Given an unsorted array. The task is to calculate the cumulative frequency of each element of the array using 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 loops picks an element from left to right that are not visited. The inner loop counts its occurrences and mark occurrences as visited. Time complexity of this solution is O(n*n) and auxiliary space required is O(n).
A better solution is to use sorting. We sort the array so that 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++ program to count cumlative // frequencies of elements in an unsorted array. #include <bits/stdc++.h> using namespace std;
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]]++;
// Declare a set
set<pair<int, int> > st;
// insert the element and
// and insert its frequency in a set
for (auto x : hm) {
st.insert({ x.first, x.second });
}
int cumul = 0;
// iterate the set and print the
// cumulative frequency
for (auto x : st) {
cumul += x.second;
cout << x.first << " " << cumul << endl;
}
} // 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 count cumlative // 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 cumlative # 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 cumlative // 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 |
1 2 2 4 3 5 4 6
Time complexity of the solution is O(n log n).
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++ 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 */ |
1->2 3->3 2->5 4->6
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