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

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:
  • Iterate over the characters of the string arr[i] using variable j and in each iteration push the pair {i+1, j+1} in the vector mp[arr[i][j]].
• 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 <bits/stdc++.h>
using namespace std;

// Function to print every occurrence
// of every characters in every string
void printOccurrences(vector<string> arr, int N)
{
    map<char, vector<pair<int, int> > > 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<string> 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<String> arr, int N) {
    Map<Character, List<Pair<Integer, Integer>>> 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<Character, List<Pair<Integer, Integer>>> 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<String> arr = Arrays.asList("geeksforgeeks", "gfg");
    int N = arr.size();

    // Function call
    printOccurrences(arr, N);
  }
}

// custom pair class
class Pair<T, U> {
    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<Tuple<int, int>>> mp = new Dictionary<char, List<Tuple<int, int>>>();

      // 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<Tuple<int, int>>();
                }
                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

<script>

// JavaScript program for the above approach

// Function to print every occurrence
// of every characters in every string
function printOccurrences(arr, N) {
  let mp = new Map();

  // Iterate over the vector arr[]
  for (let i = 0; i < N; i++) {
    // Traverse the string arr[i]
    for (let j = 0; j < arr[i].length; j++) {
      // Push the pair of{i+1, j+1}
      // in mp[arr[i][j]]
      if (mp.has(arr[i][j])) {
        let temp = mp.get(arr[i][j]);
        temp.push([i + 1, j + 1]);

        mp.set(arr[i][j], temp);
      } else {
        mp.set(arr[i][j], [[i + 1, j + 1]]);
      }
    }
  }
  // Print the occurrences of every
  // character
  for (let it of new Map([...mp.entries()].sort())) {
    document.write("Occurrences of: " + it[0] + " = ");
    for (let j = 0; j < it[1].length; j++) {
      document.write(" [" + it[1][j][0] + " " + it[1][j][1] + "] ");
    }
    document.write("<br>");
  }
}

// Driver Code

// Input
let arr = ["geeksforgeeks", "gfg"];
let N = arr.length;
// Function call
printOccurrences(arr, N);

</script>

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)