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# Sort a string according to the frequency of characters

• Difficulty Level : Medium
• Last Updated : 21 Mar, 2023

Given a string str, the task is to sort the string according to the frequency of each character, in ascending order. If two elements have the same frequency, then they are sorted in lexicographical order.
Examples:

Input: str = “geeksforgeeks”
Output: forggkksseeee
Explanation:
Frequency of characters: g2 e4 k2 s2 f1 o1 r1
Sorted characters according to frequency: f1 o1 r1 g2 k2 s2 e4
f, o, r occurs one time so they are ordered lexicographically and so are g, k and s.
Hence the final output is forggkksseeee.
Input: str = “abc”
Output: abc

Approach The idea is to store each character with its frequency in a vector of pairs and then sort the vector pairs according to the frequency stored. Finally, print the vector in order.
Below is the implementation of the above approach:

## C++

 `// C++ implementation to Sort strings``// according to the frequency of``// characters in ascending order` `#include ``using` `namespace` `std;` `// Returns count of character in the string``int` `countFrequency(string str, ``char` `ch)``{``    ``int` `count = 0;` `    ``for` `(``int` `i = 0; i < str.length(); i++)` `        ``// Check for vowel``        ``if` `(str[i] == ch)``            ``++count;` `    ``return` `count;``}` `// Function to sort the string``// according to the frequency``void` `sortArr(string str)``{``    ``int` `n = str.length();` `    ``// Vector to store the frequency of``    ``// characters with respective character``    ``vector > vp;` `    ``// Inserting frequency``    ``// with respective character``    ``// in the vector pair``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``vp.push_back(``            ``make_pair(``                ``countFrequency(str, str[i]),``                ``str[i]));``    ``}` `    ``// Sort the vector, this will sort the pair``    ``// according to the number of characters``    ``sort(vp.begin(), vp.end());` `    ``// Print the sorted vector content``    ``for` `(``int` `i = 0; i < vp.size(); i++)``        ``cout << vp[i].second;``}` `// Driver code``int` `main()``{``    ``string str = ``"geeksforgeeks"``;` `    ``sortArr(str);` `    ``return` `0;``}`

## Java

 `import` `java.util.*;` `class` `GFG {``    ``// Returns count of character in the string``    ``static` `int` `countFrequency(String str, ``char` `ch)``    ``{``        ``int` `count = ``0``;` `        ``for` `(``int` `i = ``0``; i < str.length(); i++) {``            ``// Check for character``            ``if` `(str.charAt(i) == ch) {``                ``++count;``            ``}``        ``}` `        ``return` `count;``    ``}` `    ``// Function to sort the string according to the``    ``// frequency of characters in ascending order``    ``static` `void` `sortArr(String str)``    ``{``        ``int` `n = str.length();` `        ``// Dictionary to store the frequency of characters``        ``Map freqDict``            ``= ``new` `HashMap();` `        ``// Count the frequency of each character in the``        ``// input string``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``if` `(freqDict.containsKey(str.charAt(i))) {``                ``freqDict.put(str.charAt(i),``                             ``freqDict.get(str.charAt(i))``                                 ``+ ``1``);``            ``}``            ``else` `{``                ``freqDict.put(str.charAt(i), ``1``);``            ``}``        ``}` `        ``// Sort the dictionary by value (frequency) in``        ``// ascending order``        ``List > sortedDict``            ``= ``new` `ArrayList >(``                ``freqDict.entrySet());``        ``Collections.sort(``            ``sortedDict,``            ``new` `Comparator<``                ``Map.Entry >() {``                ``public` `int` `compare(``                    ``Map.Entry o1,``                    ``Map.Entry o2)``                ``{``                    ``return` `(o1.getValue()``                            ``== (o2.getValue()))``                        ``? o1.getKey() - o2.getKey()``                        ``: o1.getValue() - o2.getValue();``                ``}``            ``});` `        ``// Print the sorted characters in the order of their``        ``// frequency``        ``for` `(Map.Entry entry :``             ``sortedDict) {``            ``for` `(``int` `i = ``0``; i < entry.getValue(); i++) {``                ``System.out.print(entry.getKey());``            ``}``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``"geeksforgeeks"``;``    ` `          ``// Driver code``        ``sortArr(str);``    ``}``}`

## Python3

 `# Python3 implementation to Sort strings``# according to the frequency of``# characters in ascending order` `# Returns count of character in the string``def` `countFrequency(string ,  ch) :` `    ``count ``=` `0``;` `    ``for` `i ``in` `range``(``len``(string)) :` `        ``# Check for vowel``        ``if` `(string[i] ``=``=` `ch) :``            ``count ``+``=` `1``;` `    ``return` `count;` `# Function to sort the string``# according to the frequency``def` `sortArr(string) :``    ``n ``=` `len``(string);` `    ``# Vector to store the frequency of``    ``# characters with respective character``    ``vp ``=` `[];` `    ``# Inserting frequency``    ``# with respective character``    ``# in the vector pair``    ``for` `i ``in` `range``(n) :` `        ``vp.append((countFrequency(string, string[i]), string[i]));``        ` `    ``# Sort the vector, this will sort the pair``    ``# according to the number of characters``    ``vp.sort();``    ` `    ``# Print the sorted vector content``    ``for` `i ``in` `range``(``len``(vp)) :``        ``print``(vp[i][``1``],end``=``"");` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``string ``=` `"geeksforgeeks"``;` `    ``sortArr(string);` `    ``# This code is contributed by Yash_R`

## C#

 `using` `System;``using` `System.Collections.Generic;``using` `System.Linq;` `public` `class` `Program``{``  ``// Returns count of character in the string``  ``static` `int` `countFrequency(``string` `str, ``char` `ch)``  ``{``    ``int` `count = 0;` `    ``for` `(``int` `i = 0; i < str.Length; i++)``    ``{``      ``// Check for character``      ``if` `(str[i] == ch)``      ``{``        ``++count;``      ``}``    ``}` `    ``return` `count;``  ``}` `  ``// Function to sort the string according to the``  ``// frequency of characters in ascending order``  ``static` `void` `sortArr(``string` `str)``  ``{``    ``int` `n = str.Length;` `    ``// Dictionary to store the frequency of characters``    ``Dictionary<``char``, ``int``> freqDict = ``new` `Dictionary<``char``, ``int``>();` `    ``// Count the frequency of each character in the input string``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``      ``if` `(freqDict.ContainsKey(str[i]))``      ``{``        ``freqDict[str[i]]++;``      ``}``      ``else``      ``{``        ``freqDict[str[i]] = 1;``      ``}``    ``}` `    ``// Sort the dictionary by value (frequency) in ascending order``    ``var` `sortedDict = freqDict.OrderBy(x => x.Value);` `    ``// Print the sorted characters in the order of their frequency``    ``foreach` `(``var` `kvp ``in` `sortedDict)``    ``{``      ``for` `(``int` `i = 0; i < kvp.Value; i++)``      ``{``        ``Console.Write(kvp.Key);``      ``}``    ``}``  ``}` `  ``// Driver code``  ``static` `void` `Main(``string``[] args)``  ``{``    ``string` `str = ``"geeksforgeeks"``;` `    ``sortArr(str);``  ``}``}`

## Javascript

 ``

Output

`forggkksseeee`

Time Complexity: O(n2)

Auxiliary Space: O(n)

### Method 2: (Optimized Approach – Min Heap Based)

Algorithm:

1. Take the frequency of each character into a map.

2 .Take a MIN Heap, store in FREQUENCY, CHAR

3. After all insertions, Topmost element is the less frequent character

4. We keep a CUSTOM COMPARATOR for LESS FREQ, WHEN SAME FREQ – Ascending Order Characters.

5. Then Pop one by one and append in ANS String for FREQ no. of times.

## C++

 `#include ``using` `namespace` `std;` `//O(N*LogN) Time, O(Distinct(N)) Space``//MIN HEAP Based - as we need less frequent element first``#define ppi pair` `//CUSTOM COMPARATOR for Heap``class` `Compare{``  ``public``:``  ``//Override``  ``bool` `operator()(pair<``int``,``char``>below, pair<``int``,``char``> above){``    ``if``(below.first == above.first){``      ``//freq same``      ``return` `below.second > above.second; ``//lexicographically smaller is TOP``    ``}``    ``return` `below.first > above.first; ``//less freq at TOP``  ``}``};` `string frequencySort(string s) {` `  ``unordered_map<``char``,``int``> mpp;``  ``priority_queue,Compare> minH; ``// freq , character` `  ``for``(``char` `ch : s){``    ``mpp[ch]++;``  ``}` `  ``for``(``auto` `m : mpp){``    ``minH.push({m.second, m.first}); ``// as freq is 1st , char is 2nd``  ``}` `  ``string ans=``""``;``  ``//Now we have in the TOP - Less Freq chars` `  ``while``(minH.size()>0){` `    ``int` `freq = minH.top().first;``    ``char` `ch = minH.top().second;``    ``for``(``int` `i=0; i

## Java

 `import` `java.util.*;``class` `Pair ``implements` `Comparable``{``    ``int` `first;``    ``char` `second;``    ``Pair(``int` `first,``char` `second)``    ``{``        ``this``.first  = first;``        ``this``.second  = second;``    ``}``    ``// Custom comparator useful for heap``   ` `    ``public` `int` `compareTo(Pair a)``    ``{``       ``// If frequencies are same for two characters``       ``// sort according to their order``        ``if``(``this``.first==a.first)``            ``return` `this``.second-a.second;``         ``return` `this``.first-a.first;``    ``}``  ` `}``class` `Main {``    ``// O(N*LogN) Time, O(Distinct(N)) Space``    ``public` `static` `String frequencySort(String s) {` `        ``// Creating a HashMap to store the frequency of characters``        ``HashMap mpp = ``new` `HashMap();` `        ``// Creating a min heap to store the frequency and corresponding character``        ``PriorityQueue min_heap = ``new` `PriorityQueue();` `        ``// Looping through the string to calculate the frequency of each character``        ``for` `(``char` `ch : s.toCharArray()) {``            ``mpp.put(ch, mpp.getOrDefault(ch, ``0``) + ``1``);``        ``}` `        ``// Adding the frequency and character to the min heap``        ``for` `(``char` `m : mpp.keySet()) {``            ``min_heap.offer(``new` `Pair(mpp.get(m), m));``        ``}` `        ``String ans = ``""``;``        ``// Now we have in the TOP - Less Freq chars``        ``while` `(!min_heap.isEmpty()) {``            ``Pair pair = min_heap.poll();``            ``int` `freq = pair.first;``            ``char` `ch = pair.second;``            ``// Append as many times of frequency``            ``for` `(``int` `i = ``0``; i < freq; i++) {``                ``ans += ch;``            ``}``        ``}``        ``return` `ans;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args) {``        ``String str = ``"geeksforgeeks"``;``        ``System.out.println(frequencySort(str));``    ``}` `}`

## Python3

 `import` `heapq` `# O(N*LogN) Time, O(Distinct(N)) Space``def` `frequencySort(s):``    ``mpp ``=` `{}``    ``min_heap ``=` `[]` `    ``for` `ch ``in` `s:``        ``if` `ch ``in` `mpp:``            ``mpp[ch] ``+``=` `1``        ``else``:``            ``mpp[ch] ``=` `1` `    ``for` `m ``in` `mpp:``        ``heapq.heappush(min_heap, (mpp[m], m)) ``# as freq is 1st , char is 2nd` `    ``ans ``=` `""``    ``#Now we have in the TOP - Less Freq chars``    ``while` `min_heap:``        ``freq, ch ``=` `heapq.heappop(min_heap)``        ``ans ``+``=` `ch ``*` `freq ``# append as many times of freq``    ``return` `ans` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``str` `=` `"geeksforgeeks"``    ``print``(frequencySort(``str``))`` ` `# This code is contributed by Prince Kumar`

## C#

 `// C# approach``using` `System;``using` `System.Collections.Generic;` `class` `Pair : IComparable {``    ``public` `int` `first;``    ``public` `char` `second;``    ``public` `Pair(``int` `first, ``char` `second)``    ``{``        ``this``.first = first;``        ``this``.second = second;``    ``}` `    ``// Custom comparator useful for heap``    ``public` `int` `CompareTo(Pair a)``    ``{``        ``// If frequencies are same for two characters``        ``// sort according to their order``        ``if` `(``this``.first == a.first)``            ``return` `this``.second - a.second;``        ``return` `this``.first - a.first;``    ``}``}``class` `Program {``    ``// O(N*LogN) Time, O(Distinct(N)) Space``    ``public` `static` `string` `frequencySort(``string` `s)``    ``{``        ``// Creating a HashMap to store the frequency of``        ``// characters``        ``Dictionary<``char``, ``int``> mpp``            ``= ``new` `Dictionary<``char``, ``int``>();` `        ``// Creating a min heap to store the frequency and``        ``// corresponding character``        ``SortedSet min_heap = ``new` `SortedSet();` `        ``// Looping through the string to calculate the``        ``// frequency of each character``        ``foreach``(``char` `ch ``in` `s.ToCharArray())``        ``{``            ``if` `(mpp.ContainsKey(ch))``                ``mpp[ch] = mpp[ch] + 1;``            ``else``                ``mpp[ch] = 1;``        ``}` `        ``// Adding the frequency and character to the min``        ``// heap``        ``foreach``(``char` `m ``in` `mpp.Keys)``        ``{``            ``min_heap.Add(``new` `Pair(mpp[m], m));``        ``}` `        ``string` `ans = ``""``;``        ``// Now we have in the TOP - Less Freq chars``        ``while` `(min_heap.Count > 0) {``            ``Pair pair = min_heap.Min;``            ``int` `freq = pair.first;``            ``char` `ch = pair.second;``            ``// Append as many times of frequency``            ``for` `(``int` `i = 0; i < freq; i++) {``                ``ans += ch;``            ``}``            ``min_heap.Remove(pair);``        ``}``        ``return` `ans;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``string` `str = ``"geeksforgeeks"``;``        ``Console.WriteLine(frequencySort(str));``    ``}``}` `// This code is contributed by Susobhan Akhuli`

Output

`forggkksseeee`

Time Complexity:

O(N+ N* Log N + N* Log N)  =  O(N* Log N)

Reason:

• 1 insertion in heap takes O(Log N), For N insertions O(N*LogN). (HEAP = Priority Queue)
• 1 deletion in heap takes O(Log N), For N insertions O(N*LogN).
• Unordered Map takes O(1) for 1 Insertion.
• N is the length of the String S (input)

Extra Space Complexity:

O(N)

Reason:

• Map takes O(Distinct(N)) Space.
• Heap also takes O(Distinct(N)) Space.
• N is the length of the String S (input)

The Code, Approach, and Idea are proposed by Balakrishnan R (rbkraj000 GFG ID)

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