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# Find frequency of each character with positions in given Array of Strings

• Last Updated : 30 Jul, 2021

Given an array, arr[] consisting of N strings where each character of the string is lower case English alphabet, the task is to store and print the occurrence of every distinct character in every string.

Examples:

Input: arr[] = { “geeksforgeeks”, “gfg” }
Output: Occurences of: e = [1 2] [1 3] [1 10] [1 11]
Occurences of: f = [1 6] [2 2]
Occurences of: g = [1 1] [1 9] [2 1] [2 3]
Occurences of: k = [1 4] [1 12]
Occurences of: o = [1 7]
Occurences of: r = [1 8]
Occurences of: s = [1 5] [1 13]

Input: arr[] = { “abc”, “ab” }
Output: Occurences of: a = [1 1] [2 1]
Occurences of: b = [1 2] [2 2]
Occurences of: c = [1 3]

Approach: The above problem can be solved using Map and Vector data structures. Follow the steps below to solve the problem:

• Initialize a map<char, vector<pair<int, int>> > say mp to store the occurrences of a character in the vector of pairs, where each pair stores the index of the string in array as the first element and the position of the character in the string as the second element.
• Traverse the vector arr using a variable i and perform the following step:
• Finally, after completing the above steps, print the occurrences of every character by iterating over the map mp.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to print every occurence``// of every characters in every string``void` `printOccurences(vector arr, ``int` `N)``{``    ``map<``char``, vector > > mp;` `    ``// Iterate over the vector arr[]``    ``for` `(``int` `i = 0; i < N; i++) {``        ``// Traverse the string arr[i]``        ``for` `(``int` `j = 0; j < arr[i].length(); j++) {` `            ``// Push the pair of{i+1, j+1}``            ``// in mp[arr[i][j]]``            ``mp[arr[i][j]].push_back(``                ``make_pair(i + 1, j + 1));``        ``}``    ``}``    ``// Print the occurences of every``    ``// character``    ``for` `(``auto` `it : mp) {``        ``cout << ``"Occurences of: "` `<< it.first << ``" = "``;``        ``for` `(``int` `j = 0; j < (it.second).size(); j++) {``            ``cout << ``"["` `<< (it.second)[j].first << ``" "``                 ``<< (it.second)[j].second << ``"] "``;``        ``}``        ``cout << endl;``    ``}``}` `// Driver Code``int` `main()``{``    ``// Input``    ``vector arr = { ``"geeksforgeeks"``, ``"gfg"` `};``    ``int` `N = arr.size();``    ``// Function call``    ``printOccurences(arr, N);``}`

## Python3

 `# Python3 program for the above approach` `# Function to print every occurence``# of every characters in every string``def` `printOccurences(arr, N):``    ` `    ``mp ``=` `[[] ``for` `i ``in` `range``(``26``)]` `    ``# Iterate over the vector arr[]``    ``for` `i ``in` `range``(N):``        ` `        ``# Traverse the string arr[i]``        ``for` `j ``in` `range``(``len``(arr[i])):``            ` `            ``# Push the pair of{i+1, j+1}``            ``# in mp[arr[i][j]]``            ``mp[``ord``(arr[i][j]) ``-` `ord``(``'a'``)].append(``                           ``(i ``+` `1``, j ``+` `1``))``            ` `    ``# print(mp)` `    ``# Print the occurences of every``    ``# character``    ``for` `i ``in` `range``(``26``):``        ``if` `len``(mp[i]) ``=``=` `0``:``            ``continue``        ` `        ``print``(``"Occurences of:"``, ``chr``(i ``+``              ``ord``(``'a'``)), ``"="``, end ``=` `" "``)``        ``for` `j ``in` `mp[i]:``            ``print``(``"["` `+` `str``(j[``0``]) ``+` `" "` `+``                        ``str``(j[``1``]) ``+` `"] "``, end ``=` `"")``        ``print``()` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Input``    ``arr``=` `[ ``"geeksforgeeks"``, ``"gfg"` `]``    ``N  ``=` `len``(arr)``    ` `    ``# Function call``    ``printOccurences(arr, N)` `# This code is contributed by mohit kumar 29`

## Javascript

 ``
Output
```Occurences of: e = [1 2] [1 3] [1 10] [1 11]
Occurences of: f = [1 6] [2 2]
Occurences of: g = [1 1] [1 9] [2 1] [2 3]
Occurences of: k = [1 4] [1 12]
Occurences of: o = [1 7]
Occurences of: r = [1 8]
Occurences of: s = [1 5] [1 13] ```

Time Complexity: O(N*M), where M is the length of the longest string.
Auxiliary Space: O(N*M)

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