Cumulative frequency of count of each element in an unsorted array
Last Updated : 14 Aug, 2024
Given an unsorted array. The task is to calculate the cumulative frequency of each element of the array using a count array.
Examples:
Input : arr[] = [1, 2, 2, 1, 3, 4]
Output :1->2
2->4
3->5
4->6
Input : arr[] = [1, 1, 1, 2, 2, 2]
Output :1->3
2->6
A simple solution is to use two nested loops. The outer loop picks an element from left to right that is not visited. The inner loop counts its occurrences and marks occurrences as visited. The time complexity of this solution is O(n*n) and the auxiliary space required is O(n).
A better solution is to use sorting. We sort the array so that the same elements come together. After sorting, we linearly traverse elements and count their frequencies.
An efficient solution is to use hashing. Insert the element and its frequency in a set of pairs. As the set stores unique values in a sorted order, it will store all the elements with their frequencies in a sorted order. Iterate in the set and print the frequencies by adding the previous ones at every step.
Below is the implementation of the above approach:
C++ #include <bits/stdc++.h> using namespace std; // Function to print the cumulative frequency according to // the order given void countFreq(int a[], int n) { // Declaring a map so values get inserted in a sorted // manner map<int, int> m; // Inserting values into the map for (int i = 0; i < n; i++) { m[a[i]]++; } // Variable to store the count of previous number // cumulative frequency int cumul = 0; for (auto v : m) { cout << v.first << " " << v.second + cumul << endl; cumul += v.second; } } int main() { int arr[] = { 1, 3, 2, 4, 2, 1 }; int n = sizeof(arr) / sizeof(arr[0]); countFreq(arr, n); return 0; } // This code is contributed by Vinayak Pareek (Kargil) // Java program to count cumulative // frequencies of elements in an unsorted array. import java.util.*; class GFG { static void countFreq(int[] a, int n) { // Insert elements and their // frequencies in hash map. HashMap<Integer, Integer> hm = new HashMap<>(); for (int i = 0; i < n; i++) hm.put(a[i], hm.get(a[i]) == null ? 1 : hm.get(a[i]) + 1); // Declare a Map SortedMap<Integer, Integer> st = new TreeMap<>(); // insert the element and // and insert its frequency in a set for (HashMap.Entry<Integer, Integer> x : hm.entrySet()) { st.put(x.getKey(), x.getValue()); } int cumul = 0; // iterate the set and print the // cumulative frequency for (SortedMap.Entry<Integer, Integer> x : st.entrySet()) { cumul += x.getValue(); System.out.println(x.getKey() + " " + cumul); } } // Driver Code public static void main(String[] args) { int[] a = { 1, 3, 2, 4, 2, 1 }; int n = a.length; countFreq(a, n); } } // This code is contributed by // sanjeev2552 # Python3 program to count cumulative # frequencies of elements in an unsorted array. def countFreq(a, n): # Insert elements and their # frequencies in hash map. hm = {} for i in range(0, n): hm[a[i]] = hm.get(a[i], 0) + 1 # Declare a set st = set() # Insert the element and # its frequency in a set for x in hm: st.add((x, hm[x])) cumul = 0 # Iterate the set and print # the cumulative frequency for x in sorted(st): cumul += x[1] print(x[0], cumul) # Driver Code if __name__ == "__main__": a = [1, 3, 2, 4, 2, 1] n = len(a) countFreq(a, n) # This code is contributed by Rituraj Jain // C# program to count cumulative // frequencies of elements in an // unsorted array. using System; using System.Collections.Generic; using System.Linq; class GFG{ static void countFreq(int[] a, int n) { // Insert elements and their // frequencies in hash map. Dictionary<int, int> hm = new Dictionary<int, int>(); for(int i = 0; i < n; i++) { if (hm.ContainsKey(a[i])) { hm[a[i]]++; } else { hm[a[i]] = 1; } } int cumul = 0; // Iterate the set and print the // cumulative frequency foreach(KeyValuePair<int, int> x in hm.OrderBy(key => key.Key)) { cumul += x.Value; Console.Write(x.Key + " " + cumul + "\n"); } } // Driver Code public static void Main(string[] args) { int[] a = { 1, 3, 2, 4, 2, 1 }; int n = a.Length; countFreq(a, n); } } // This code is contributed by rutvik_56 <script> // Javascript program to count cumulative // frequencies of elements in an unsorted array. function countFreq(a,n) { // Insert elements and their // frequencies in hash map. let hm = new Map(); for (let i = 0; i < n; i++) hm.set(a[i], hm.get(a[i]) == null ? 1 : hm.get(a[i]) + 1); // Declare a Map let st = new Set(); // insert the element and // and insert its frequency in a set for (let [key, value] of hm.entries()) { st.add([key, value]); } st=[...st.entries()].sort() let cumul = 0; // iterate the set and print the // cumulative frequency for (let [key, value] of st.entries()) { cumul += value[1][1]; document.write(value[1][0] + " " + cumul+"<br>"); } } // Driver Code let a=[1, 3, 2, 4, 2, 1]; let n = a.length; countFreq(a, n); // This code is contributed by unknown2108 </script> Time Complexity: O(n log n), where n is the size of the given array
Auxiliary Space: O(n), as extra space of size n is used to create a map
What if we need frequencies of elements according to the order of the first occurrence?
For example, an array [2, 4, 1, 2, 1, 3, 4], the frequency of 2 should be printed first, then of 4, then 1, and finally 3.
Approach: Hash the count of occurrences of an element. Traverse in the array and print the cumulative frequency. Once the element and its cumulative frequency has been printed, hash the occurrence of that element as 0 so that it not printed again if it appears in the latter half of array while traversal.
Below is the implementation of the above approach:
C++ // C++ program to print the cumulative frequency // according to the order given #include <bits/stdc++.h> using namespace std; // Function to print the cumulative frequency // according to the order given void countFreq(int a[], int n) { // Insert elements and their // frequencies in hash map. unordered_map<int, int> hm; for (int i=0; i<n; i++) hm[a[i]]++; int cumul = 0; // traverse in the array for(int i=0;i<n;i++) { // add the frequencies cumul += hm[a[i]]; // if the element has not been // visited previously if(hm[a[i]]) { cout << a[i] << "->" << cumul << endl; } // mark the hash 0 // as the element's cumulative frequency // has been printed hm[a[i]]=0; } } // Driver Code int main() { int a[] = {1, 3, 2, 4, 2, 1}; int n = sizeof(a)/sizeof(a[0]); countFreq(a, n); return 0; } // Java program to print the cumulative frequency // according to the order given class GFG { // Function to print the cumulative frequency // according to the order given static void countFreq(int a[], int n) { // Insert elements and their // frequencies in hash map. int hm[] = new int[n]; for (int i = 0; i < n; i++) hm[a[i]]++; int cumul = 0; // traverse in the array for(int i = 0; i < n; i++) { // add the frequencies cumul += hm[a[i]]; // if the element has not been // visited previously if(hm[a[i]] != 0) { System.out.println(a[i] + "->" + cumul); } // mark the hash 0 // as the element's cumulative frequency // has been printed hm[a[i]] = 0; } } // Driver Code public static void main(String[] args) { int a[] = {1, 3, 2, 4, 2, 1}; int n = a.length; countFreq(a, n); } } // This code has been contributed by 29AjayKumar # Python3 program to print the cumulative # frequency according to the order given # Function to print the cumulative frequency # according to the order given def countFreq(a, n): # Insert elements and their # frequencies in hash map. hm = dict() for i in range(n): hm[a[i]] = hm.get(a[i], 0) + 1 cumul = 0 # traverse in the array for i in range(n): # add the frequencies cumul += hm[a[i]] # if the element has not been # visited previously if(hm[a[i]] > 0): print(a[i], "->", cumul) # mark the hash 0 # as the element's cumulative # frequency has been printed hm[a[i]] = 0 # Driver Code a = [1, 3, 2, 4, 2, 1] n = len(a) countFreq(a, n) # This code is contributed by mohit kumar
// C# program to print the cumulative frequency // according to the order given using System; class GFG { // Function to print the cumulative frequency // according to the order given static void countFreq(int []a, int n) { // Insert elements and their // frequencies in hash map. int []hm = new int[n]; for (int i = 0; i < n; i++) hm[a[i]]++; int cumul = 0; // traverse in the array for(int i = 0; i < n; i++) { // add the frequencies cumul += hm[a[i]]; // if the element has not been // visited previously if(hm[a[i]] != 0) { Console.WriteLine(a[i] + "->" + cumul); } // mark the hash 0 // as the element's cumulative frequency // has been printed hm[a[i]] = 0; } } // Driver Code public static void Main(String[] args) { int []a = {1, 3, 2, 4, 2, 1}; int n = a.Length; countFreq(a, n); } } /* This code contributed by PrinciRaj1992 */ <script> // Javascript program to print the cumulative frequency // according to the order given // Function to print the cumulative frequency // according to the order given function countFreq(a,n) { // Insert elements and their // frequencies in hash map. let hm = new Array(n); for(let i=0;i<hm.length;i++) { hm[i]=0; } for (let i = 0; i < n; i++) hm[a[i]]++; let cumul = 0; // traverse in the array for(let i = 0; i < n; i++) { // add the frequencies cumul += hm[a[i]]; // if the element has not been // visited previously if(hm[a[i]] != 0) { document.write(a[i] + "->" + cumul+"<br>"); } // mark the hash 0 // as the element's cumulative frequency // has been printed hm[a[i]] = 0; } } // Driver Code let a=[1, 3, 2, 4, 2, 1]; let n = a.length; countFreq(a, n); // This code is contributed by patel2127 </script> Output1->2 3->3 2->5 4->6
Time Complexity: O(n), where n is the size of the given array
Auxiliary Space: O(n), as extra space of size n is used to create a map
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