# Count of subsequences consisting of the same element

Given an array A[] consisting of N integers, the task is to find the total number of subsequence which contain only one distinct number repeated throughout the subsequence.

Examples:

Input: A[] = {1, 2, 1, 5, 2}
Output:
Explanation:
Subsequences {1}, {2}, {1}, {5}, {2}, {1, 1} and {2, 2} satisfy the required conditions.

Input: A[] = {5, 4, 4, 5, 10, 4}
Output: 11
Explanation:
Subsequences {5}, {4}, {4}, {5}, {10}, {4}, {5, 5}, {4, 4}, {4, 4}, {4, 4} and {4, 4, 4} satisfy the required conditions.

Approach:
Follow the steps below to solve the problem:

• Iterate over the array and calculate the frequency of each element in a HashMap
• Traverse the HashMap. For each element, calculate the number of desired subsequences possible by the equation:

Number of subsequences possible by arr[i] = 2freq[arr[i]] – 1

• Calculate the total possible subsequences from the given array.

Below is the implementation of the above approach:

 `// C++ program to implement ` `// the above approach ` `#include ` `using` `namespace` `std; `   `// Function to count subsequences in ` `// array containing same element ` `void` `CountSubSequence(``int` `A[], ``int` `N) ` `{ ` `    ``// Stores the count ` `    ``// of subsequences ` `    ``int` `result = 0; `   `    ``// Stores the frequency ` `    ``// of array elements ` `    ``map<``int``, ``int``> mp; `   `    ``for` `(``int` `i = 0; i < N; i++) { `   `        ``// Update frequency of A[i] ` `        ``mp[A[i]]++; ` `    ``} `   `    ``for` `(``auto` `it : mp) { `   `        ``// Calculate number of subsequences ` `        ``result ` `            ``= result + ``pow``(2, it.second) - 1; ` `    ``} `   `    ``// Print the result ` `    ``cout << result << endl; ` `} `   `// Driver code ` `int` `main() ` `{ ` `    ``int` `A[] = { 5, 4, 4, 5, 10, 4 }; `   `    ``int` `N = ``sizeof``(A) / ``sizeof``(A); `   `    ``CountSubSequence(A, N); `   `    ``return` `0; ` `} `

 `// Java program to implement ` `// the above approach ` `import` `java.util.*;`   `class` `GFG{`   `// Function to count subsequences in` `// array containing same element` `static` `void` `CountSubSequence(``int` `A[], ``int` `N)` `{` `    `  `    ``// Stores the count` `    ``// of subsequences` `    ``int` `result = ``0``;`   `    ``// Stores the frequency` `    ``// of array elements` `    ``Map mp = ``new` `HashMap();`   `    ``for``(``int` `i = ``0``; i < N; i++)` `    ``{` `        `  `        ``// Update frequency of A[i]` `        ``mp.put(A[i], mp.getOrDefault(A[i], ``0``) + ``1``);` `    ``}`   `    ``for``(Integer it : mp.values())` `    ``{` `        `  `        ``// Calculate number of subsequences` `        ``result = result + (``int``)Math.pow(``2``, it) - ``1``;` `    ``}` `    `  `    ``// Print the result` `    ``System.out.println(result);` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `A[] = { ``5``, ``4``, ``4``, ``5``, ``10``, ``4` `};` `    ``int` `N = A.length;` `    `  `    ``CountSubSequence(A, N);` `}` `}`   `// This code is contributed by offbeat`

 `# Python3 program to implement  ` `# the above approach  `   `# Function to count subsequences in  ` `# array containing same element  ` `def` `CountSubSequence(A, N):` `    `  `    ``# Stores the frequency  ` `    ``# of array elements  ` `    ``mp ``=` `{}` `    `  `    ``for` `element ``in` `A:` `        ``if` `element ``in` `mp:` `            ``mp[element] ``+``=` `1` `        ``else``:` `            ``mp[element] ``=` `1` `            `  `    ``result ``=` `0` `    `  `    ``for` `key, value ``in` `mp.items():` `        `  `        ``# Calculate number of subsequences  ` `        ``result ``+``=` `pow``(``2``, value) ``-` `1` `        `  `    ``# Print the result      ` `    ``print``(result)`   `# Driver code` `A ``=` `[ ``5``, ``4``, ``4``, ``5``, ``10``, ``4` `]` `N ``=` `len``(A)`   `CountSubSequence(A, N)`   `# This code is contributed by jojo9911`

 `// C# program to implement ` `// the above approach ` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG{ `   `// Function to count subsequences in ` `// array containing same element ` `public` `static` `void` `CountSubSequence(``int` `[]A, ``int` `N) ` `{ ` `    `  `    ``// Stores the count ` `    ``// of subsequences ` `    ``int` `result = 0; `   `    ``// Stores the frequency ` `    ``// of array elements ` `    ``var` `mp = ``new` `Dictionary<``int``, ``int``>();`   `    ``for``(``int` `i = 0; i < N; i++) ` `    ``{ ` `        `  `        ``// Update frequency of A[i] ` `        ``if``(mp.ContainsKey(A[i]))` `            ``mp[A[i]] += 1;` `        ``else` `            ``mp.Add(A[i], 1);` `    ``} `   `    ``foreach``(``var` `it ``in` `mp) ` `    ``{ ` `        `  `        ``// Calculate number of subsequences ` `        ``result = result + ` `                 ``(``int``)Math.Pow(2, it.Value) - 1; ` `    ``} ` `    `  `    ``// Print the result ` `    ``Console.Write(result); ` `} `   `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `[]A = { 5, 4, 4, 5, 10, 4 }; ` `    ``int` `N = A.Length; ` `    `  `    ``CountSubSequence(A, N); ` `} ` `} `   `// This code is contributed by grand_master`

Output:
```11

```

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

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Improved By : offbeat, grand_master, jojo9911

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