Count subarrays having a single distinct element that can be obtained from a given array

Last Updated : 04 Mar, 2021

Given an array arr[] of size N, the task is to count the number of subarrays consisting of a single distinct element that can be obtained from a given array.

Examples:

Input: N = 4, arr[] = { 2, 2, 2, 2 }
Output: 7
Explanation: All such subarrays {{2}, {2}, {2}, {2}, {2, 2}, {2, 2, 2}, {2, 2, 2, 2}}. Therefore, total count of such subarrays are 7.
Input: N = 3, arr[] = { 1, 1, 3 }
Output: 4

Approach: Follow the steps below to solve the problem:

Initialize a Map to store the frequency of array elements.
Traverse the array and update frequency of each element in the Map.
Traverse the Map and check if frequency of current element is greater than 1.
If found to be true, then increment count of subarrays by frequency of current element – 1.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count subarrays of
// single distinct element into
// which given array can be split
void divisionalArrays(int arr[3], int N)
{
    // Stores the count
    int sum = N;
 
    // Stores frequency of
    // array elements
    unordered_map<int, int> mp;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
        mp[arr[i]]++;
    }
 
    // Traverse the map
    for (auto x : mp) {
        if (x.second > 1) {
 
            // Increase count of
            // subarrays by (frequency-1)
            sum += x.second - 1;
        }
    }
    cout << sum << endl;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 1, 3 };
    int N = sizeof(arr)
            / sizeof(arr[0]);
    divisionalArrays(arr, N);
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG
{
 
  // Function to count subarrays of
  // single distinct element into
  // which given array can be split
  static void divisionalArrays(int arr[], int N)
  {
 
    // Stores the count
    int sum = N;
 
    // Stores frequency of
    // array elements
    HashMap<Integer, Integer> mp
      = new HashMap<Integer, Integer>();
 
    // Traverse the array
    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 the map
    for (Map.Entry x : mp.entrySet())
    {
 
      if ((int)x.getValue() > 1)
      {
 
        // Increase count of
        // subarrays by (frequency-1)
        sum += (int)x.getValue() - 1;
      }
    }
 
    System.out.println(sum);
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int arr[] = { 1, 1, 3 };
    int N = arr.length;
    divisionalArrays(arr, N);
  }
}
 
// This code is contributed by Dharanendra L V

Python3

# python 3 program for the above approach
from collections import defaultdict
 
# Function to count subarrays of
# single distinct element into
# which given array can be split
def divisionalArrays(arr, N):
 
    # Stores the count
    sum = N
 
    # Stores frequency of
    # array elements
    mp = defaultdict(int)
 
    # Traverse the array
    for i in range(N):
        mp[arr[i]] += 1
 
    # Traverse the map
    for x in mp:
        if (mp[x] > 1):
 
            # Increase count of
            # subarrays by (frequency-1)
            sum += mp[x] - 1
    print(sum)
 
# Driver Code
if __name__ == "__main__":
 
    arr = [1, 1, 3]
    N = len(arr)
    divisionalArrays(arr, N)
 
    # This code is contributed by ukasp.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to count subarrays of
// single distinct element into
// which given array can be split
static void divisionalArrays(int []arr, int N)
{
     
    // Stores the count
    int sum = N;
     
    // Stores frequency of
    // array elements
    Dictionary<int,
               int> mp = new Dictionary<int,
                                        int>();
     
    // Traverse the array
    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);
        }
    }
     
    // Traverse the map
    foreach(KeyValuePair<int, int> x in mp)
    {
        if ((int)x.Value > 1)
        {
             
            // Increase count of
            // subarrays by (frequency-1)
            sum += (int)x.Value - 1;
        }
    }
     
    Console.WriteLine(sum);
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 1, 3 };
    int N = arr.Length;
     
    divisionalArrays(arr, N);
}
}
 
// This code is contributed by shikhasingrajput

Output:

Time Complexity: O(N)
Auxiliary Space: O(N)

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