Find frequency of each element in a limited range array in less than O(n) time

Given a sorted array of positive integers, the count number of occurrences for each element in the array. Assume all elements in the array are less than some constant M.

Do this without traversing the complete array. i.e. expected time complexity is less than O(n).

Examples:

Input: arr[] = [1, 1, 1, 2, 3, 3, 5,
               5, 8, 8, 8, 9, 9, 10] 
Output:
Element 1 occurs 3 times
Element 2 occurs 1 times
Element 3 occurs 2 times
Element 5 occurs 2 times
Element 8 occurs 3 times
Element 9 occurs 2 times
Element 10 occurs 1 times

Input: arr[] = [2, 2, 6, 6, 7, 7, 7, 11] 
Output:
Element 2 occurs 2 times
Element 6 occurs 2 times
Element 7 occurs 3 times
Element 11 occurs 1 times



Method 1: This method uses the technique of Linear Search to solve the following.

  • Approach: The idea is to traverse the input array and for each distinct element of the array, store its frequency in a HashMap and finally print the HashMap.
  • Algorithm:
    1. Create a HashMap to map the frequency to element, i.e to store element-frequency pair.
    2. Traverse the array from start to end.
    3. For each element in the array update the frequency, i.e hm[array[i]]++
    4. Traverse the HashMap and print the element frequency pair.
  • Implementation:

    C++

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    // C++ program to count number of occurrences of
    // each element in the array #include <iostream>
    #include <bits/stdc++.h>
    using namespace std;
      
    // It prints number of
    // occurrences of each element in the array.
    void findFrequency(int arr[], int n)
    {
        // HashMap to store frequencies
        unordered_map<int, int> mp;
      
        // traverse the array
        for (int i = 0; i < n; i++) {
            // update the frequency
            mp[arr[i]]++;
        }
      
        // traverse the hashmap
        for (auto i : mp) {
            cout << "Element " << i.first << " occurs "
                 << i.second << " times" << endl;
        }
    }
      
    // Driver function
    int main()
    {
        int arr[] = { 1, 1, 1, 2, 3, 3, 5, 5,
                      8, 8, 8, 9, 9, 10 };
        int n = sizeof(arr) / sizeof(arr[0]);
      
        findFrequency(arr, n);
      
        return 0;
    }

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    Python3

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    # Python program to count number of occurrences of
    # each element in the array #include <iostream>
      
    # It prints number of
    # occurrences of each element in the array.
    def findFrequency(arr, n):
          
        # HashMap to store frequencies
        mp = {}
      
        # traverse the array
        for i in range(n):
      
            # update the frequency
            if arr[i] not in mp:
                mp[arr[i]] = 0
            mp[arr[i]] += 1
              
        # traverse the hashmap
        for i in mp:
            print("Element", i, "occurs", mp[i], "times")
          
    # Driver function
    arr = [1, 1, 1, 2, 3, 3, 5, 5,8, 8, 8, 9, 9, 10]
    n = len(arr)
      
    findFrequency(arr, n)
          
    # This code is contributed by shubhamsingh10

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    Output:

    Element 1 occurs 3 times
    Element 2 occurs 1 times
    Element 3 occurs 2 times
    Element 5 occurs 2 times
    Element 8 occurs 3 times
    Element 9 occurs 2 times
    Element 10 occurs 1 times
    
  • Complexity Analysis:

    • Time Complexity:O(n), only one traversal of the array is needed.
    • Space Complexity:O(n), to store the elements in the HashMap O(n) extra space is needed.

Method 2: This method uses the technique of Binary Search to arrive at the solution.

  • Approach:The problem can be solved in less than O(n) time if all its elements are sorted, i.e. if similar elements exist in the array then the elements are in a contiguous subarray or it can be said that if ends of a subarray are same then all the elements inside the subarray are equal. So the count of that element is the size of the subarray and all the elements of that subarray need not be counted.
  • Algorithm:
    1. Create a HashMap (hm) to store the frequency of elements.
    2. Create a recursive function that accepts a array and size.
    3. Check if the first element of the array is equal to the last element. If equal then all the elements are same and update the frequency by hm[array[0]+=size
    4. Else divide the array into two equal halves and call the function recursively for both the halves.
    5. Traverse the hashmap and print the element frequency pair.
  • Implementation:

    C++

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    // C++ program to count number of occurrences of
    // each element in the array in less than O(n) time
    #include <iostream>
    #include <vector>
    using namespace std;
      
    // A recursive function to count number of occurrences
    // for each element in the array without traversing
    // the whole array
    void findFrequencyUtil(int arr[], int low, int high,
                           vector<int>& freq)
    {
        // If element at index low is equal to element
        // at index high in the array
        if (arr[low] == arr[high]) {
            // increment the frequency of the element
            // by count of elements between high and low
            freq[arr[low]] += high - low + 1;
        }
        else {
            // Find mid and recurse for left and right
            // subarray
            int mid = (low + high) / 2;
            findFrequencyUtil(arr, low, mid, freq);
            findFrequencyUtil(arr, mid + 1, high, freq);
        }
    }
      
    // A wrapper over recursive function
    // findFrequencyUtil(). It print number of
    // occurrences of each element in the array.
    void findFrequency(int arr[], int n)
    {
        // create a empty vector to store frequencies
        // and initialize it by 0. Size of vector is
        // maximum value (which is last value in sorted
        // array) plus 1.
        vector<int> freq(arr[n - 1] + 1, 0);
      
        // Fill the vector with frequency
        findFrequencyUtil(arr, 0, n - 1, freq);
      
        // Print the frequencies
        for (int i = 0; i <= arr[n - 1]; i++)
            if (freq[i] != 0)
                cout << "Element " << i << " occurs "
                     << freq[i] << " times" << endl;
    }
      
    // Driver function
    int main()
    {
        int arr[] = { 1, 1, 1, 2, 3, 3, 5, 5,
                      8, 8, 8, 9, 9, 10 };
        int n = sizeof(arr) / sizeof(arr[0]);
      
        findFrequency(arr, n);
      
        return 0;
    }

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    Java

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    // Java program to count number of occurrences of
    // each element in the array in less than O(n) time
    import java.util.*;
      
    class GFG {
      
        // A recursive function to count number of occurrences
        // for each element in the array without traversing
        // the whole array
        static void findFrequencyUtil(int arr[], int low,
                                      int high, int[] freq)
        {
            // If element at index low is equal to element
            // at index high in the array
            if (arr[low] == arr[high]) {
                // increment the frequency of the element
                // by count of elements between high and low
                freq[arr[low]] += high - low + 1;
            }
            else {
                // Find mid and recurse for left and right
                // subarray
                int mid = (low + high) / 2;
                findFrequencyUtil(arr, low, mid, freq);
                findFrequencyUtil(arr, mid + 1, high, freq);
            }
        }
      
        // A wrapper over recursive function
        // findFrequencyUtil(). It print number of
        // occurrences of each element in the array.
        static void findFrequency(int arr[], int n)
        {
            // create a empty vector to store frequencies
            // and initialize it by 0. Size of vector is
            // maximum value (which is last value in sorted
            // array) plus 1.
            int[] freq = new int[arr[n - 1] + 1];
      
            // Fill the vector with frequency
            findFrequencyUtil(arr, 0, n - 1, freq);
      
            // Print the frequencies
            for (int i = 0; i <= arr[n - 1]; i++)
                if (freq[i] != 0)
                    System.out.println("Element " + i + " occurs " + freq[i] + " times");
        }
      
        // Driver Code
        public static void main(String[] args)
        {
            int arr[] = { 1, 1, 1, 2, 3, 3, 5,
                          5, 8, 8, 8, 9, 9, 10 };
            int n = arr.length;
      
            findFrequency(arr, n);
        }
    }
      
    // This code is contributed by 29AjayKumar

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    Python3

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    # Python 3 program to count number of occurrences of
    # each element in the array in less than O(n) time
      
    # A recursive function to count number of occurrences 
    # for each element in the array without traversing 
    # the whole array
    def findFrequencyUtil(arr, low, high, freq):
          
        # If element at index low is equal to element 
        # at index high in the array
        if (arr[low] == arr[high]):
              
            # increment the frequency of the element
            # by count of elements between high and low
            freq[arr[low]] += high - low + 1
          
        else:
              
            # Find mid and recurse for left  
            # and right subarray
            mid = int((low + high) / 2)
            findFrequencyUtil(arr, low, mid, freq)
            findFrequencyUtil(arr, mid + 1, high, freq)
          
    # A wrapper over recursive function 
    # findFrequencyUtil(). It print number of 
    # occurrences of each element in the array.
    def findFrequency(arr, n):
          
        # create a empty vector to store frequencies
        # and initialize it by 0. Size of vector is 
        # maximum value (which is last value in sorted
        # array) plus 1.
        freq = [0 for i in range(n - 1 + 1)]
          
        # Fill the vector with frequency
        findFrequencyUtil(arr, 0, n - 1, freq)
      
        # Print the frequencies
        for i in range(0, arr[n - 1] + 1, 1):
            if (freq[i] != 0):
                print("Element", i, "occurs"
                            freq[i], "times")
      
    # Driver Code
    if __name__ == '__main__':
        arr = [1, 1, 1, 2, 3, 3, 5,
               5, 8, 8, 8, 9, 9, 10]
        n = len(arr)
        findFrequency(arr, n)
          
    # This code is contributed by
    # Surendra_Gangwar

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    C#

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    // C# program to count number of occurrences of
    // each element in the array in less than O(n) time
    using System;
      
    class GFG {
      
        // A recursive function to count number of occurrences
        // for each element in the array without traversing
        // the whole array
        static void findFrequencyUtil(int[] arr, int low,
                                      int high, int[] freq)
        {
            // If element at index low is equal to element
            // at index high in the array
            if (arr[low] == arr[high]) {
                // increment the frequency of the element
                // by count of elements between high and low
                freq[arr[low]] += high - low + 1;
            }
            else {
                // Find mid and recurse for left and right
                // subarray
                int mid = (low + high) / 2;
                findFrequencyUtil(arr, low, mid, freq);
                findFrequencyUtil(arr, mid + 1, high, freq);
            }
        }
      
        // A wrapper over recursive function
        // findFrequencyUtil(). It print number of
        // occurrences of each element in the array.
        static void findFrequency(int[] arr, int n)
        {
            // create a empty vector to store frequencies
            // and initialize it by 0. Size of vector is
            // maximum value (which is last value in sorted
            // array) plus 1.
            int[] freq = new int[arr[n - 1] + 1];
      
            // Fill the vector with frequency
            findFrequencyUtil(arr, 0, n - 1, freq);
      
            // Print the frequencies
            for (int i = 0; i <= arr[n - 1]; i++)
                if (freq[i] != 0)
                    Console.WriteLine("Element " + i + " occurs " + freq[i] + " times");
        }
      
        // Driver Code
        public static void Main(String[] args)
        {
            int[] arr = { 1, 1, 1, 2, 3, 3, 5,
                          5, 8, 8, 8, 9, 9, 10 };
            int n = arr.Length;
      
            findFrequency(arr, n);
        }
    }
      
    // This code is contributed by Princi Singh

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    Output:

    Element 1 occurs 3 times
    Element 2 occurs 1 times
    Element 3 occurs 2 times
    Element 5 occurs 2 times
    Element 8 occurs 3 times
    Element 9 occurs 2 times
    Element 10 occurs 1 times
    
  • Complexity Analysis:
    • Time Complexity: O(m log n).
      Where m is number of distinct elements in the array of size n. Since m <= M (a constant) (elements are in a limited range), the time complexity of this solution is O(log n).
    • Space Complexity:O(n).
      To store the elements in the HashMap O(n) extra space is needed.

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