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 implementation to Sort strings // according to the frequency of // characters in ascending order #include <stdio.h> #include <string.h> // Returns count of character in the string int countFrequency( char str[], char ch)
{ int count = 0;
for ( int i = 0; i < strlen (str); i++) {
// Check for vowel
if (str[i] == ch)
++count;
}
return count;
} // Function to sort the string // according to the frequency void sortArr( char str[])
{ int n = strlen (str);
// Array of pairs to store the frequency of
// characters with respective character
struct Pair {
int count;
char ch;
} pairs[n];
int numOfPairs = 0;
// Inserting frequency
// with respective character
// in the array of pairs
for ( int i = 0; i < n; i++) {
int count = countFrequency(str, str[i]);
int found = 0;
for ( int j = 0; j < numOfPairs; j++) {
if (pairs[j].ch == str[i]) {
pairs[j].count = count;
found = 1;
break ;
}
}
if (!found) {
pairs[numOfPairs].count = count;
pairs[numOfPairs].ch = str[i];
numOfPairs++;
}
}
// Sort the array of pairs, this will sort the pair
// according to the number of characters
for ( int i = 0; i < numOfPairs - 1; i++) {
for ( int j = i + 1; j < numOfPairs; j++) {
if (pairs[i].count > pairs[j].count) {
struct Pair temp = pairs[i];
pairs[i] = pairs[j];
pairs[j] = temp;
}
}
}
// Print the sorted array of pairs content
for ( int i = 0; i < numOfPairs; i++) {
for ( int j = 0; j < pairs[i].count; j++) {
printf ( "%c" , pairs[i].ch);
}
}
} // Driver code int main()
{ char str[] = "geeksforgeeks" ;
sortArr(str);
return 0;
} |
// 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;
} |
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<Character, Integer> freqDict
= new HashMap<Character, Integer>();
// 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<Map.Entry<Character, Integer> > sortedDict
= new ArrayList<Map.Entry<Character, Integer> >(
freqDict.entrySet());
Collections.sort(
sortedDict,
new Comparator<
Map.Entry<Character, Integer> >() {
public int compare(
Map.Entry<Character, Integer> o1,
Map.Entry<Character, Integer> 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<Character, Integer> 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 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
|
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);
}
} |
<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:
#include <stdbool.h> #include <stdio.h> #include <stdlib.h> #include <string.h> // O(N*LogN) Time, O(Distinct(N)) Space // MIN HEAP Based - as we need less frequent element first typedef struct {
int first;
char second;
} ppi; // CUSTOM COMPARATOR for Heap bool compare(ppi below, ppi 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
} char * frequencySort( char * s)
{ int i;
int n = strlen (s);
char * ans = ( char *) malloc ( sizeof ( char ) * (n + 1));
memset (ans, '\0' , sizeof ( char ) * (n + 1));
int * mpp = ( int *) calloc ( sizeof ( int ), 256);
ppi* arr = (ppi*) malloc ( sizeof (ppi) * 256);
int k = 0;
for (i = 0; i < n; i++) {
mpp[s[i]]++;
}
for (i = 0; i < 256; i++) {
if (mpp[i]) {
arr[k].first = mpp[i];
arr[k].second = i;
k++;
}
}
ppi* heap = (ppi*) malloc ( sizeof (ppi) * k);
for (i = 0; i < k; i++) {
heap[i] = arr[i];
}
int heapSize = k;
for (i = heapSize / 2; i >= 0; i--) {
int parent = i;
int child = 2 * parent + 1;
while (child < heapSize) {
if (child + 1 < heapSize
&& compare(heap[child], heap[child + 1])) {
child++;
}
if (compare(heap[parent], heap[child])) {
ppi temp = heap[parent];
heap[parent] = heap[child];
heap[child] = temp;
parent = child;
child = 2 * parent + 1;
}
else {
break ;
}
}
}
while (heapSize > 0) {
ppi top = heap[0];
for (i = 0; i < top.first; i++) {
ans[ strlen (ans)] = top.second;
}
heap[0] = heap[heapSize - 1];
heapSize--;
int parent = 0;
int child = 2 * parent + 1;
while (child < heapSize) {
if (child + 1 < heapSize
&& compare(heap[child], heap[child + 1])) {
child++;
}
if (compare(heap[parent], heap[child])) {
ppi temp = heap[parent];
heap[parent] = heap[child];
heap[child] = temp;
parent = child;
child = 2 * parent + 1;
}
else {
break ;
}
}
}
free (arr);
free (heap);
free (mpp);
return ans;
} // Driver code int main()
{ char str[] = "geeksforgeeks" ;
printf ( "%s\n" , frequencySort(str));
return 0;
} |
#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) |
import java.util.*;
class Pair implements Comparable<Pair>
{ 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<Character, Integer> mpp = new HashMap<Character, Integer>();
// Creating a min heap to store the frequency and corresponding character
PriorityQueue<Pair> min_heap = new PriorityQueue<Pair>();
// 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));
}
} |
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# approach using System;
using System.Collections.Generic;
class Pair : IComparable<Pair> {
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<Pair> min_heap = new SortedSet<Pair>();
// 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 |
// JavaScript approach class Pair { constructor(first, second){
this .first = first;
this .second = second;
}
// Custom comparator useful for heap
compareTo(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;
}
} // O(N*LogN) Time, O(Distinct(N)) Space function frequencySort(s) {
// Creating a HashMap to store the frequency of characters
let mpp = new Map();
// Creating a min heap to store the frequency and corresponding character
let min_heap = [];
// Looping through the string to calculate the frequency of each character
for (let i = 0; i < s.length; i++) {
let ch = s.charAt(i);
let count = mpp.get(ch) || 0;
mpp.set(ch, count + 1);
}
// Adding the frequency and character to the min heap
for (let [key, value] of mpp.entries()) {
min_heap.push( new Pair(value, key));
}
min_heap.sort( function (a, b){ return a.compareTo(b); });
let ans = "" ;
// Now we have in the TOP - Less Freq chars
while (min_heap.length > 0) {
let pair = min_heap.shift();
let freq = pair.first;
let ch = pair.second;
// Append as many times of frequency
for (let i = 0; i < freq; i++) {
ans += ch;
}
}
return ans;
} // Driver code let str = "geeksforgeeks" ;
console.log(frequencySort(str)); // This Code is Contributed by Susobhan Akhuli. |
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)