# 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 the strings have equal number of pairs.
2. The first vowels of both the strings are same.
3. The last vowel of both the strings are 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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, 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, ` `                           ``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; ` `} `

## 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 `

Output:

```1
```

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Improved By : mohit kumar 29