# Sort a string according to the frequency of characters

• Difficulty Level : Medium
• Last Updated : 19 Jul, 2022

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;` `}`

## 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`

## Javascript

 ``

Output

`forggkksseeee`

Time Complexity: O(n2)

Auxiliary Space: O(n)

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

Algorithm:

1. Take the 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; ``//lexicographicaly 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

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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