Computer Science, asked by rawatprahalad222, 9 months ago

write a program in c++ to print the sorted list of all palindromes that can be constructed from the alphabets of the given palindrome

Answers

Answered by sharu848
0
#include
using namespace std;
const char MAX_CHAR = 26;
void countFreq(char str[], int freq[], int n){
for (int i = 0; i < n; i++)
freq[str[i] - 'a']++;
}
bool canMakePalindrome(int freq[], int n){
int count_odd = 0;
for (int i = 0; i < 26; i++)
if (freq[i] % 2 != 0)
count_odd++;
if (n % 2 == 0) {
if (count_odd > 0)
return false;
else
return true;
}
if (count_odd != 1)
return false;
return true;
}
bool isPalimdrome(char str[], int n){
int freq[26] = { 0 };
countFreq(str, freq, n);
if (!canMakePalindrome(freq, n))
return false;
char odd_char;
for (int i = 0; i < 26; i++) {
if (freq[i] % 2 != 0) {
freq[i]--;
odd_char = (char)(i + 'a');
break;
}
}
int front_index = 0, rear_index = n - 1;
for (int i = 0; i < 26; i++) {
if (freq[i] != 0) {
char ch = (char)(i + 'a');
for (int j = 1; j <= freq[i] / 2; j++) {
str[front_index++] = ch;
str[rear_index--] = ch;
}
}
}
if (front_index == rear_index)
str[front_index] = odd_char;
return true;
}
void reverse(char str[], int i, int j){
while (i < j) {
swap(str[i], str[j]);
i++;
j--;
}
}
bool nextPalindrome(char str[], int n){
if (n <= 3)
return false;
int mid = n / 2 - 1;
int i, j;
for (i = mid - 1; i >= 0; i--)
if (str[i] < str[i + 1])
break;
if (i < 0)
return false;
int smallest = i + 1;
for (j = i + 2; j <= mid; j++)
if (str[j] > str[i] && str[j] < str[smallest])
smallest = j;
swap(str[i], str[smallest]);
swap(str[n - i - 1], str[n - smallest - 1]);
reverse(str, i + 1, mid);
if (n % 2 == 0)
reverse(str, mid + 1, n - i - 2);
else
reverse(str, mid + 2, n - i - 2);
return true;
}
void printAllPalindromes(char str[], int n){
if (!(isPalimdrome(str, n))) {
cout<<"-1";
return;
}
do {
cout< } while (nextPalindrome(str, n));
}
int main(){
char str[] = "abccba";
int n = strlen(str);
cout<<”The list of palindromes possible is :\n”;
printAllPalindromes(str, n);
return 0;
}
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