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# Count number of rotated strings which have more number of vowels in the first half than second half

• Difficulty Level : Medium
• Last Updated : 21 Apr, 2021

Given string str of even size N consisting of lowercase English alphabets. The task is to find the number of rotated strings of str which have more vowels in the first half than the second half.

Examples:

Input: str = “abcd”
Output:
Explanation:
All rotated string are “abcd”, “dabc”, “cdab”, “bcda”.
The first two rotated strings have more vowels in
the first half than the second half.

Input: str = “abecidft”
Output:
Explanation:
There are 4 possible strings with rotation where there are more vowels in first half than in the second half.

Approach: An efficient approach is to make string s = str + str then the size of the s will be 2 * N. Now, make a prefix array to store the number of vowels present from the 0th index to the ith index. Then run a loop from N – 1 to 2 * N – 2 to get all the rotated strings of str. Find the count of required rotated strings.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the count of rotated``// strings which have more number of vowels in``// the first half than the second half``int` `cntRotations(string s, ``int` `n)``{``    ``// Create a new string``    ``string str = s + s;` `    ``// Pre array to store count of all vowels``    ``int` `pre[2 * n] = { 0 };` `    ``// Compute the prefix array``    ``for` `(``int` `i = 0; i < 2 * n; i++) {``        ``if` `(i != 0)``            ``pre[i] += pre[i - 1];` `        ``if` `(str[i] == ``'a'` `|| str[i] == ``'e'``            ``|| str[i] == ``'i'` `|| str[i] == ``'o'``            ``|| str[i] == ``'u'``) {``            ``pre[i]++;``        ``}``    ``}` `    ``// To store the required answer``    ``int` `ans = 0;` `    ``// Find all rotated strings``    ``for` `(``int` `i = n - 1; i < 2 * n - 1; i++) {` `        ``// Right and left index of the string``        ``int` `r = i, l = i - n;` `        ``// x1 stores the number of vowels``        ``// in the rotated string``        ``int` `x1 = pre[r];``        ``if` `(l >= 0)``            ``x1 -= pre[l];``        ``r = i - n / 2;` `        ``// Left stores the number of vowels``        ``// in the first half of rotated string``        ``int` `left = pre[r];``        ``if` `(l >= 0)``            ``left -= pre[l];` `        ``// Right stores the number of vowels``        ``// in the second half of rotated string``        ``int` `right = x1 - left;` `        ``// If the count of vowels in the first half``        ``// is greater than the count in the second half``        ``if` `(left > right) {``            ``ans++;``        ``}``    ``}` `    ``// Return the required answer``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``string s = ``"abecidft"``;``    ``int` `n = s.length();` `    ``cout << cntRotations(s, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{` `// Function to return the count of rotated``// Strings which have more number of vowels in``// the first half than the second half``static` `int` `cntRotations(String s, ``int` `n)``{``    ``// Create a new String``    ``String str = s + s;` `    ``// Pre array to store count of all vowels``    ``int` `pre[]=``new` `int``[``2` `* n] ;` `    ``// Compute the prefix array``    ``for` `(``int` `i = ``0``; i < ``2` `* n; i++)``    ``{``        ``if` `(i != ``0``)``            ``pre[i] += pre[i - ``1``];` `        ``if` `(str.charAt(i) == ``'a'` `|| str.charAt(i) == ``'e'` `||``            ``str.charAt(i) == ``'i'` `|| str.charAt(i) == ``'o'` `||``            ``str.charAt(i) == ``'u'``)``        ``{``            ``pre[i]++;``        ``}``    ``}` `    ``// To store the required answer``    ``int` `ans = ``0``;` `    ``// Find all rotated Strings``    ``for` `(``int` `i = n - ``1``; i < ``2` `* n - ``1``; i++)``    ``{` `        ``// Right and left index of the String``        ``int` `r = i, l = i - n;` `        ``// x1 stores the number of vowels``        ``// in the rotated String``        ``int` `x1 = pre[r];``        ``if` `(l >= ``0``)``            ``x1 -= pre[l];``        ``r = i - n / ``2``;` `        ``// Left stores the number of vowels``        ``// in the first half of rotated String``        ``int` `left = pre[r];``        ``if` `(l >= ``0``)``            ``left -= pre[l];` `        ``// Right stores the number of vowels``        ``// in the second half of rotated String``        ``int` `right = x1 - left;` `        ``// If the count of vowels in the first half``        ``// is greater than the count in the second half``        ``if` `(left > right)``        ``{``            ``ans++;``        ``}``    ``}` `    ``// Return the required answer``    ``return` `ans;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``String s = ``"abecidft"``;``    ``int` `n = s.length();` `    ``System.out.println( cntRotations(s, n));``}``}` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 implementation of the approach` `# Function to return the count of rotated``# strings which have more number of vowels in``# the first half than the second half``def` `cntRotations(s, n):` `    ``# Create a new string``    ``str` `=` `s ``+` `s;` `    ``# Pre array to store count of all vowels``    ``pre ``=` `[``0``] ``*` `(``2` `*` `n);` `    ``# Compute the prefix array``    ``for` `i ``in` `range``(``2` `*` `n):``        ``if` `(i !``=` `0``):``            ``pre[i] ``+``=` `pre[i ``-` `1``];` `        ``if` `(``str``[i] ``=``=` `'a'` `or` `str``[i] ``=``=` `'e'` `or``            ``str``[i] ``=``=` `'i'` `or` `str``[i] ``=``=` `'o'` `or``            ``str``[i] ``=``=` `'u'``):``            ``pre[i] ``+``=` `1``;``        ` `    ``# To store the required answer``    ``ans ``=` `0``;` `    ``# Find all rotated strings``    ``for` `i ``in` `range``(n ``-` `1``, ``2` `*` `n ``-` `1``, ``1``):` `        ``# Right and left index of the string``        ``r ``=` `i; l ``=` `i ``-` `n;` `        ``# x1 stores the number of vowels``        ``# in the rotated string``        ``x1 ``=` `pre[r];``        ``if` `(l >``=` `0``):``            ``x1 ``-``=` `pre[l];``        ``r ``=` `(``int``)(i ``-` `n ``/` `2``);` `        ``# Left stores the number of vowels``        ``# in the first half of rotated string``        ``left ``=` `pre[r];``        ``if` `(l >``=` `0``):``            ``left ``-``=` `pre[l];` `        ``# Right stores the number of vowels``        ``# in the second half of rotated string``        ``right ``=` `x1 ``-` `left;` `        ``# If the count of vowels in the first half``        ``# is greater than the count in the second half``        ``if` `(left > right):``            ``ans ``+``=` `1``;``        ` `    ``# Return the required answer``    ``return` `ans;` `# Driver code``s ``=` `"abecidft"``;``n ``=` `len``(s);` `print``(cntRotations(s, n));` `# This code is contributed by Rajput-Ji`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `    ``// Function to return the count of rotated``    ``// Strings which have more number of vowels in``    ``// the first half than the second half``    ``static` `int` `cntRotations(``string` `s, ``int` `n)``    ``{``        ``// Create a new String``        ``string` `str = s + s;``    ` `        ``// Pre array to store count of all vowels``        ``int` `[]pre = ``new` `int``[2 * n];``    ` `        ``// Compute the prefix array``        ``for` `(``int` `i = 0; i < 2 * n; i++)``        ``{``            ``if` `(i != 0)``                ``pre[i] += pre[i - 1];``    ` `            ``if` `(str[i] == ``'a'` `|| str[i] == ``'e'` `||``                ``str[i] == ``'i'` `|| str[i] == ``'o'` `||``                ``str[i] == ``'u'``)``            ``{``                ``pre[i]++;``            ``}``        ``}``    ` `        ``// To store the required answer``        ``int` `ans = 0;``    ` `        ``// Find all rotated Strings``        ``for` `(``int` `i = n - 1; i < 2 * n - 1; i++)``        ``{``    ` `            ``// Right and left index of the String``            ``int` `r = i, l = i - n;``    ` `            ``// x1 stores the number of vowels``            ``// in the rotated String``            ``int` `x1 = pre[r];``            ``if` `(l >= 0)``                ``x1 -= pre[l];``            ``r = i - n / 2;``    ` `            ``// Left stores the number of vowels``            ``// in the first half of rotated String``            ``int` `left = pre[r];``            ``if` `(l >= 0)``                ``left -= pre[l];``    ` `            ``// Right stores the number of vowels``            ``// in the second half of rotated String``            ``int` `right = x1 - left;``    ` `            ``// If the count of vowels in the first half``            ``// is greater than the count in the second half``            ``if` `(left > right)``            ``{``                ``ans++;``            ``}``        ``}``    ` `        ``// Return the required answer``        ``return` `ans;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``String s = ``"abecidft"``;``        ``int` `n = s.Length;``    ` `        ``Console.WriteLine( cntRotations(s, n));``    ``}``}` `// This code is contributed by AnkitRai01`

## Javascript

 ``
Output
`4`

Time Complexity: O(2 * N)
Auxiliary Space: O(2 * N)

Efficient Approach: To reduce the Space complexity to constant for the above approach, store the number of vowels in both halves in two variables and iterate through all rotation by changing the index.

• In each rotation, the first element of the first-half gets removed and inserted into the second-half and if this element is a vowel then decrease the number of vowels in the first-half by 1 and increase the number of vowels in the second-half by 1.
• The first element of the second-half gets removed and inserted into the first-half and if this element is a vowel then increase the number of vowels in the first-half by 1 and decrease the number of vowels in the second-half by 1.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the count of``// rotated strings which have more``// number of vowels in the first``// half than the second half``int` `cntRotations(``char` `s[], ``int` `n)``{``    ``int` `lh = 0, rh = 0, i, ans = 0;` `    ``// Compute the number of``    ``// vowels in first-half``    ``for``(i = 0; i < n / 2; ++i)``        ``if` `(s[i] == ``'a'` `|| s[i] == ``'e'` `||``            ``s[i] == ``'i'` `|| s[i] == ``'o'` `||``            ``s[i] == ``'u'``)``        ``{``            ``lh++;``        ``}` `    ``// Compute the number of``    ``// vowels in second-half``    ``for``(i = n / 2; i < n; ++i)``        ``if` `(s[i] == ``'a'` `|| s[i] == ``'e'` `||``            ``s[i] == ``'i'` `|| s[i] == ``'o'` `||``            ``s[i] == ``'u'``)``        ``{``            ``rh++;``        ``}` `    ``// Check if first-half``    ``// has more vowels``    ``if` `(lh > rh)``        ``ans++;` `    ``// Check for all possible rotations``    ``for``(i = 1; i < n; ++i)``    ``{``        ``if` `(s[i - 1] == ``'a'` `|| s[i - 1] == ``'e'` `||``            ``s[i - 1] == ``'i'` `|| s[i - 1] == ``'o'` `||``            ``s[i - 1] == ``'u'``)``        ``{``            ``rh++;``            ``lh--;``        ``}``        ``if` `(s[(i - 1 + n / 2) % n] == ``'a'` `||``            ``s[(i - 1 + n / 2) % n] == ``'e'` `||``            ``s[(i - 1 + n / 2) % n] == ``'i'` `||``            ``s[(i - 1 + n / 2) % n] == ``'o'` `||``            ``s[(i - 1 + n / 2) % n] == ``'u'``)``        ``{``            ``rh--;``            ``lh++;``        ``}``        ``if` `(lh > rh)``            ``ans++;``    ``}` `    ``// Return the answer``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``char` `s[] = ``"abecidft"``;` `    ``int` `n = ``strlen``(s);` `    ``// Function call``    ``cout << ``" "` `<< cntRotations(s, n);` `    ``return` `0;``}` `// This code is contributed by shivanisinghss2110`

## C

 `// C implementation of the approach``#include ``#include ` `// Function to return the count of``// rotated strings which have more``// number of vowels in the first``// half than the second half``int` `cntRotations(``char` `s[], ``int` `n)``{``    ``int` `lh = 0, rh = 0, i, ans = 0;` `    ``// Compute the number of``    ``// vowels in first-half``    ``for` `(i = 0; i < n / 2; ++i)``        ``if` `(s[i] == ``'a'` `|| s[i] == ``'e'` `|| s[i] == ``'i'``            ``|| s[i] == ``'o'` `|| s[i] == ``'u'``)``        ``{``            ``lh++;``        ``}` `    ``// Compute the number of``    ``// vowels in second-half``    ``for` `(i = n / 2; i < n; ++i)``        ``if` `(s[i] == ``'a'` `|| s[i] == ``'e'` `|| s[i] == ``'i'``            ``|| s[i] == ``'o'` `|| s[i] == ``'u'``) {``            ``rh++;``        ``}` `    ``// Check if first-half``    ``// has more vowels``    ``if` `(lh > rh)``        ``ans++;` `    ``// Check for all possible rotations``    ``for` `(i = 1; i < n; ++i) {``        ``if` `(s[i - 1] == ``'a'` `|| s[i - 1] == ``'e'``            ``|| s[i - 1] == ``'i'` `|| s[i - 1] == ``'o'``            ``|| s[i - 1] == ``'u'``) {``            ``rh++;``            ``lh--;``        ``}``        ``if` `(s[(i - 1 + n / 2) % n] == ``'a'``            ``|| s[(i - 1 + n / 2) % n] == ``'e'``            ``|| s[(i - 1 + n / 2) % n] == ``'i'``            ``|| s[(i - 1 + n / 2) % n] == ``'o'``            ``|| s[(i - 1 + n / 2) % n] == ``'u'``) {``            ``rh--;``            ``lh++;``        ``}``        ``if` `(lh > rh)``            ``ans++;``    ``}` `    ``// Return the answer``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``char` `s[] = ``"abecidft"``;` `    ``int` `n = ``strlen``(s);` `    ``// Function call``    ``printf``(``"%d"``, cntRotations(s, n));` `    ``return` `0;``}`

## Java

 `// Java implementation of``// the approach``class` `GFG{``    ` `// Function to return the count of``// rotated strings which have more``// number of vowels in the first``// half than the second half``public` `static` `int` `cntRotations(``char` `s[],``                               ``int` `n)``{``  ``int` `lh = ``0``, rh = ``0``, i, ans = ``0``;` `  ``// Compute the number of``  ``// vowels in first-half``  ``for` `(i = ``0``; i < n / ``2``; ++i)``    ``if` `(s[i] == ``'a'` `|| s[i] == ``'e'` `||``        ``s[i] == ``'i'` `|| s[i] == ``'o'` `||``        ``s[i] == ``'u'``)``    ``{``      ``lh++;``    ``}` `  ``// Compute the number of``  ``// vowels in second-half``  ``for` `(i = n / ``2``; i < n; ++i)``    ``if` `(s[i] == ``'a'` `|| s[i] == ``'e'` `||``        ``s[i] == ``'i'` `|| s[i] == ``'o'` `||``        ``s[i] == ``'u'``)``    ``{``      ``rh++;``    ``}` `  ``// Check if first-half``  ``// has more vowels``  ``if` `(lh > rh)``    ``ans++;` `  ``// Check for all possible``  ``// rotations``  ``for` `(i = ``1``; i < n; ++i)``  ``{``    ``if` `(s[i - ``1``] == ``'a'` `|| s[i - ``1``] == ``'e'` `||``        ``s[i - ``1``] == ``'i'` `|| s[i - ``1``] == ``'o'` `||``        ``s[i - ``1``] == ``'u'``)``    ``{``      ``rh++;``      ``lh--;``    ``}``    ``if` `(s[(i - ``1` `+ n / ``2``) % n] == ``'a'` `||``        ``s[(i - ``1` `+ n / ``2``) % n] == ``'e'` `||``        ``s[(i - ``1` `+ n / ``2``) % n] == ``'i'` `||``        ``s[(i - ``1` `+ n / ``2``) % n] == ``'o'` `||``        ``s[(i - ``1` `+ n / ``2``) % n] == ``'u'``)``    ``{``      ``rh--;``      ``lh++;``    ``}``    ``if` `(lh > rh)``      ``ans++;``  ``}` `  ``// Return the answer``  ``return` `ans;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``  ``char` `s[] = {``'a'``,``'b'``,``'e'``,``'c'``,``              ``'i'``,``'d'``,``'f'``,``'t'``};``  ``int` `n = s.length;``  ` `  ``// Function call``  ``System.out.println(``         ``cntRotations(s, n));``}``}` `// This code is contributed by divyeshrabadiya07`

## Python3

 `# Python3 implementation of the approach` `# Function to return the count of``# rotated strings which have more``# number of vowels in the first``# half than the second half``def` `cntRotations(s, n):` `    ``lh, rh, ans ``=` `0``, ``0``, ``0` `    ``# Compute the number of``    ``# vowels in first-half``    ``for` `i ``in` `range` `(n ``/``/` `2``):``        ``if` `(s[i] ``=``=` `'a'` `or` `s[i] ``=``=` `'e'` `or``            ``s[i] ``=``=` `'i'` `or` `s[i] ``=``=` `'o'` `or``            ``s[i] ``=``=` `'u'``):``            ``lh ``+``=` `1` `    ``# Compute the number of``    ``# vowels in second-half``    ``for` `i ``in` `range` `(n ``/``/` `2``, n):``        ``if` `(s[i] ``=``=` `'a'` `or` `s[i] ``=``=` `'e'` `or``            ``s[i] ``=``=` `'i'` `or` `s[i] ``=``=` `'o'` `or``            ``s[i] ``=``=` `'u'``):``            ``rh ``+``=` `1` `    ``# Check if first-half``    ``# has more vowels``    ``if` `(lh > rh):``        ``ans ``+``=` `1` `    ``# Check for all possible rotations``    ``for` `i ``in` `range` `(``1``, n):``        ``if` `(s[i ``-` `1``] ``=``=` `'a'` `or` `s[i ``-` `1``] ``=``=` `'e'` `or``            ``s[i ``-` `1``] ``=``=` `'i'` `or` `s[i ``-` `1``] ``=``=` `'o'` `or``            ``s[i ``-` `1``] ``=``=` `'u'``):``            ``rh ``+``=` `1``            ``lh ``-``=` `1``        ` `        ``if` `(s[(i ``-` `1` `+` `n ``/``/` `2``) ``%` `n] ``=``=` `'a'` `or``            ``s[(i ``-` `1` `+` `n ``/``/` `2``) ``%` `n] ``=``=` `'e'` `or``            ``s[(i ``-` `1` `+` `n ``/``/` `2``) ``%` `n] ``=``=` `'i'` `or``            ``s[(i ``-` `1` `+` `n ``/``/` `2``) ``%` `n] ``=``=` `'o'` `or``            ``s[(i ``-` `1` `+` `n ``/``/` `2``) ``%` `n] ``=``=` `'u'``):``            ``rh ``-``=` `1``            ``lh ``+``=` `1``        ` `        ``if` `(lh > rh):``            ``ans ``+``=` `1``   ` `    ``# Return the answer``    ``return` `ans` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``  ` `    ``s ``=` `"abecidft"``    ``n ``=` `len``(s)` `    ``# Function call``    ``print``(cntRotations(s, n))` `# This code is contributed by Chitranayal`

## C#

 `// C# implementation of``// the approach``using` `System;``class` `GFG``{``    ` `    ``// Function to return the count of``    ``// rotated strings which have more``    ``// number of vowels in the first``    ``// half than the second half``    ``static` `int` `cntRotations(``char``[] s, ``int` `n)``    ``{``      ``int` `lh = 0, rh = 0, i, ans = 0;``     ` `      ``// Compute the number of``      ``// vowels in first-half``      ``for` `(i = 0; i < n / 2; ++i)``        ``if` `(s[i] == ``'a'` `|| s[i] == ``'e'` `||``            ``s[i] == ``'i'` `|| s[i] == ``'o'` `||``            ``s[i] == ``'u'``)``        ``{``          ``lh++;``        ``}``     ` `      ``// Compute the number of``      ``// vowels in second-half``      ``for` `(i = n / 2; i < n; ++i)``        ``if` `(s[i] == ``'a'` `|| s[i] == ``'e'` `||``            ``s[i] == ``'i'` `|| s[i] == ``'o'` `||``            ``s[i] == ``'u'``)``        ``{``          ``rh++;``        ``}``     ` `      ``// Check if first-half``      ``// has more vowels``      ``if` `(lh > rh)``        ``ans++;``     ` `      ``// Check for all possible``      ``// rotations``      ``for` `(i = 1; i < n; ++i)``      ``{``        ``if` `(s[i - 1] == ``'a'` `|| s[i - 1] == ``'e'` `||``            ``s[i - 1] == ``'i'` `|| s[i - 1] == ``'o'` `||``            ``s[i - 1] == ``'u'``)``        ``{``          ``rh++;``          ``lh--;``        ``}``        ``if` `(s[(i - 1 + n / 2) % n] == ``'a'` `||``            ``s[(i - 1 + n / 2) % n] == ``'e'` `||``            ``s[(i - 1 + n / 2) % n] == ``'i'` `||``            ``s[(i - 1 + n / 2) % n] == ``'o'` `||``            ``s[(i - 1 + n / 2) % n] == ``'u'``)``        ``{``          ``rh--;``          ``lh++;``        ``}``        ``if` `(lh > rh)``          ``ans++;``      ``}``     ` `      ``// Return the answer``      ``return` `ans;``    ``}``  ` `  ``// Driver code    ``  ``static` `void` `Main()``  ``{``      ``char``[] s = {``'a'``,``'b'``,``'e'``,``'c'``,``              ``'i'``,``'d'``,``'f'``,``'t'``};``      ``int` `n = s.Length;``       ` `      ``// Function call``      ``Console.WriteLine(cntRotations(s, n));``  ``}``}` `// This code is contributed by divyesh072019`

## Javascript

 ``
Output
`4`

Time Complexity: O(n)
Space Complexity: O(1)

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