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:
- String with k distinct characters and no same characters adjacent
- Maximum number of characters between any two same character in a string
- String with maximum number of unique characters
- Rearrange characters in a string such that no two adjacent are same
- Minimum replacements to make adjacent characters unequal in a ternary string | Set-2
- Minimum replacements to make adjacent characters unequal in a ternary string
- Min flips of continuous characters to make all characters same in a string
- Minimum number of operations to move all uppercase characters before all lower case characters
- Replace minimal number of characters to make all characters pair wise distinct
- Length of the smallest sub-string consisting of maximum distinct characters
- Maximum length of balanced string after swapping and removal of characters
- Minimize number of unique characters in string
- Count of strings where adjacent characters are of difference one
- Number of ways to remove a sub-string from S such that all remaining characters are same
- Sub-strings having exactly k characters that have ASCII value greater than p
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