Algorithm
problem Link : https://onlinejudge.org/index.php?option=onlinejudge&Itemid=8&page=show_problem&problem=1394
By definition palindrome is a string which is not changed when reversed. “MADAM” is a nice example of palindrome. It is an easy job to test whether a given string is a palindrome or not. But it may not be so easy to generate a palindrome. Here we will make a palindrome generator which will take an input string and return a palindrome. You can easily verify that for a string of length n, no more than (n−1) characters are required to make it a palindrome. Consider ‘abcd’ and its palindrome ‘abcdcba’ or ‘abc’ and its palindrome ‘abcba’. But life is not so easy for programmers!! We always want optimal cost. And you have to find the minimum number of characters required to make a given string to a palindrome if you are allowed to insert characters at any position of the string. Input Each input line consists only of lower case letters. The size of input string will be at most 1000. Input is terminated by EOF. Output For each input print the minimum number of characters and such a palindrome separated by one space in a line. There may be many such palindromes. Any one will be accepted. Sample Input abcd aaaa abc aab abababaabababa pqrsabcdpqrs
Sample Output 3 abcdcba 0 aaaa 2 abcba 1 baab 0 abababaabababa 9 pqrsabcdpqrqpdcbasrqp
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <bits/stdc++.h>
using namespace std;
int main()
{
string in;
while(getline(cin,in)){
vector < vector<int>> dp(in.length(),vector<int>(in.length(),1e6));
vector < vector<int>> decision(in.length(),vector<int>(in.length()));
for(int len=0;len<in.length();len++){
for(int left=0;left+len<in.length();left++){
int right = left+len;
if(in[left] == in[right]){
dp[left][right] = left+1>=right-1 ? 0 : dp[left+1][right-1];
} else {
int leftCost = dp[left][right-1]; // insert on left
int rightCost = dp[left+1][right]; // insert on right
if(leftCost > rightCost) {
dp[left][right] = rightCost+1;
decision[left][right] = 1;
} else {
dp[left][right] = leftCost+1;
decision[left][right] = 2;
}
}
}
}
int left=0,right=in.length()-1;
string leftStr="",rightStr="";
while(left<=right){
if(in[left] == in[right]){
if(left==right) leftStr += in[left];
else {
leftStr += in[left];
rightStr += in[left];
}
left++; right--;
} else if(decision[left][right] == 1){
// insert left character on right
rightStr += in[left];
leftStr += in[left];
left++;
} else if(decision[left][right] == 2){
// insert right character on left
leftStr += in[right];
rightStr += in[right];
right--;
}
}
reverse(rightStr.begin(),rightStr.end());
cout << dp[0][in.length(>-1] << " " << leftStr << rightStr << endl;
}
}
Input
aaaa
abc
aab
abababaabababa
pqrsabcdpqrs
Output
0 aaaa
2 abcba
1 baab
0 abababaabababa
9 pqrsabcdpqrqpdcbasrqp
Demonstration
UVA Online Judge solution - 10453-Make Palindrome - UVA Online Judge solution in C,C++,java