Problem:
Find the length and/or string of the longest palindrome of a string.
The problem appeared as an Interview question in Amazon, Microsoft.
Solution:
One important property of a palindrome is if we remove the first and last characters of it , it would be another palindrome.
So A string S[a..b] will be a palindrome if and only if S[(a+1)…(b-1)] is a palindrome.
We can use DP to solve the problem.
- Keep a 2D array of ARRAY [String.SIZE] [String.SIZE] for keeping the length of palindrome found from the string.
- Each char of the string is 1 length palindrome sub string. So ARRAY[i][i] = 1 , here i=0….String.SIZE
- Now find the length of possible palindrome for 2 char length strings starting from each index of the string S.
- similarly find palindrome for 3,4,…..char length string of the provided String S.
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#include <stdio.h> #include <string.h> #include <string> #include <iostream> using namespace std; int _max(int a,int b) { if(a<=b) return b; return a; } int findPalindrome(string s,int M,int *st,int *ed) { int arr[M][M]; int j; int max_len=1; (*st)=0; (*ed)=0; for(int i=0;i<M;i++) for(int j=0;j<M;j++) arr[i][j]=0; for(int i=0;i<M;i++) { arr[i][i]=1; } for(int len=2;len<=M;len++) { for(int i=0;(i+len-1)<M;i++) { j=i+len-1; if(s[i]==s[j] && len==2) { arr[i][j]=2; if(max_len<arr[i][j]) { max_len = arr[i][j]; (*st)=i; (*ed)=j; } } if(s[i]==s[j] && len>2 && arr[i+1][j-1]!=0) { arr[i][j]=arr[i+1][j-1]+2; if(max_len<arr[i][j]) { max_len = arr[i][j]; (*st)=i; (*ed)=j; } //cout<<s[i]<<s[j]<<" "<<i<<","<<j<<"\n"; } /*if(s[i]!=s[j]) { arr[i][j]=_max(arr[i+1][j],arr[i][j-1]); }*/ } } return max_len;//arr[0][M-1]; } int main() { int t,n,M; string s; scanf("%d",&t); while(t--) { cin>>s; M=s.size(); int st,ed; //cout<< int length = findPalindrome(s,M,&st,&ed);//<<endl; //cout<<length<<"\n"; for(int i=st;i<=ed;i++)cout<<s[i];cout<<endl; } return 0; } |