Shortest Palindromic Substring
Last Updated :
29 Dec, 2022
Given a string you need to find the shortest palindromic substring of the string. If there are multiple answers output the lexicographically smallest.
Examples:
Input: zyzz
Output:y
Input: abab
Output: a
Naive Approach:
- The approach is similar to finding the longest palindromic substring. We keep track of even and odd lengths substring and keep storing it in a vector.
- After that, we will sort the vector and print the lexicographically smallest substring. This may also include empty substrings but we need to ignore them.
Below is the implementation of the above approach:
C++
// C++ program to find the shortest
// palindromic substring
#include <bits/stdc++.h>
using namespace std;
// Function return the shortest
// palindromic substring
string ShortestPalindrome(string s)
{
int n = s.length();
vector<string> v;
// One by one consider every character
// as center point of even and length
// palindromes
for (int i = 0; i < n; i++)
{
int l = i;
int r = i;
string ans1 = "";
string ans2 = "";
// Find the longest odd length palindrome
// with center point as i
while (l >= 0 && r < n && s[l] == s[r])
{
ans1 += s[l];
l--;
r++;
}
l = i - 1;
r = i;
// Find the even length palindrome
// with center points as i-1 and i.
while (l >= 0 && r < n && s[l] == s[r])
{
ans2 += s[l];
l--;
r++;
}
v.push_back(ans1);
v.push_back(ans2);
}
string ans = v[0];
// Smallest substring which is
// not empty
for (int i = 0; i < v.size(); i++)
{
if (v[i] != "")
{
ans = min(ans, v[i]);
}
}
return ans;
}
// Driver code
int main()
{
string s = "geeksforgeeks";
cout << ShortestPalindrome(s);
return 0;
}
// Java program to find the shortest
// palindromic substring
import java.util.*;
import java.io.*;
class GFG{
// Function return the shortest
// palindromic substring
public static String ShortestPalindrome(String s)
{
int n = s.length();
Vector<String> v = new Vector<String>();
// One by one consider every character
// as center point of even and length
// palindromes
for(int i = 0; i < n; i++)
{
int l = i;
int r = i;
String ans1 = "";
String ans2 = "";
// Find the longest odd length palindrome
// with center point as i
while (l >= 0 && r < n &&
s.charAt(l) == s.charAt(r))
{
ans1 += s.charAt(l);
l--;
r++;
}
l = i - 1;
r = i;
// Find the even length palindrome
// with center points as i-1 and i.
while (l >= 0 && r < n &&
s.charAt(l) == s.charAt(r))
{
ans2 += s.charAt(l);
l--;
r++;
}
v.add(ans1);
v.add(ans2);
}
String ans = v.get(0);
// Smallest substring which is
// not empty
for(int i = 0; i < v.size(); i++)
{
if (v.get(i) != "")
{
if (ans.charAt(0) >=
v.get(i).charAt(0))
{
ans = v.get(i);
}
}
}
return ans;
}
// Driver code
public static void main(String []args)
{
String s = "geeksforgeeks";
System.out.println(ShortestPalindrome(s));
}
}
// This code is contributed by rag2127
# Python3 program to find the shortest
# palindromic substring
# Function return the shortest
# palindromic substring
def ShortestPalindrome(s) :
n = len(s)
v = []
# One by one consider every character
# as center point of even and length
# palindromes
for i in range(n) :
l = i
r = i
ans1 = ""
ans2 = ""
# Find the longest odd length palindrome
# with center point as i
while ((l >= 0) and (r < n) and (s[l] == s[r])) :
ans1 += s[l]
l -= 1
r += 1
l = i - 1
r = i
# Find the even length palindrome
# with center points as i-1 and i.
while ((l >= 0) and (r < n) and (s[l] == s[r])) :
ans2 += s[l]
l -= 1
r += 1
v.append(ans1)
v.append(ans2)
ans = v[0]
# Smallest substring which is
# not empty
for i in range(len(v)) :
if (v[i] != "") :
ans = min(ans, v[i])
return ans
s = "geeksforgeeks"
print(ShortestPalindrome(s))
# This code is contributed by divyesh072019
// C# program to find the shortest
// palindromic substring
using System;
using System.Collections.Generic;
class GFG
{
// Function return the shortest
// palindromic substring
static string ShortestPalindrome(string s)
{
int n = s.Length;
List<string> v = new List<string>();
// One by one consider every character
// as center point of even and length
// palindromes
for(int i = 0; i < n; i++)
{
int l = i;
int r = i;
string ans1 = "";
string ans2 = "";
// Find the longest odd length palindrome
// with center point as i
while(l >= 0 && r < n && s[l] == s[r])
{
ans1 += s[l];
l--;
r++;
}
l = i - 1;
r = i;
// Find the even length palindrome
// with center points as i-1 and i.
while(l >= 0 && r < n && s[l] == s[r])
{
ans2 += s[l];
l--;
r++;
}
v.Add(ans1);
v.Add(ans2);
}
string ans = v[0];
// Smallest substring which is
// not empty
for(int i = 0; i < v.Count; i++)
{
if(v[i] != "")
{
if(ans[0] >= v[i][0])
{
ans = v[i];
}
}
}
return ans;
}
// Driver code
static public void Main ()
{
string s = "geeksforgeeks";
Console.WriteLine(ShortestPalindrome(s));
}
}
// This code is contributed by avanitrachhadiya2155
<script>
// Javascript program to find the shortest
// palindromic substring
// Function return the shortest
// palindromic substring
function ShortestPalindrome(s)
{
let n = s.length;
let v = [];
// One by one consider every character
// as center point of even and length
// palindromes
for(let i = 0; i < n; i++)
{
let l = i;
let r = i;
let ans1 = "";
let ans2 = "";
// Find the longest odd length palindrome
// with center point as i
while(l >= 0 && r < n && s[l] == s[r])
{
ans1 += s[l];
l--;
r++;
}
l = i - 1;
r = i;
// Find the even length palindrome
// with center points as i-1 and i.
while(l >= 0 && r < n && s[l] == s[r])
{
ans2 += s[l];
l--;
r++;
}
v.push(ans1);
v.push(ans2);
}
let ans = v[0];
// Smallest substring which is
// not empty
for(let i = 0; i < v.length; i++)
{
if(v[i] != "")
{
if(ans[0] >= v[i][0])
{
ans = v[i];
}
}
}
return ans;
}
let s = "geeksforgeeks";
document.write(ShortestPalindrome(s));
// This code is contributed by decode2207.
</script>
Time Complexity: O(N^2), where N is the length of the string.
Auxiliary Space: O(N^2)
Efficient Approach: An observation here is that a single character is also a palindrome. So, we just need to print the lexicographically smallest character present in the string.
Below is the implementation of the above approach:
C++
// C++ program to find the shortest
// palindromic substring
#include <bits/stdc++.h>
using namespace std;
// Function return the shortest
// palindromic substring
char ShortestPalindrome(string s)
{
int n = s.length();
char ans = s[0];
// Finding the smallest character
// present in the string
for(int i = 1; i < n ; i++)
{
ans = min(ans, s[i]);
}
return ans;
}
// Driver code
int main()
{
string s = "geeksforgeeks";
cout << ShortestPalindrome(s);
return 0;
}
// Java program to find the shortest
// palindromic subString
class GFG{
// Function return the shortest
// palindromic subString
static char ShortestPalindrome(String s)
{
int n = s.length();
char ans = s.charAt(0);
// Finding the smallest character
// present in the String
for(int i = 1; i < n; i++)
{
ans = (char) Math.min(ans, s.charAt(i));
}
return ans;
}
// Driver code
public static void main(String[] args)
{
String s = "geeksforgeeks";
System.out.print(ShortestPalindrome(s));
}
}
// This code is contributed by Rajput-Ji
# Python3 program to find the shortest
# palindromic substring
# Function return the shortest
# palindromic substring
def ShortestPalindrome(s):
n = len(s)
ans = s[0]
# Finding the smallest character
# present in the string
for i in range(1, n):
ans = min(ans, s[i])
return ans
# Driver code
s = "geeksforgeeks"
print(ShortestPalindrome(s))
# This code is contributed by divyeshrabadiya07
// C# program to find the shortest
// palindromic subString
using System;
class GFG{
// Function return the shortest
// palindromic subString
static char ShortestPalindrome(String s)
{
int n = s.Length;
char ans = s[0];
// Finding the smallest character
// present in the String
for(int i = 1; i < n; i++)
{
ans = (char) Math.Min(ans, s[i]);
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
String s = "geeksforgeeks";
Console.Write(ShortestPalindrome(s));
}
}
// This code is contributed by 29AjayKumar
<script>
// Javascript program to find the shortest
// palindromic subString
// Function return the shortest
// palindromic subString
function ShortestPalindrome(s)
{
let n = s.length;
let ans = s[0].charCodeAt();
// Finding the smallest character
// present in the String
for(let i = 1; i < n; i++)
{
ans = Math.min(ans, s[i].charCodeAt());
}
return String.fromCharCode(ans);
}
// Driver code
let s = "geeksforgeeks";
document.write(ShortestPalindrome(s));
// This code is contributed by suresh07
</script>
Time Complexity: O(N), where N is the length of the string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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