# 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:**

- First find the total number of sub-arrays that can be formed from the array and denote this by
*total*then*total = (n*(n+1))/2*. - Now find the sub-arrays that have all the elements distinct (can be found out using window sliding technique) and denote this by
*unique*. - Finally, the number of sub-arrays that have at least one element duplicate are
*(total – unique)*

Below is the implementation of the above approach:

## C++

`// 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; ` `} ` |

*chevron_right*

*filter_none*

## Python3

# 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

**Output:**

3

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