Count pairs of strings that satisfy the given conditions

Given an array arr[] of N strings consisting of lowercase characters, the task is to count the pairs in the array which satisfy the given conditions:

1. Both strings have an equal number of pairs.
2. The first vowels of both the strings are same.
3. The last vowel of both strings is the same.

Note that a string can only be used in a single pair. Examples:

Input: arr[] = {“geeks”, “for”, “geeks”, “geek”} Output: 1 The only valid pair is (“geeks”, “geeks”). “geek” could also be paired with “geeks” but both the “geeks” have already been paired. Input: arr[] = {“code”, “shoot”, “mode”} Output: 1

Approach: We will store all the vowels that appear in a word for each word and we make a tuple of first vowel, last vowel and the total count of vowels and store corresponding index regarding that tuple using map. At last, we will go through the map and count number of pairs that can be formed using the tuple values stored in the map. Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function that returns true if c is vowel
bool is_vowel(char c)
{
    return (c == 'a' || c == 'e' || c == 'i'
            || c == 'o' || c == 'u');
}

// Function to return the count of required pairs
int count(string s[], int n)
{

    map<tuple<char, char, int>, vector<int> > map;

    // For every string of the array
    for (int i = 0; i < n; i++) {

        // Vector to store the vowels
        // of the current string
        vector<char> vowel;
        for (int j = 0; j < s[i].size(); j++) {

            // If current character is a vowel
            if (is_vowel(s[i][j]))
                vowel.push_back(s[i][j]);
        }

        // If current string contains vowels
        if (vowel.size() > 0) {
            int len = vowel.size();

            // Create tuple (first vowel,
            // last vowel, total vowels)
            map[make_tuple(vowel[0],
                           vowel[len - 1], len)]
                .push_back(i);
        }
    }

    int count = 0;
    for (auto i : map) {

        // v stores the indices for which
        // the given condition satisfies
        // Total valid pairs will be half the size
        vector<int> v = i.second;
        count += v.size() / 2;
    }

    return count;
}

// Driver code
int main()
{
    string s[] = { "geeks", "for", "geeks" };
    int n = sizeof(s) / sizeof(string);

    cout << count(s, n);

    return 0;
}

Java

import java.util.*;

class GFG
{

  // Returns true when it is vowel
  static boolean is_vowel(char c)
  {
    return (c == 'a' || c == 'e' || c == 'i' || c == 'o'
            || c == 'u');
  }

  // Function to return the count of required pairs
  static int count(String s[], int n)
  {
    Map<String, List<Integer> > map = new HashMap<>();

    for (int i = 0; i < n; i++) {
      List<Character> vowels = new ArrayList<>();
      for (int j = 0; j < s[i].length(); j++) {
        if (is_vowel(s[i].charAt(j))) {
          vowels.add(s[i].charAt(j));
        }
      }

      if (vowels.size() > 0) {
        int len = vowels.size();
        String key = vowels.get(0) + ""
          + vowels.get(len - 1) + ""
          + len;
        if (!map.containsKey(key)) {
          map.put(key, new ArrayList<Integer>());
        }
        map.get(key).add(i);
      }
    }

    int count = 0;
    for (List<Integer> value : map.values())
    {

      // value stores the indices for which
      // the given condition satisfies
      // Total valid pairs will be half the size
      count += value.size() / 2;
    }

    return count;
  }

  public static void main(String[] args)
  {
    String s[] = { "geeks", "for", "geeks" };
    int n = s.length;

    System.out.println(count(s, n));
  }
}

// This code is contributed by abn95knd1.

Python3

# Python3 implementation of the approach

# Function that returns true if c is vowel
def is_vowel(c):
    return (c == 'a' or c == 'e' or c == 'i'
            or c == 'o' or c == 'u')


# Function to return the count of required pairs
def count(s, n):


    map=dict()

    # For every of the array
    for i in range(n):

        # Vector to store the vowels
        # of the current string
        vowel=[]
        for j in range(len(s[i])):

            # If current character is a vowel
            if (is_vowel(s[i][j])):
                vowel.append(s[i][j])


        # If current contains vowels
        if (len(vowel) > 0):
            Len = len(vowel)

            # Create tuple (first vowel,
            # last vowel, total vowels)
            if (vowel[0],vowel[Len - 1], Len) in map.keys():
                map[(vowel[0],vowel[Len - 1], Len)].append(i)
            else:
                map[(vowel[0],vowel[Len - 1], Len)]=[i,]


    count = 0
    for i in map:

        # v stores the indices for which
        # the given condition satisfies
        # Total valid pairs will be half the size
        v = map[i]
        count += len(v)// 2


    return count

# Driver code
s = ["geeks", "for", "geeks"]
n = len(s)

print(count(s, n))

# This code is contributed by mohit kumar 29

Javascript

// Returns true when it is vowel
function is_vowel(c) {
  return (c === 'a' || c === 'e' || c === 'i' || c === 'o' || c === 'u');
}

// Function to return the count of required pairs
function count(s, n) {
  const map = new Map();

  for (let i = 0; i < n; i++) {
    const vowels = [];
    for (let j = 0; j < s[i].length; j++) {
      if (is_vowel(s[i].charAt(j))) {
        vowels.push(s[i].charAt(j));
      }
    }

    if (vowels.length > 0) {
      const len = vowels.length;
      const key = vowels[0] + "" + vowels[len - 1] + "" + len;
      if (!map.has(key)) {
        map.set(key, []);
      }
      map.get(key).push(i);
    }
  }

  let count = 0;
  for (const value of map.values()) {

    // value stores the indices for which
    // the given condition satisfies
    // Total valid pairs will be half the size
    count += Math.floor(value.length / 2);
  }

  return count;
}

const s = ["geeks", "for", "geeks"];
const n = s.length;

console.log(count(s, n));

C#

using System;
using System.Collections.Generic;

class GFG
{
// Returns true when it is vowel
static bool IsVowel(char c)
{
return (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u');
}
// Function to return the count of required pairs
static int Count(string[] s, int n)
{
    Dictionary<string, List<int>> map = new Dictionary<string, List<int>>();

    for (int i = 0; i < n; i++)
    {
        List<char> vowels = new List<char>();
        for (int j = 0; j < s[i].Length; j++)
        {
            if (IsVowel(s[i][j]))
            {
                vowels.Add(s[i][j]);
            }
        }

        if (vowels.Count > 0)
        {
            int len = vowels.Count;
            string key = vowels[0] + "" + vowels[len - 1] + "" + len;
            if (!map.ContainsKey(key))
            {
                map.Add(key, new List<int>());
            }
            map[key].Add(i);
        }
    }

    int count = 0;
    foreach (List<int> value in map.Values)
    {
        // value stores the indices for which
        // the given condition satisfies
        // Total valid pairs will be half the size
        count += value.Count / 2;
    }

    return count;
}

static void Main(string[] args)
{
    string[] s = { "geeks", "for", "geeks" };
    int n = s.Length;

    Console.WriteLine(Count(s, n));
}
}

Output

1

Time Complexity: O(n * l), where n is the length of str array and l is the maximum length of a string in the array.
Auxiliary Space: O(n), where n is the length of the given string.