## Sunday, April 23, 2017

### Lexicographically first palindromic string - GeeksforGeeks

Lexicographically first palindromic string - GeeksforGeeks
Rearrange the characters of the given string to form a lexicographically first palindromic string. If no such string exists display message "no palindromic string".

Properties for palindromic string:
1. If length of string is even, then the frequency of each character in the string must be even.
2. If the length is odd then there should be one character whose frequency is odd and all other chars must have even frequency and at-least one occurrence of the odd character must be present in the middle of the string.
Algorithm
1. Store frequency of each character in the given string
2. Check whether a palindromic string can be formed or not using the properties of palindromic string mentioned above.
3. If palindromic string cannot be formed, return “No Palindromic String”.
4. Else we create three strings and then return front_str + odd_str + rear_str.
• odd_str : It is empty if there is no character with odd frequency. Else it contains all occurrences of odd character.
• front_str : Contains half occurrences of all even occurring characters of string in increasing order.
• rear_str Contains half occurrences of all even occurring characters of string in reverse order of front_str.
`void` `countFreq(string str, ``int` `freq[], ``int` `len)`
`{`
`    ``for` `(``int` `i=0; i<len; i++)`
`        ``freq[str.at(i) - ``'a'``]++;`
`}`

`// Cases to check whether a palindr0mic`
`// string can be formed or not`
`bool` `canMakePalindrome(``int` `freq[], ``int` `len)`
`{`
`    ``// count_odd to count no of`
`    ``// chars with odd frequency`
`    ``int` `count_odd = 0;`
`    ``for` `(``int` `i=0; i<MAX_CHAR; i++)`
`        ``if` `(freq[i]%2 != 0)`
`            ``count_odd++;`

`    ``// For even length string`
`    ``// no odd freq character`
`    ``if` `(len%2 == 0)`
`    ``{`
`        ``if` `(count_odd > 0)`
`            ``return` `false``;`
`        ``else`
`            ``return` `true``;`
`    ``}`

`    ``// For odd length string`
`    ``// one odd freq character`
`    ``if` `(count_odd != 1)`
`        ``return` `false``;`

`    ``return` `true``;`
`}`

`// Function to find odd freq char and`
`// reducing its freq by 1returns "" if odd freq`
`// char is not present`
`string findOddAndRemoveItsFreq(``int` `freq[])`
`{`
`    ``string odd_str = ``""``;`
`    ``for` `(``int` `i=0; i<MAX_CHAR; i++)`
`    ``{`
`        ``if` `(freq[i]%2 != 0)`
`        ``{`
`            ``freq[i]--;`
`            ``odd_str = odd_str + (``char``)(i+``'a'``);`
`            ``return` `odd_str;`
`        ``}`
`    ``}`
`    ``return` `odd_str;`
`}`

`// To find lexicographically first palindromic`
`// string.`
`string findPalindromicString(string str)`
`{`
`    ``int` `len = str.length();`

`    ``int` `freq[MAX_CHAR] = {0};`
`    ``countFreq(str, freq, len);`

`    ``if` `(!canMakePalindrome(freq, len))`
`        ``return` `"No Palindromic String"``;`

`    ``// Assigning odd freq character if present`
`    ``// else empty string.`
`    ``string odd_str = findOddAndRemoveItsFreq(freq);`

`    ``string front_str = ``""``, rear_str = ``" "``;`

`    ``// Traverse characters in increasing order`
`    ``for` `(``int` `i=0; i<MAX_CHAR; i++)`
`    ``{`
`        ``string temp = ``""``;`
`        ``if` `(freq[i] != 0)`
`        ``{`
`            ``char` `ch = (``char``)(i + ``'a'``);`

`            ``// Divide all occurrences into two`
`            ``// halves. Note that odd character`
`            ``// is removed by findOddAndRemoveItsFreq()`
`            ``for` `(``int` `j=1; j<=freq[i]/2; j++)`
`                ``temp = temp + ch;`

`            ``// creating front string`
`            ``front_str = front_str + temp;`

`            ``// creating rear string`
`            ``rear_str = temp + rear_str;`
`        ``}`
`    ``}`

`    ``// Final palindromic string which is`
`    ``// lexicographically first`
`    ``return` `(front_str + odd_str + rear_str);`
`}`