# 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}

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ implementation of the approach ` `#include ` `#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 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; ` `} `

 `// 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 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 `

 `# 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 `

 `// 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 `

Output:
```3
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :