 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;
}

Java

// java Program to implement
// the above approach
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
public class FormString {
    static class Pair implements Comparable<Pair> {
        char first;
        int second;

        Pair(char first, int second) {
            this.first = first;
            this.second = second;
        }
        // custom comparator for sorting
        @Override
        public int compareTo(Pair o) {
             // is frequency same
            if (this.second == o.second) {
                // Character with lower ASCII
                // value appears first
                return this.first - o.first;
            }
            // Otherwise character with higher
            // frequency appears first
            return o.second - this.second;
        }
    }
// Function to sort map accordingly
    static String sort(Map<Character, Integer> map) {
        // Declaring vector of pairs
        List<Pair> list = new ArrayList<>();
        // Traversing map and pushing
        // pairs to List Integerface
        for (Map.Entry<Character, Integer> entry : map.entrySet()) {
            list.add(new Pair(entry.getKey(), entry.getValue()));
        }
        // Sorting using Collections sort
        Collections.sort(list);
        StringBuilder sb = new StringBuilder();
        for (Pair pair : list) {
             // Get the possible digit
            // from assigned value and
            // Ensuring k does not exceed 9
            int k = (pair.first - 'a' + 1) % 10;
            // append each digit
            sb.append(k);
        }
        // Returning final result
        return sb.toString();
    }

    static String formString(String s) {
        // map are used to store freq
        Map<Character, Integer> map = new HashMap<>();
        for (int i = 0; i < s.length(); i++) {
            map.put(s.charAt(i), map.getOrDefault(s.charAt(i), 0) + 1);
        }
        // sort map and return
        return sort(map);
    }
   // Main function (Driver code)
    public static void main(String[] args) {
        String Str = "akkzzz";
        System.out.println(formString(Str));
    }
}

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

Javascript

<script>

// JavaScript Program to implement
// the above approach

// Custom comparator for sorting
function comp(p1,p2)
{

    // If frequency is same
    if (p1[1] == p2[1])

        // Character with lower ASCII
        // value appears first
        return p2.charCodeAt(0) - p1.charCodeAt(0);

    // Otherwise character with higher
    // frequency appears first
    return p2[1] - p1[1];
}

// Function to sort map accordingly
function sort(m)
{

    // Declaring vector of pairs
    let a = [];

    // Output string to store the result
    let out="";

    // Traversing map and pushing
    // pairs to vector
    for (let [x,y] of m) {

        a.push([ x, y ]);
    }

    // Using custom comparator
    a.sort(comp);

    // Traversing the Vector
    for (let x of a) {

        // Get the possible digit
        // from assigned value
        let k = (x[0].charCodeAt(0) - 97 + 1)%10;


        // Append each digit
        out += k.toString();
    }

    // Returning final result
    return out;
}

// Function to generate and return
// the required number
function formString(s)
{

    // Stores the frequencies
    let mp = new Map();

    for (let i = 0; i < s.length; i++){
        if(mp.has(s[i])){
            mp.set(s[i],mp.get(s[i])+1);
        }
        else{
            mp.set(s[i],1);
        }
    }

    // Sort map in required order
    let res = sort(mp);

    // Return the final result
    return res;
}

// Driver Code
let N = 4;
let Str = "akkzzz";
document.write(formString(Str),"</br>");

// This code is contributed by shinjanpatra

</script>

C#

// C# program to implement the above approach
using System;
using System.Collections.Generic;
using System.Linq;

public class Program
{
    // Function to sort dictionary
    // accordingly
    static string SortDictionary(Dictionary<char, int> dict)
    {
        // Declaring dictionary of pairs
        Dictionary<char, int> sortedDict = new Dictionary<char, int>();

        // Traversing dictionary and
        // pushing pairs to dictionary of pairs
        foreach (var pair in dict)
        {
            sortedDict.Add(pair.Key, pair.Value);
        }

        // Sorting dictionary by value in descending order
        sortedDict = sortedDict.OrderByDescending(pair => pair.Value)
                             .ThenBy(pair => pair.Key)
                             .ToDictionary(pair => pair.Key, pair => pair.Value);

        // Building output string
        string output = "";
        foreach (var pair in sortedDict)
        {
            // Get the possible digit
            // from assigned value and
            // ensuring k does not exceed 9
            int k = (pair.Key - 'a' + 1) % 10;

            // Append each digit
            output += k.ToString();
        }

        // Returning final result
        return output;
    }

    // Function to generate and return
    // the required number
    static string GetFormedString(string str)
    {
        // Stores the frequencies
        Dictionary<char, int> freq = new Dictionary<char, int>();
        foreach (char c in str)
        {
            if (freq.ContainsKey(c))
            {
                freq++;
            }
            else
            {
                freq.Add(c, 1);
            }
        }

        // Sort dictionary in
        // required order
        string formedStr = SortDictionary(freq);

        // Return the
        // final result
        return formedStr;
    }

    // Main function (Driver code)
    public static void Main(string[] args)
    {
        string str = "akkzzz";
        Console.WriteLine(GetFormedString(str));
    }
}

// Contributed by adiyasha4x71

Output

611

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

Approach 2:

1. Initialize an array count of size 26 to store the frequency of each character in the string.
2. Traverse the string and update the frequency count of each character in the array.
3. Traverse the count array and create a vector of pairs where each pair consists of a character and its frequency.
4. Sort the vector of pairs in decreasing order of frequency. For pairs with the same frequency, sort them in increasing order of ASCII value.
5. Traverse the sorted vector and generate the digit corresponding to each character. Append each digit to the result string.
6. Return the result string.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;

string formString(string s) {
    // Step 1: Initialize count array
    int count[26] = {0};
    for (int i = 0; i < s.length(); i++) {
        count[s[i] - 'a']++;
    }

    // Step 2: Create vector of pairs
    vector<pair<char, int>> v;
    for (int i = 0; i < 26; i++) {
        if (count[i] > 0) {
            v.push_back(make_pair('a' + i, count[i]));
        }
    }

    // Step 3: Sort vector of pairs
    sort(v.begin(), v.end(), [](pair<char, int>& p1, pair<char, int>& p2) {
        if (p1.second == p2.second) {
            return p1.first < p2.first;
        }
        return p1.second > p2.second;
    });

    // Step 4: Generate result string
    string res;
    for (auto& p : v) {
        int digit = (p.first - 'a' + 1) % 10;
        res += to_string(digit);
    }

    // Step 5: Return result string
    return res;
}

int main() {
    int N = 6;
    string Str = "akkzzz";
    cout << formString(Str); // Output: 611
    return 0;
}

Java

import java.util.*;

public class Main {
    // Pair class implementation
    static class Pair<K, V> {
        public K key;
        public V value;

        public Pair(K key, V value) {
            this.key = key;
            this.value = value;
        }

        public K getKey() {
            return key;
        }

        public V getValue() {
            return value;
        }
    }

    public static String formString(String s) {
        // Step 1: Initialize count array
        int[] count = new int[26];
        for (int i = 0; i < s.length(); i++) {
            count[s.charAt(i) - 'a']++;
        }

        // Step 2: Create list of pairs
        List<Pair<Character, Integer>> v = new ArrayList<>();
        for (int i = 0; i < 26; i++) {
            if (count[i] > 0) {
                v.add(new Pair<>((char)('a' + i), count[i]));
            }
        }

        // Step 3: Sort list of pairs
        Collections.sort(v, new Comparator<Pair<Character, Integer>>() {
            @Override
            public int compare(Pair<Character, Integer> p1, Pair<Character, Integer> p2) {
                if (p1.getValue().equals(p2.getValue())) {
                    return Character.compare(p1.getKey(), p2.getKey());
                }
                return Integer.compare(p2.getValue(), p1.getValue());
            }
        });

        // Step 4: Generate result string
        StringBuilder res = new StringBuilder();
        for (Pair<Character, Integer> p : v) {
            int digit = (p.getKey() - 'a' + 1) % 10;
            res.append(digit);
        }

        // Step 5: Return result string
        return res.toString();
    }

    public static void main(String[] args) {
        int N = 6;
        String Str = "akkzzz";
        System.out.println(formString(Str)); // Output: 611
    }
}

Python3

def form_string(s):
    # Step 1: Initialize count array
    count = [0]*26
    for i in range(len(s)):
        count[ord(s[i]) - ord('a')] += 1

    # Step 2: Create list of tuples
    v = []
    for i in range(26):
        if count[i] > 0:
            v.append((chr(i + ord('a')), count[i]))

    # Step 3: Sort list of tuples
    v.sort(key=lambda x: (-x[1], x[0]))

    # Step 4: Generate result string
    res = ""
    for p in v:
        digit = (ord(p[0]) - ord('a') + 1) % 10
        res += str(digit)

    # Step 5: Return result string
    return res
s = "akkzzz"

print(form_string(s))  # Output: 611

Javascript

// Javascript code addition

function formString(s) {
  // Step 1: Initialize count array
  const count = new Array(26).fill(0);
  for (let i = 0; i < s.length; i++) {
    count[s.charCodeAt(i) - 97]++;
  }

  // Step 2: Create list of pairs
  const v = [];
  for (let i = 0; i < 26; i++) {
    if (count[i] > 0) {
      v.push([String.fromCharCode(i + 97), count[i]]);
    }
  }

  // Step 3: Sort list of pairs
  v.sort((p1, p2) => {
    if (p1[1] === p2[1]) {
      return p1[0].localeCompare(p2[0]);
    }
    return p2[1] - p1[1];
  });

  // Step 4: Generate result string
  let res = '';
  for (let i = 0; i < v.length; i++) {
    const digit = (v[i][0].charCodeAt(0) - 97 + 1) % 10;
    res += digit.toString();
  }

  // Step 5: Return result string
  return res;
}

const N = 6;
const Str = 'akkzzz';
console.log(formString(Str)); // Output: 611

// The code  is contributed by Arushi Goel.

Output

611

Time Complexity: O(nlogn)
Auxiliary Space: O(1)