Sort a string according to the frequency of characters
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 <bits/stdc++.h> 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<pair< int , char > > 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
<script> // JavaScript implementation of the above approach // Returns count of character in the string function countFrequency(str, ch) { var count = 0; for ( var 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 function sortArr(str) { var n = str.length; // Vector to store the frequency of // characters with respective character vp = new Array(n); // Inserting frequency // with respective character // in the vector pair for ( var i = 0; i < n; i++) { vp[i] = [countFrequency(str, str[i]), str[i]]; } // Sort the vector, this will sort the pair // according to the number of characters vp.sort(); // Print the sorted vector content for ( var i = 0; i < n; i++) document.write(vp[i][1]); } // Driver Code // Array of points let str = "geeksforgeeks" ; sortArr(str); </script> |
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.
CODE:
C++
#include <bits/stdc++.h> using namespace std; //O(N*LogN) Time, O(Distinct(N)) Space //MIN HEAP Based - as we need less frequent element first #define ppi pair<int,char> //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<ppi,vector<ppi>,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<freq; i++){ ans+=ch; // append as many times of freq } minH.pop(); //Heapify happens } return ans; } // Driver code int main() { string str = "geeksforgeeks" ; cout<<frequencySort(str)<< "\n" ; return 0; } //Code is Contributed by Balakrishnan R (rbkraj000) |
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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