Problem Description
In this 3 Palindrome, Given an input string word, split the string into exactly 3 palindromic substrings. Working from left to right, choose the smallest split for the first substring that still allows the remaining word to be split into 2 palindromes. Similarly, choose the smallest second palindromic substring that leaves a third palindromic substring.
If there is no way to split the word into exactly three palindromic substrings, print “Impossible” (without quotes). Every character of the string needs to be consumed.
Cases not allowed –
After finding 3 palindromes using above instructions, if any character of the original string remains unconsumed.
No character may be shared in forming 3 palindromes.
Constraints
1 <= the length of input sting <= 1000
Input
First line contains the input string consisting of characters between [a-z].
Output
Print 3 substrings one on each line.
Time Limit
1
Example 1
Input
nayannamantenet
Output
nayan
naman
tenet
Explanation
The original string can be split into 3 palindromes as mentioned in the output.
However, if the input was nayanamantenet, then the answer would be “Impossible”.
Example 2
Input
aaaaa
Output
a
a
aaa
Explanation
The other ways to split the given string into 3 palindromes are as follows –
[a, aaa, a], [aaa, a, a], [aa, aa, a], etc.
Since we want to minimize the length of the first palindromic substring using left to right processing, the correct way to split is [a, a, aaa].
Example 3
Input
aaaabaaaa
Output
a
aaabaaa
a
Explanation
The other ways to split the given string into 3 palindromes are as follows –
[aaaa, b, aaaa], [aa, aabaa, aa], etc.
Since we want to minimize the length of the first palindromic substring using left to right processing, the correct way to split is [a, aaabaaa, a].
Solution in C:
#include <stdio.h>
#define mod 1000000007
#define MAX_LENGTH 1000
typedef long long int lld;
int isPalindrome(char str[], int start, int end) {
while (start < end) {
if (str[start] != str[end]) {
return 0; // Not a palindrome
}
start++;
end--;
}
return 1; // Palindrome
}
int main() {
char s[MAX_LENGTH], s1[MAX_LENGTH], s2[MAX_LENGTH], s3[MAX_LENGTH];
scanf("%s", s);
int l = 0;
while (s[l] != '\0') {
l++;
}
for (int i = 1; i < l - 1; i++) {
for (int k = 0; k < i; k++) {
s1[k] = s[k];
}
s1[i] = '\0';
if (isPalindrome(s, 0, i - 1)) {
for (int j = 1; j < l - i; j++) {
for (int k = 0; k < j; k++) {
s2[k] = s[i + k];
}
s2[j] = '\0';
for (int k = 0; k < l - i - j; k++) {
s3[k] = s[i + j + k];
}
s3[l - i - j] = '\0';
if (isPalindrome(s2, 0, j - 1) && isPalindrome(s3, 0, l - i - j - 1)) {
printf("%s\n%s\n%s\n", s1, s2, s3);
return 0;
}
}
}
}
printf("Impossible\n");
return 0;
}
You can also run it on an online IDE:
https://ide.geeksforgeeks.org/online-python3-compiler/294545fa-8a51-46da-ab50-8fb862187677
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