Count subarrays having each distinct element occuring at least twice

Given an array arr[] of size N, the task is to count the number of subarrays from the given array, such that each distinct element in these subarray occurs at least twice.

Examples:

Input: arr[] = {1, 1, 2, 2, 2}
Output: 6
Explanation: Subarrays having each element occuring at least twice are :{{1, 1}, {1, 1, 2, 2}, {1, 1, 2, 2, 2}, {2, 2}, {2, 2, 2}, {2, 2}}.
Therefore, the required output is 6.

Input: arr[] = {1, 2, 1, 2, 3}
Output: 1

Naive Approach: The simplest approach to solve this problem is to traverse the array and generate all possible subarrays of the given array and for each subarray, check if all elements in the subarray occurs at least twice or not. If found to be true, then increment the count. Finally, print the count obtained.



Time Complexity: O(N3)
Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach the idea is to use Hashing. Follow the steps below to solve the problem:

  • Initialize a variable, say cntSub to store the count of subarrays such that each element in the subarray occurs at least twice.
  • Create a Map, say cntFreq, to store the frequency of elements of each subarray.
  • Initialize a variable, say cntUnique, to store the count of elements in a subarray whose frequency is 1.
  • Traverse the array and generate all possible subarrays. For each possible subarray, store the frequency of each element of the array and check if the value of cntUnique is 0 or not. If found to be true, then increment the value of cntSub.
  • Finally, print the value of cntSub.

Below is the implementation of the above approach:

C++

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// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to get the count
// of subarrays having each
// element occuring at least twice
int cntSubarrays(int arr[], int N)
{
    // Stores count of subarrays
    // having each distinct element
    // occuring at least twice
    int cntSub = 0;
 
    // Stores count of unique
    // elements in a subarray
    int cntUnique = 0;
 
    // Store frequency of
    // each element of a subarray
    unordered_map<int, int> cntFreq;
 
    // Traverse the given
    // array
    for (int i = 0; i < N;
         i++) {
 
        // Count frequency and
        // check conditions for
        // each subarray
        for (int j = i; j < N;
             j++) {
 
            // Update frequency
            cntFreq[arr[j]]++;
 
            // Check if frequency of
            // arr[j] equal to 1
            if (cntFreq[arr[j]]
                == 1) {
 
                // Update Count of
                // unique elements
                cntUnique++;
            }
            else if (cntFreq[arr[j]]
                     == 2) {
 
                // Update count of
                // unique elements
                cntUnique--;
            }
 
            // If each element of subarray
            // occurs at least twice
            if (cntUnique == 0) {
 
                // Update cntSub
                cntSub++;
            }
        }
 
        // Remove all elements
        // from the subarray
        cntFreq.clear();
 
        // Update cntUnique
        cntUnique = 0;
    }
    return cntSub;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 1, 2, 2, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << cntSubarrays(arr, N);
}

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Java

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// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
   
// Function to get the count
// of subarrays having each
// element occuring at least twice
static int cntSubarrays(int arr[], int N)
{
     
    // Stores count of subarrays
    // having each distinct element
    // occuring at least twice
    int cntSub = 0;
 
    // Stores count of unique
    // elements in a subarray
    int cntUnique = 0;
 
    // Store frequency of
    // each element of a subarray
     Map<Integer,
         Integer> cntFreq = new HashMap<Integer,
                                        Integer>();
                                         
    // Traverse the given
    // array
    for(int i = 0; i < N; i++)
    {
         
        // Count frequency and
        // check conditions for
        // each subarray
        for(int j = i; j < N; j++)
        {
             
            // Update frequency
            cntFreq.put(arr[j],
                        cntFreq.getOrDefault(
                        arr[j], 0) + 1);
 
            // Check if frequency of
            // arr[j] equal to 1
            if (cntFreq.get(arr[j]) == 1)
            {
                 
                // Update Count of
                // unique elements
                cntUnique++;
            }
            else if (cntFreq.get(arr[j]) == 2)
            {
                 
                // Update count of
                // unique elements
                cntUnique--;
            }
 
            // If each element of subarray
            // occurs at least twice
            if (cntUnique == 0)
            {
                 
                // Update cntSub
                cntSub++;
            }
        }
 
        // Remove all elements
        // from the subarray
        cntFreq.clear();
 
        // Update cntUnique
        cntUnique = 0;
    }
    return cntSub;
}
 
// Driver Code
public static void main(String args[])
{
    int arr[] = { 1, 1, 2, 2, 2 };
    int N = arr.length;
     
    System.out.println(cntSubarrays(arr, N));
}
}
 
// This code is contributed by SURENDRA_GANGWAR

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Python3

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# Python3 program to implement
# the above appraoch
from collections import defaultdict
 
# Function to get the count
# of subarrays having each
# element occuring at least twice
def cntSubarrays(arr, N):
 
    # Stores count of subarrays
    # having each distinct element
    # occuring at least twice
    cntSub = 0
 
    # Stores count of unique
    # elements in a subarray
    cntUnique = 0
 
    # Store frequency of
    # each element of a subarray
    cntFreq = defaultdict(lambda : 0)
 
    # Traverse the given
    # array
    for i in range(N):
         
        # Count frequency and
        # check conditions for
        # each subarray
        for j in range(i, N):
             
            # Update frequency
            cntFreq[arr[j]] += 1
 
            # Check if frequency of
            # arr[j] equal to 1
            if (cntFreq[arr[j]] == 1):
 
                # Update Count of
                # unique elements
                cntUnique += 1
 
            elif (cntFreq[arr[j]] == 2):
                 
                # Update count of
                # unique elements
                cntUnique -= 1
 
            # If each element of subarray
            # occurs at least twice
            if (cntUnique == 0):
                 
                # Update cntSub
                cntSub += 1
 
        # Remove all elements
        # from the subarray
        cntFreq.clear()
 
        # Update cntUnique
        cntUnique = 0
 
    return cntSub
 
# Driver code
if __name__ == '__main__':
 
    arr = [ 1, 1, 2, 2, 2 ]
    N = len(arr)
     
    print(cntSubarrays(arr, N))
 
# This code is contributed by Shivam Singh

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Output: 

6







 

Time Complexity: O(N2)
Auxiliary Space: O(N)

 

 

 

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