Number of sub-arrays that have at least one duplicate

Given an array arr of n elements, the task is to find the number of the sub-arrays of the given array that contain at least one duplicate element.

Examples:

Input: arr[] = {1, 2, 3}
Output: 0
There is no sub-array with duplicate elements.

Input: arr[] = {4, 3, 4, 3}
Output: 3
Possible sub-arrays are {4, 3, 4}, {4, 3, 4, 3} and {3, 4, 3}

Approach:



Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
  
// Function to return the count of the
// sub-arrays that have at least one duplicate
ll count(ll arr[], ll n)
{
    ll unique = 0;
  
    // two pointers
    ll i = -1, j = 0;
  
    // to store frequencies of the numbers
    unordered_map<ll, ll> freq;
    for (j = 0; j < n; j++) {
        freq[arr[j]]++;
  
        // number is not distinct
        if (freq[arr[j]] >= 2) {
            i++;
            while (arr[i] != arr[j]) {
                freq[arr[i]]--;
                i++;
            }
            freq[arr[i]]--;
            unique = unique + (j - i);
        }
        else
            unique = unique + (j - i);
    }
  
    ll total = n * (n + 1) / 2;
  
    return total - unique;
}
  
// Driver code
int main()
{
    ll arr[] = { 4, 3, 4, 3 };
    ll n = sizeof(arr) / sizeof(arr[0]);
    cout << count(arr, n) << endl;
    return 0;
}
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// Java implementation of the approach
import java.util.*;
  
class GFG 
{
  
// Function to return the count of the
// sub-arrays that have at least one duplicate
static Integer count(Integer arr[], Integer n)
{
    Integer unique = 0;
  
    // two pointers
    Integer i = -1, j = 0;
  
    // to store frequencies of the numbers
    Map<Integer, Integer> freq = new HashMap<>();
    for (j = 0; j < n; j++) 
    {
        if(freq.containsKey(arr[j]))
        {
            freq.put(arr[j], freq.get(arr[j]) + 1);
        }
        else
        {
            freq.put(arr[j], 1);
        }
  
        // number is not distinct
        if (freq.get(arr[j]) >= 2
        {
            i++;
            while (arr[i] != arr[j])
            {
                freq.put(arr[i], freq.get(arr[i]) - 1);
                i++;
            }
            freq.put(arr[i], freq.get(arr[i]) - 1);
            unique = unique + (j - i);
        }
        else
            unique = unique + (j - i);
    }
  
    Integer total = n * (n + 1) / 2;
  
    return total - unique;
}
  
// Driver code
public static void main(String[] args)
{
    Integer arr[] = { 4, 3, 4, 3 };
    Integer n = arr.length;
    System.out.println(count(arr, n));
}
}
  
// This code is contributed by 29AjayKumar
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# Python3 implementation of the approach 
from collections import defaultdict
  
# Function to return the count of the 
# sub-arrays that have at least one duplicate 
def count(arr, n): 
  
    unique = 0
  
    # two pointers 
    i, j = -1, 0
  
    # to store frequencies of the numbers 
    freq = defaultdict(lambda:0
    for j in range(0, n): 
        freq[arr[j]] += 1
  
        # number is not distinct 
        if freq[arr[j]] >= 2
            i += 1
              
            while arr[i] != arr[j]: 
                freq[arr[i]] -= 1
                i += 1
              
            freq[arr[i]] -= 1
            unique = unique + (j - i) 
          
        else:
            unique = unique + (j - i) 
      
    total = (n * (n + 1)) // 2
  
    return total - unique 
  
# Driver Code
if __name__ == "__main__":
  
    arr = [4, 3, 4, 3
    n = len(arr) 
    print(count(arr, n))
  
# This code is contributed 
# by Rituraj Jain
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// C# implementation of the approach
using System;
using System.Collections.Generic;             
  
class GFG 
{
  
// Function to return the count of the
// sub-arrays that have at least one duplicate
static int count(int []arr, int n)
{
    int unique = 0;
  
    // two pointers
    int i = -1, j = 0;
  
    // to store frequencies of the numbers
    Dictionary<int,
               int> freq = new Dictionary<int
                                          int>();
    for (j = 0; j < n; j++) 
    {
        if(freq.ContainsKey(arr[j]))
        {
            freq[arr[j]] = freq[arr[j]] + 1;
        }
        else
        {
            freq.Add(arr[j], 1);
        }
  
        // number is not distinct
        if (freq[arr[j]] >= 2) 
        {
            i++;
            while (arr[i] != arr[j])
            {
                freq[arr[i]] = freq[arr[i]] - 1;
                i++;
            }
            freq[arr[i]] = freq[arr[i]] - 1;
            unique = unique + (j - i);
        }
        else
            unique = unique + (j - i);
    }
  
    int total = n * (n + 1) / 2;
  
    return total - unique;
}
  
// Driver code
public static void Main(String[] args)
{
    int []arr = { 4, 3, 4, 3 };
    int n = arr.Length;
    Console.WriteLine(count(arr, n));
}
}
  
// This code is contributed by PrinciRaj1992
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Output:
3

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