Open in App
Not now

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

• Last Updated : 01 Feb, 2023

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

Input: arr[] = { “abc”, “ab” }
Output: Occurrences of: a = [1 1] [2 1]
Occurrences of: b = [1 2] [2 2]
Occurrences 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 occurrence` `// of every characters in every string` `void` `printOccurrences(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 occurrences of every` `    ``// character` `    ``for` `(``auto` `it : mp) {` `        ``cout << ``"Occurrences 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` `    ``printOccurrences(arr, N);` `}`

## Java

 `import` `java.util.*;` `import` `java.util.Map.Entry;`   `class` `GFG {` `  ``public` `static` `void` `printOccurrences(List arr, ``int` `N) {` `    ``Map>> mp = ``new` `HashMap<>();`   `    ``// Iterate over the List arr[]` `    ``for` `(``int` `i = ``0``; i < N; i++) ` `    ``{` `      `  `      ``// Traverse the string arr[i]` `      ``for` `(``int` `j = ``0``; j < arr.get(i).length(); j++)` `      ``{` `        `  `        ``// Push the pair of{i+1, j+1}` `        ``// in mp[arr[i][j]]` `        ``if` `(!mp.containsKey(arr.get(i).charAt(j))) {` `          ``mp.put(arr.get(i).charAt(j), ``new` `ArrayList<>());` `        ``}` `        ``mp.get(arr.get(i).charAt(j)).add(``new` `Pair<>(i + ``1``, j + ``1``));` `      ``}` `    ``}` `    `  `    ``// Print the occurrences of every` `    ``// character` `    ``for` `(Entry>> it : mp.entrySet()) {` `      ``System.out.print(``"Occurrences of: "` `+ it.getKey() + ``" = "``);` `      ``for` `(``int` `j = ``0``; j < it.getValue().size(); j++) {` `        ``System.out.print(``"["` `+ it.getValue().get(j).getKey() + ``" "` `+ it.getValue().get(j).getValue() + ``"] "``);` `      ``}` `      ``System.out.println();` `    ``}` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `main(String[] args) ` `  ``{` `    `  `    ``// Input` `    ``List arr = Arrays.asList(``"geeksforgeeks"``, ``"gfg"``);` `    ``int` `N = arr.size();` `    `  `    ``// Function call` `    ``printOccurrences(arr, N);` `  ``}` `}`   `// custom pair class` `class` `Pair {` `    ``private` `T key;` `    ``private` `U value;`   `    ``public` `Pair(T key, U value) {` `        ``this``.key = key;` `        ``this``.value = value;` `    ``}`   `    ``public` `T getKey() {` `        ``return` `key;` `    ``}`   `    ``public` `U getValue() {` `        ``return` `value;` `    ``}` `}`   `// This code is contributed by aadityapburujwale`

## Python3

 `# Python3 program for the above approach`   `# Function to print every occurrence` `# of every characters in every string` `def` `printOccurrences(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 occurrences of every` `    ``# character` `    ``for` `i ``in` `range``(``26``):` `        ``if` `len``(mp[i]) ``=``=` `0``:` `            ``continue` `        `  `        ``print``(``"Occurrences 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` `    ``printOccurrences(arr, N)`   `# This code is contributed by mohit kumar 29`

## C#

 `using` `System;` `using` `System.Collections.Generic;`   `class` `Program` `{` `    ``static` `void` `Main(``string``[] args)` `    ``{` `        ``string``[] arr = { ``"geeksforgeeks"``, ``"gfg"` `};` `        ``int` `N = arr.Length;`   `        ``Dictionary<``char``, List>> mp = ``new` `Dictionary<``char``, List>>();`   `      ``// Push the pair of{i+1, j+1}` `        ``// in mp[arr[i][j]]` `        ``for` `(``int` `i = 0; i < N; i++)` `        ``{` `            ``for` `(``int` `j = 0; j < arr[i].Length; j++)` `            ``{` `                ``if` `(!mp.ContainsKey(arr[i][j]))` `                ``{` `                    ``mp[arr[i][j]] = ``new` `List>();` `                ``}` `                ``mp[arr[i][j]].Add(Tuple.Create(i + 1, j + 1));` `            ``}` `        ``}`   `      ``// Print the occurrences of every` `    ``// character` `        ``foreach` `(``var` `item ``in` `mp)` `        ``{` `            ``Console.WriteLine(``"Occurrences of: "` `+ item.Key + ``" = "``);` `            ``foreach` `(``var` `pair ``in` `item.Value)` `            ``{` `                ``Console.Write(``"["` `+ pair.Item1 + ``" "` `+ pair.Item2 + ``"] "``);` `            ``}` `            ``Console.WriteLine();` `        ``}` `    ``}` `}`   `// This code is contributed by divyansh2212`

## Javascript

 ``

Output

```Occurrences of: e = [1 2] [1 3] [1 10] [1 11]
Occurrences of: f = [1 6] [2 2]
Occurrences of: g = [1 1] [1 9] [2 1] [2 3]
Occurrences of: k = [1 4] [1 12]
Occurrences of: o = [1 7]
Occurrences of: r = [1 8]
Occurrences 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)

My Personal Notes arrow_drop_up
Related Articles