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++

 // 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);     cout << count(arr, n) << endl;     return 0; }

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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