Measuring the frequency of elements in an array is a really handy skill and is required a lot of competitive coding problems. We, in a lot of problems, are required to measure the frequency of various elements like numbers, alphabets, symbols, etc. as a part of our problem.
Examples:
Input: arr[] = {10, 20, 20, 10, 10, 20, 5, 20}
Output: 10 3
20 4
5 1
Input: arr[] = {10, 20, 20}
Output: 10 2
20 1
Naive method: We run two loops. For every item count number of times, it occurs. To avoid duplicate printing, keep track of processed items.
Implementation:
C++
// C++ program to count frequencies of array items#include <bits/stdc++.h>using namespace std; void countFreq(int arr[], int n){ // Mark all array elements as not visited vector<int> visited(n, false); // Traverse through array elements and // count frequencies for (int i = 0; i < n; i++) { // Skip this element if already processed if (visited[i] == true) continue; // Count frequency int count = 1; for (int j = i + 1; j < n; j++) { if (arr[i] == arr[j]) { visited[j] = true; count++; } } cout << arr[i] << " " << count << endl; }} int main(){ int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 }; int n = sizeof(arr) / sizeof(arr[0]); countFreq(arr, n); return 0;} |
Java
// Java program to count frequencies // of array itemsimport java.util.*;class GFG{static void countFreq(int arr[], int n){ // Mark all array elements as not visited boolean []visited = new boolean[n]; // Traverse through array elements and // count frequencies for (int i = 0; i < n; i++) { // Skip this element if already processed if (visited[i] == true) continue; // Count frequency int count = 1; for (int j = i + 1; j < n; j++) { if (arr[i] == arr[j]) { visited[j] = true; count++; } } System.out.println(arr[i] + " " + count); }}// Driver Codepublic static void main(String[] args){ int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 }; int n = arr.length; countFreq(arr, n);}}// This code is contributed by 29AjayKumar |
Python3
# Python program to count frequencies # of array itemsdef countFreq(arr, n): # mark all array elements as not visited visited = [False for i in range(n)] # Traverse through array elements # and count frequencies for i in range(n): # Skip this element if already processed if visited[i] == True: continue # count frequency count = 1 for j in range(i + 1, n): if arr[i] == arr[j]: visited[j] = True count += 1 print(arr[i], count) # Driver codea = [10, 20, 20, 10, 10, 20, 5, 20]n = len(a)countFreq(a, n)# This code is contributed # by Mohit kumar 29 |
C#
// C# program to count frequencies // of array itemsusing System; class GFG{static void countFreq(int []arr, int n){ // Mark all array elements as not visited Boolean []visited = new Boolean[n]; // Traverse through array elements and // count frequencies for (int i = 0; i < n; i++) { // Skip this element if already processed if (visited[i] == true) continue; // Count frequency int count = 1; for (int j = i + 1; j < n; j++) { if (arr[i] == arr[j]) { visited[j] = true; count++; } } Console.WriteLine(arr[i] + " " + count); }}// Driver Codepublic static void Main(String[] args){ int []arr = { 10, 20, 20, 10, 10, 20, 5, 20 }; int n = arr.Length; countFreq(arr, n);}}// This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript program to count frequencies of array items function countFreq(arr, n) { // Mark all array elements as not visited let visited = new Array(n); visited.fill(false); // Traverse through array elements and // count frequencies for (let i = 0; i < n; i++) { // Skip this element if already processed if (visited[i] == true) continue; // Count frequency let count = 1; for (let j = i + 1; j < n; j++) { if (arr[i] == arr[j]) { visited[j] = true; count++; } } document.write(arr[i] + " " + count + "</br>"); } } let arr = [ 10, 20, 20, 10, 10, 20, 5, 20 ]; let n = arr.length; countFreq(arr, n); // This code is contributed by mukesh07.</script> |
Output
10 3 20 4 5 1
Time Complexity: O(n*n)
Auxiliary Space: O(n)
Optimized methods :
Measuring frequencies when elements are limited by value: If our input array has small values, we can use array elements as index in a count array and increment count. In below example, elements are maximum 10.
Input : arr[] = {5, 5, 6, 6, 5, 6, 1, 2, 3, 10, 10}
limit = 10
Output : 1 1
2 1
3 1
5 3
6 3
10 2
Implementation:
C++
// C++ program to count frequencies of array items// having small values.#include <bits/stdc++.h>using namespace std; void countFreq(int arr[], int n, int limit){ // Create an array to store counts. The size // of array is limit+1 and all values are // initially 0 vector<int> count(limit+1, 0); // Traverse through array elements and // count frequencies (assuming that elements // are limited by limit) for (int i = 0; i < n; i++) count[arr[i]]++; for (int i = 0; i <= limit; i++) if (count[i] > 0) cout << i << " " << count[i] << endl; } int main(){ int arr[] = {5, 5, 6, 6, 5, 6, 1, 2, 3, 10, 10}; int n = sizeof(arr) / sizeof(arr[0]); int limit = 10; countFreq(arr, n, limit); return 0;} |
Java
// Java program to count frequencies of array items// having small values.class GFG{ static void countFreq(int arr[], int n, int limit){ // Create an array to store counts. The size // of array is limit+1 and all values are // initially 0 int []count = new int[limit + 1]; // Traverse through array elements and // count frequencies (assuming that elements // are limited by limit) for (int i = 0; i < n; i++) count[arr[i]]++; for (int i = 0; i <= limit; i++) if (count[i] > 0) System.out.println(i + " " + count[i]);}// Driver Codepublic static void main(String[] args) { int arr[] = {5, 5, 6, 6, 5, 6, 1, 2, 3, 10, 10}; int n = arr.length; int limit = 10; countFreq(arr, n, limit);}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 program to count frequencies of # array items having small values.def countFreq(arr, n, limit): # Create an array to store counts. # The size of array is limit+1 and # all values are initially 0 count = [0 for i in range(limit + 1)] # Traverse through array elements and # count frequencies (assuming that # elements are limited by limit) for i in range(n): count[arr[i]] += 1 for i in range(limit + 1): if (count[i] > 0): print(i, count[i])# Driver Codearr = [ 5, 5, 6, 6, 5, 6, 1, 2, 3, 10, 10 ]n = len(arr)limit = 10countFreq(arr, n, limit)# This code is contributed by avanitrachhadiya2155 |
C#
// C# program to count frequencies of // array items having small values.using System; class GFG{ static void countFreq(int []arr, int n, int limit){ // Create an array to store counts. // The size of array is limit+1 and // all values are initially 0 int []count = new int[limit + 1]; // Traverse through array elements and // count frequencies (assuming that // elements are limited by limit) for (int i = 0; i < n; i++) count[arr[i]]++; for (int i = 0; i <= limit; i++) if (count[i] > 0) Console.WriteLine(i + " " + count[i]);}// Driver Codepublic static void Main(String[] args) { int []arr = {5, 5, 6, 6, 5, 6, 1, 2, 3, 10, 10}; int n = arr.Length; int limit = 10; countFreq(arr, n, limit);}}// This code is contributed // by Princi Singh |
Javascript
<script>// Javascript program to count frequencies// of array items having small values.function countFreq(arr, n, limit){ // Create an array to store counts. // The size of array is limit+1 and // all values are initially 0 let count = new Array(limit + 1); count.fill(0); // Traverse through array elements and // count frequencies (assuming that // elements are limited by limit) for(let i = 0; i < n; i++) count[arr[i]]++; for(let i = 0; i <= limit; i++) if (count[i] > 0) document.write(i + " " + count[i] + "</br>");}// Driver codelet arr = [ 5, 5, 6, 6, 5, 6, 1, 2, 3, 10, 10 ];let n = arr.length;let limit = 10;countFreq(arr, n, limit);// This code is contributed by rameshtravel07 </script> |
Output
1 1 2 1 3 1 5 3 6 3 10 2
Time Complexity: O(n+k) where n is the size of the input array and k is the limit value.
Auxiliary Space: O(k).
Implementation:
C++
// C++ program to count frequencies of array items#include <bits/stdc++.h>using namespace std;const int limit = 255; void countFreq(string str){ // Create an array to store counts. The size // of array is limit+1 and all values are // initially 0 vector<int> count(limit+1, 0); // Traverse through string characters and // count frequencies for (int i = 0; i < str.length(); i++) count[str[i]]++; for (int i = 0; i <= limit; i++) if (count[i] > 0) cout << (char)i << " " << count[i] << endl; } int main(){ string str = "neveropen"; countFreq(str); return 0;} |
Java
// Java program to count frequencies of array itemsclass GFG {static int limit = 255; static void countFreq(String str){ // Create an array to store counts. The size // of array is limit+1 and all values are // initially 0 int []count= new int[limit + 1]; // Traverse through string characters and // count frequencies for (int i = 0; i < str.length(); i++) count[str.charAt(i)]++; for (int i = 0; i <= limit; i++) if (count[i] > 0) System.out.println((char)i + " " + count[i]);}// Driver Codepublic static void main(String[] args){ String str = "neveropen"; countFreq(str);}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 program to count frequencies of array itemslimit = 255def countFreq(Str) : # Create an array to store counts. The size # of array is limit+1 and all values are # initially 0 count = [0] * (limit + 1) # Traverse through string characters and # count frequencies for i in range(len(Str)) : count[ord(Str[i])] += 1 for i in range(limit + 1) : if (count[i] > 0) : print(chr(i), count[i]) Str = "neveropen"countFreq(Str)# This code is contributed by divyeshrabadiya07 |
C#
// C# program to count frequencies// of array itemsusing System; class GFG {static int limit = 25; static void countFreq(String str){ // Create an array to store counts. // The size of array is limit+1 and // all values are initially 0 int []count = new int[limit + 1]; // Traverse through string characters // and count frequencies for (int i = 0; i < str.Length; i++) count[str[i] - 'A']++; for (int i = 0; i <= limit; i++) if (count[i] > 0) Console.WriteLine((char)(i + 'A') + " " + count[i]);}// Driver Codepublic static void Main(String[] args) { String str = "GEEKSFORGEEKS"; countFreq(str);}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript program to count frequencies of array items let limit = 255; function countFreq(str){ // Create an array to store counts. The size // of array is limit+1 and all values are // initially 0 let count= new Array(limit + 1); for(let i=0;i<count.length;i++) { count[i]=0; } // Traverse through string characters and // count frequencies for (let i = 0; i < str.length; i++) count[str[i].charCodeAt(0)]++; for (let i = 0; i <= limit; i++) { if (count[i] > 0) document.write(String.fromCharCode(i) + " " + count[i]+"<br>"); }}// Driver Codelet str = "neveropen";countFreq(str); // This code is contributed by unknown2108</script> |
Output
G 2 e 4 f 1 k 2 o 1 r 1 s 2
Measuring frequencies when elements are in limited range: For example, consider a string containing only upper case alphabets. Elements of string are limited in range from ‘A’ to ‘Z’. The idea is to subtract smallest element (‘A’ in this example) to get the index of the element.
Implementation:
C++
// C++ program to count frequencies of array items#include <bits/stdc++.h>using namespace std;const int limit = 25; void countFreq(string str){ // Create an array to store counts. The size // of array is limit+1 and all values are // initially 0 vector<int> count(limit+1, 0); // Traverse through string characters and // count frequencies for (int i = 0; i < str.length(); i++) count[str[i] - 'A']++; for (int i = 0; i <= limit; i++) if (count[i] > 0) cout << (char)(i + 'A') << " " << count[i] << endl; } int main(){ string str = "GEEKSFORGEEKS"; countFreq(str); return 0;} |
Java
// Java program to count frequencies// of array itemsimport java.util.*;class GFG {static int limit = 25; static void countFreq(String str){ // Create an array to store counts. // The size of array is limit+1 and // all values are initially 0 int []count = new int[limit + 1]; // Traverse through string characters // and count frequencies for (int i = 0; i < str.length(); i++) count[str.charAt(i) - 'A']++; for (int i = 0; i <= limit; i++) if (count[i] > 0) System.out.println((char)(i + 'A') + " " + count[i]);}// Driver Codepublic static void main(String[] args) { String str = "GEEKSFORGEEKS"; countFreq(str);}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 program to count frequencies of array itemslimit = 25; def countFreq(str): # Create an array to store counts. The size # of array is limit+1 and all values are # initially 0 count = [0 for i in range(limit + 1)] # Traverse through string characters and # count frequencies for i in range(len(str)): count[ord(str[i]) - ord('A')] += 1 for i in range(limit + 1): if (count[i] > 0): print(chr(i + ord('A')), count[i])# Driver codeif __name__=='__main__': str = "GEEKSFORGEEKS"; countFreq(str); # This code is contributed by Pratham76 |
C#
// C# program to count frequencies// of array itemsusing System; class GFG {static int limit = 25; static void countFreq(String str){ // Create an array to store counts. // The size of array is limit+1 and // all values are initially 0 int []count = new int[limit + 1]; // Traverse through string characters // and count frequencies for (int i = 0; i < str.Length; i++) count[str[i] - 'A']++; for (int i = 0; i <= limit; i++) if (count[i] > 0) Console.WriteLine((char)(i + 'A') + " " + count[i]);} // Driver Codepublic static void Main(String[] args) { String str = "GEEKSFORGEEKS"; countFreq(str);}} // This code contributed by PrinciRaj1992 |
Javascript
<script>// JavaScript program to count frequencies// of array itemslet limit = 25;function countFreq(str){ // Create an array to store counts. // The size of array is limit+1 and // all values are initially 0 let count = new Array(limit + 1); for(let i=0;i<limit+1;i++) { count[i]=0; } // Traverse through string characters // and count frequencies for (let i = 0; i < str.length; i++) count[str[i].charCodeAt(0) - 'A'.charCodeAt(0)]++; for (let i = 0; i <= limit; i++) if (count[i] > 0) document.write(String.fromCharCode(i + 'A'.charCodeAt(0)) + " " + count[i]+"<br>");}// Driver Codelet str = "GEEKSFORGEEKS";countFreq(str);// This code is contributed by rag2127</script> |
Output
E 4 F 1 G 2 K 2 O 1 R 1 S 2
Measuring frequencies if no range and no limit: The idea is to use hashing (unordered_map in C++ and HashMap in Java) to get frequencies.
Implementation:
C++
// C++ program to count frequencies of array items#include <bits/stdc++.h>using namespace std;void countFreq(int arr[], int n){ unordered_map<int, int> mp; // Traverse through array elements and // count frequencies for (int i = 0; i < n; i++) mp[arr[i]]++; // Traverse through map and print frequencies for (auto x : mp) cout << x.first << " " << x.second << endl;}int main(){ int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 }; int n = sizeof(arr) / sizeof(arr[0]); countFreq(arr, n); return 0;} |
Java
// Java program to count frequencies of array itemsimport java.util.*;class GFG { static void countFreq(int arr[], int n){ HashMap<Integer, Integer>mp = new HashMap<Integer, Integer>(); // Traverse through array elements and // count frequencies for (int i = 0 ; i < n; i++) { if(mp.containsKey(arr[i])) { mp.put(arr[i], mp.get(arr[i]) + 1); } else { mp.put(arr[i], 1); } } // Traverse through map and print frequencies for (Map.Entry<Integer, Integer> entry : mp.entrySet()) System.out.println(entry.getKey() + " " + entry.getValue());}// Driver Codepublic static void main(String[] args) { int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 }; int n = arr.length; countFreq(arr, n); }}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to count frequencies of array itemsdef countFreq(arr, n): mp = {} # Traverse through array elements and # count frequencies for i in range(n): if arr[i] in mp: mp[arr[i]] += 1 else: mp[arr[i]] = 1 # Traverse through map and print frequencies for x in sorted(mp): print(x, mp[x])# Driver Codearr = [ 10, 20, 20, 10, 10, 20, 5, 20 ]n = len(arr)countFreq(arr, n)# This code is contributed by divyesh072019 |
C#
// C# program to count frequencies of array itemsusing System;using System.Collections.Generic;class GFG { static void countFreq(int []arr, int n){ Dictionary<int, int> mp = new Dictionary<int, int>(); // Traverse through array elements and // count frequencies for (int i = 0 ; i < n; i++) { if(mp.ContainsKey(arr[i])) { var val = mp[arr[i]]; mp.Remove(arr[i]); mp.Add(arr[i], val + 1); } else { mp.Add(arr[i], 1); } } // Traverse through map and print frequencies foreach(KeyValuePair<int, int> entry in mp) Console.WriteLine(entry.Key + " " + entry.Value);}// Driver Codepublic static void Main(String[] args) { int []arr = { 10, 20, 20, 10, 10, 20, 5, 20 }; int n = arr.Length; countFreq(arr, n); }}// This code is contributed by Princi Singh |
Javascript
<script>// Javascript program to count frequencies of array itemsfunction countFreq(arr, n){ let mp = new Map(); // Traverse through array elements and // count frequencies for (let i = 0 ; i < n; i++) { if(mp.has(arr[i])) { mp.set(arr[i], mp.get(arr[i]) + 1); } else { mp.set(arr[i], 1); } } // Traverse through map and print frequencies for (let [key, value] of mp.entries()) document.write(key + " " + value+"<br>");}// Driver Codelet arr=[10, 20, 20, 10, 10, 20, 5, 20];let n = arr.length;countFreq(arr, n); // This code is contributed by ab2127</script> |
Output
5 1 20 4 10 3
Complexity Analysis:
- Time Complexity : O(n)
- Auxiliary Space : O(n)
In above efficient solution, how to print elements in same order as they appear in input?
Implementation:
C++
// C++ program to count frequencies of array items#include <bits/stdc++.h>using namespace std;void countFreq(int arr[], int n){ unordered_map<int, int> mp; // Traverse through array elements and // count frequencies for (int i = 0; i < n; i++) mp[arr[i]]++; // To print elements according to first // occurrence, traverse array one more time // print frequencies of elements and mark // frequencies as -1 so that same element // is not printed multiple times. for (int i = 0; i < n; i++) { if (mp[arr[i]] != -1) { cout << arr[i] << " " << mp[arr[i]] << endl; mp[arr[i]] = -1; } }}int main(){ int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 }; int n = sizeof(arr) / sizeof(arr[0]); countFreq(arr, n); return 0;} |
Java
// Java program to count frequencies of array itemsimport java.util.*;class GFG {static void countFreq(int arr[], int n){ HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>(); // Traverse through array elements and // count frequencies for (int i = 0 ; i < n; i++) { if(mp.containsKey(arr[i])) { mp.put(arr[i], mp.get(arr[i]) + 1); } else { mp.put(arr[i], 1); } } // To print elements according to first // occurrence, traverse array one more time // print frequencies of elements and mark // frequencies as -1 so that same element // is not printed multiple times. for (int i = 0; i < n; i++) { if (mp.get(arr[i]) != -1) { System.out.println(arr[i] + " " + mp.get(arr[i])); mp. put(arr[i], -1); } }}// Driver Codepublic static void main(String[] args) { int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 }; int n = arr.length; countFreq(arr, n);}}// This code is contributed by Princi Singh |
Python3
# Python3 program to count frequencies of array items def countFreq(arr, n): mp = {} # Traverse through array elements and # count frequencies for i in range(n): if arr[i] not in mp: mp[arr[i]] = 1 else: mp[arr[i]] += 1 # To print elements according to first # occurrence, traverse array one more time # print frequencies of elements and mark # frequencies as -1 so that same element # is not printed multiple times. for i in range(n): if(mp[arr[i]] != -1): print(arr[i], mp[arr[i]]) mp[arr[i]] = -1# Driver codearr = [10, 20, 20, 10, 10, 20, 5, 20 ]n = len(arr)countFreq(arr, n)# This code is contributed by rag2127 |
C#
// C# program to count frequencies of array itemsusing System;using System.Collections.Generic; class GFG {static void countFreq(int []arr, int n){ Dictionary<int, int> mp = new Dictionary<int, int>(); // Traverse through array elements and // count frequencies for (int i = 0 ; i < n; i++) { if(mp.ContainsKey(arr[i])) { mp[arr[i]] = mp[arr[i]] + 1; } else { mp.Add(arr[i], 1); } } // To print elements according to first // occurrence, traverse array one more time // print frequencies of elements and mark // frequencies as -1 so that same element // is not printed multiple times. for (int i = 0; i < n; i++) { if (mp[arr[i]] != -1) { Console.WriteLine(arr[i] + " " + mp[arr[i]]); mp[arr[i]] = - 1; } }}// Driver Codepublic static void Main(String[] args) { int []arr = { 10, 20, 20, 10, 10, 20, 5, 20 }; int n = arr.Length; countFreq(arr, n);}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript program to count // frequencies of array itemsfunction countFreq(arr, n){ let mp = new Map(); // Traverse through array elements and // count frequencies for(let i = 0 ; i < n; i++) { if (mp.has(arr[i])) { mp.set(arr[i], mp.get(arr[i]) + 1); } else { mp.set(arr[i], 1); } } // To print elements according to first // occurrence, traverse array one more time // print frequencies of elements and mark // frequencies as -1 so that same element // is not printed multiple times. for(let i = 0; i < n; i++) { if (mp.get(arr[i]) != -1) { document.write(arr[i] + " " + mp.get(arr[i]) + "<br>"); mp.set(arr[i], -1); } }}// Driver Codelet arr = [ 10, 20, 20, 10, 10, 20, 5, 20 ];let n = arr.length; countFreq(arr, n);// This code is contributed by patel2127</script> |
Output
10 3 20 4 5 1
Complexity Analysis
- Time Complexity : O(n)
- Auxiliary Space : O(n)
In Java, we can get elements in same order using LinkedHashMap. Therefore we do not need an extra loop.
Lot of problems are based on frequency measurement and will be a cheesecake if we know how to calculate frequency of various elements in a given array. For example try the given below problems which are based on frequency measurement:
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