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

• Difficulty Level : Easy
• Last Updated : 01 Sep, 2021

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:

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`

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

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