Given a string str, the task is to print the maximum count of characters which are greater than both its left and right character in any permutation of the string.
Input: str = “abc”
Permutations of the string with the count of maximal character in each string are:
abc – 0
acb – 1 Here a < c > b
bac – 0
bca – 1 Here b < c > a
cab – 0
cba – 0
Input: str = “geeks”
The string will be “egesk”
- If the string’ length is less than 3 then the answer will be 0 because no permutation is possible which satisfies the given condition.
- If the length of the given string is greater than or equal to 3 then assume that in the resulting string every other character is maximal character, that is there is exactly one character between any two consecutive maximal characters (otherwise we can remove all but the lowest one and add them to the end of the string).
- Assume for simplicity that this number is odd. Then, ideally, the string can have maximal characters in even positions i.e.at most (n-1)/2, where n is the length of the given string while the rest of the remaining characters in odd positions.
- First arrange all the characters in ascending order, place the first half characters at odd positions, and then fill the remaining even positions with the rest of the characters.
In this way, all the characters in even positions will be those characters from which character at the left and the right position are smaller if there is no frequency of smaller character that is too high, start placing a character from an odd position, continue with the same character to even positions, and eventually reach a position next to the odd position from which we started to place the character. Here if the frequency of some smaller character in the string is too high then the count of maximal character will always be less than (n-1)/2.
- Calculate the frequency of each character in the given string.
- Check the character which has the maximum frequency.
- If the maximum frequency element is the smallest element in the given string then mark the flag as 0 otherwise mark the value of flag equal to 1.
- The answer will the minimum of ((n – 1) / 2, n – max_freq – flag).
Below is the implementation of the above approach:
- Minimum number of adjacent swaps required to convert a permutation to another permutation by given condition
- Count of Array elements greater than all elements on its left and at least K elements on its right
- Count of Array elements greater than all elements on its left and next K elements on its right
- Smallest subarray of size greater than K with sum greater than a given value
- Largest number with maximum trailing nines which is less than N and greater than N-D
- Find a string such that every character is lexicographically greater than its immediate next character
- Find the minimum permutation of A greater than B
- Count numbers with difference between number and its digit sum greater than specific value
- Largest number not greater than N which can become prime after rearranging its digits
- Maximize the number of indices such that element is greater than element to its left
- Minimum string such that every adjacent character of given string is still adjacent
- Count of array elements which is smaller than both its adjacent elements
- Count the number of words having sum of ASCII values less than and greater than k
- Largest number less than N with digit sum greater than the digit sum of N
- Count of array elements which are greater than all elements on its left
- Reduce the array by deleting elements which are greater than all elements to its left
- Minimum number of given operations required to convert a permutation into an identity permutation
- Minimum number of adjacent swaps to convert a string into its given anagram
- String with k distinct characters and no same characters adjacent
- Rearrange the characters of the string such that no two adjacent characters are consecutive English alphabets
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