# Number of times the given string occurs in the array in the range [l, r]

Given an array of strings arr[] and two integers l and r, the task is to find the number of times the given string str occurs in the array in the range [l, r] (1-based indexing). Note that the strings contain only lowercase letters.

Examples:

Input: arr[] = {“abc”, “def”, “abc”}, L = 1, R = 2, str = “abc”
Output: 1

Input: arr[] = {“abc”, “def”, “abc”}, L = 1, R = 3, str = “ghf”
Output: 0

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

Approach: The idea is to use an unordered_map to store the indices in which the ith string of array occurs. If the given string is not present in the map then answer is zero otherwise perform binary search on the indices of the given string present in the map, and find the number of occurrences of the string in the range [L, R].

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the number of occurrences of ` `int` `NumOccurrences(string arr[], ``int` `n, string str, ``int` `L, ``int` `R) ` `{ ` `    ``// To store the indices of strings in the array ` `    ``unordered_map > M; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``string temp = arr[i]; ` `        ``auto` `it = M.find(temp); ` ` `  `        ``// If current string doesn't ` `        ``// have an entry in the map ` `        ``// then create the entry ` `        ``if` `(it == M.end()) { ` `            ``vector<``int``> A; ` `            ``A.push_back(i + 1); ` `            ``M.insert(make_pair(temp, A)); ` `        ``} ` `        ``else` `{ ` `            ``it->second.push_back(i + 1); ` `        ``} ` `    ``} ` ` `  `    ``auto` `it = M.find(str); ` ` `  `    ``// If the given string is not ` `    ``// present in the array ` `    ``if` `(it == M.end()) ` `        ``return` `0; ` ` `  `    ``// If the given string is present ` `    ``// in the array ` `    ``vector<``int``> A = it->second; ` `    ``int` `y = upper_bound(A.begin(), A.end(), R) - A.begin(); ` `    ``int` `x = upper_bound(A.begin(), A.end(), L - 1) - A.begin(); ` ` `  `    ``return` `(y - x); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string arr[] = { ``"abc"``, ``"abcabc"``, ``"abc"` `}; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(string); ` `    ``int` `L = 1; ` `    ``int` `R = 3; ` `    ``string str = ``"abc"``; ` ` `  `    ``cout << NumOccurrences(arr, n, str, L, R); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python implementation of the approach ` `from` `bisect ``import` `bisect_right as upper_bound ` `from` `collections ``import` `defaultdict ` ` `  `# Function to return the number of occurrences of ` `def` `numOccurences(arr: ``list``, n: ``int``, string: ``str``, L: ``int``, R: ``int``) ``-``> ``int``: ` ` `  `    ``# To store the indices of strings in the array ` `    ``M ``=` `defaultdict(``lambda``: ``list``) ` `    ``for` `i ``in` `range``(n): ` `        ``temp ``=` `arr[i] ` ` `  `        ``# If current string doesn't ` `        ``# have an entry in the map ` `        ``# then create the entry ` `        ``if` `temp ``not` `in` `M: ` `            ``A ``=` `[] ` `            ``A.append(i ``+` `1``) ` `            ``M[temp] ``=` `A ` `        ``else``: ` `            ``M[temp].append(i ``+` `1``) ` ` `  `    ``# If the given string is not ` `    ``# present in the array ` `    ``if` `string ``not` `in` `M: ` `        ``return` `0` ` `  `    ``# If the given string is present ` `    ``# in the array ` `    ``A ``=` `M[string] ` `    ``y ``=` `upper_bound(A, R) ` `    ``x ``=` `upper_bound(A, L ``-` `1``) ` ` `  `    ``return` `(y ``-` `x) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``arr ``=` `[``"abc"``, ``"abcabc"``, ``"abc"``] ` `    ``n ``=` `len``(arr) ` `    ``L ``=` `1` `    ``R ``=` `3` `    ``string ``=` `"abc"` ` `  `    ``print``(numOccurences(arr, n, string, L, R)) ` ` `  `# This code is contributed by ` `# sanjeev2552 `

Output:

```2
```

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.

My Personal Notes arrow_drop_up 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.

Improved By : sanjeev2552