Minimum length of substring whose rotation generates a palindromic substring
Last Updated :
16 Jun, 2022
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Given a string str, the task is to find the minimum length of substring required to rotate that generates a palindromic substring from the given string.
Examples:
Input: str = "abcbd"
Output: 0
Explanation: No palindromic substring can be generated. There is no repeated character in the string.
Input: str = "abcdeba"
Output: 3
Explanation: Rotate substring "deb" to convert the given string to abcbeda with a palindromic substring "bcb".
Approach:
- If no character repeats in the string, then no palindromic substring can be generated.
- For every repeating character, check if the index of its previous occurrence is within one or two indices from the current index. If so, then a palindromic substring already exists.
- Otherwise, calculate the length of (current index - index of the previous occurrence - 1).
- Return the minimum of all such lengths as the answer
Below is the implementation of the above approach:
// C++ Program to find the minimum
// length of substring whose rotation
// generates a palindromic substring
#include <bits/stdc++.h>
using namespace std;
// Function to return the
// minimum length of substring
int count_min_length(string s)
{
// Store the index of
// previous occurrence
// of the character
int hash[26];
// Variable to store
// the maximum length
// of substring
int ans = INT_MAX;
for (int i = 0; i < 26; i++)
hash[i] = -1;
for (int i = 0; i < s.size(); i++) {
// If the current character
// hasn't appeared yet
if (hash[s[i] - 'a'] == -1)
hash[s[i] - 'a'] = i;
else {
// If the character has occurred
// within one or two previous
// index, a palindromic substring
// already exists
if (hash[s[i] - 'a'] == i - 1
|| hash[s[i] - 'a'] == i - 2)
return 0;
// Update the maximum
ans = min(ans,
i - hash[s[i] - 'a'] - 1);
// Replace the previous
// index of the character by
// the current index
hash[s[i] - 'a'] = i;
}
}
// If character appeared
// at least twice
if (ans == INT_MAX)
return -1;
return ans;
}
// Driver Code
int main()
{
string str = "abcdeba";
cout << count_min_length(str);
}
// Java Program to find the minimum
// length of substring whose rotation
// generates a palindromic substring
import java.util.*;
import java.lang.*;
class GFG{
// Function to return the
// minimum length of substring
static int count_min_length(String s)
{
// Store the index of
// previous occurrence
// of the character
int[] hash = new int[26];
// Variable to store
// the maximum length
// of substring
int ans = Integer.MAX_VALUE;
for (int i = 0; i < 26; i++)
hash[i] = -1;
for (int i = 0; i < s.length(); i++)
{
// If the current character
// hasn't appeared yet
if (hash[s.charAt(i) - 'a'] == -1)
hash[s.charAt(i) - 'a'] = i;
else
{
// If the character has occurred
// within one or two previous
// index, a palindromic substring
// already exists
if (hash[s.charAt(i) - 'a'] == i - 1 ||
hash[s.charAt(i) - 'a'] == i - 2)
return 0;
// Update the maximum
ans = Math.min(ans,
i - hash[s.charAt(i) - 'a'] - 1);
// Replace the previous
// index of the character by
// the current index
hash[s.charAt(i) - 'a'] = i;
}
}
// If character appeared
// at least twice
if (ans == Integer.MAX_VALUE)
return -1;
return ans;
}
// Driver code
public static void main(String[] args)
{
String str = "abcdeba";
System.out.println(count_min_length(str));
}
}
// This code is contributed by offbeat
# Python3 program to find the minimum
# length of substring whose rotation
# generates a palindromic substring
import sys
INT_MAX = sys.maxsize;
# Function to return the
# minimum length of substring
def count_min_length(s):
# Store the index of
# previous occurrence
# of the character
hash = [0] * 26;
# Variable to store
# the maximum length
# of substring
ans = sys.maxsize;
for i in range(26):
hash[i] = -1;
for i in range(len(s)):
# If the current character
# hasn't appeared yet
if (hash[ord(s[i]) - ord('a')] == -1):
hash[ord(s[i]) - ord('a')] = i;
else :
# If the character has occurred
# within one or two previous
# index, a palindromic substring
# already exists
if (hash[ord(s[i]) - ord('a')] == i - 1 or
hash[ord(s[i]) - ord('a')] == i - 2) :
return 0;
# Update the maximum
ans = min(ans, i - hash[ord(s[i]) -
ord('a')] - 1);
# Replace the previous
# index of the character by
# the current index
hash[ord(s[i]) - ord('a')] = i;
# If character appeared
# at least twice
if (ans == INT_MAX):
return -1;
return ans;
# Driver Code
if __name__ == "__main__":
string = "abcdeba";
print(count_min_length(string));
# This code is contributed by AnkitRai01
// C# Program to find the minimum
// length of substring whose rotation
// generates a palindromic substring
using System;
class GFG{
// Function to return the
// minimum length of substring
static int count_min_length(string s)
{
// Store the index of
// previous occurrence
// of the character
int[] hash = new int[26];
// Variable to store
// the maximum length
// of substring
int ans = int.MaxValue;
for (int i = 0; i < 26; i++)
hash[i] = -1;
for (int i = 0; i < s.Length; i++)
{
// If the current character
// hasn't appeared yet
if (hash[s[i] - 'a'] == -1)
hash[s[i] - 'a'] = i;
else
{
// If the character has occurred
// within one or two previous
// index, a palindromic substring
// already exists
if (hash[s[i] - 'a'] == i - 1 ||
hash[s[i] - 'a'] == i - 2)
return 0;
// Update the maximum
ans = Math.Min(ans,
i - hash[s[i] - 'a'] - 1);
// Replace the previous
// index of the character by
// the current index
hash[s[i] - 'a'] = i;
}
}
// If character appeared
// at least twice
if (ans == int.MaxValue)
return -1;
return ans;
}
// Driver code
public static void Main(string[] args)
{
string str = "abcdeba";
Console.WriteLine(count_min_length(str));
}
}
// This code is contributed by AnkitRai01
<script>
// JavaScript Program to find the minimum
// length of substring whose rotation
// generates a palindromic substring
// Function to return the
// minimum length of substring
function count_min_length(s) {
// Store the index of
// previous occurrence
// of the character
var hash = new Array(26).fill(0);
// Variable to store
// the maximum length
// of substring
var ans = 2147483648;
for (var i = 0; i < 26; i++)
hash[i] = -1;
for (var i = 0; i < s.length; i++) {
// If the current character
// hasn't appeared yet
if (hash[s[i].charCodeAt(0) - "a".charCodeAt(0)] == -1)
hash[s[i].charCodeAt(0) - "a".charCodeAt(0)] = i;
else {
// If the character has occurred
// within one or two previous
// index, a palindromic substring
// already exists
if (
hash[s[i].charCodeAt(0) - "a".charCodeAt(0)] == i - 1 ||
hash[s[i].charCodeAt(0) - "a".charCodeAt(0)] == i - 2
)
return 0;
// Update the maximum
ans = Math.min(
ans,
i - hash[s[i].charCodeAt(0) - "a".charCodeAt(0)] - 1
);
// Replace the previous
// index of the character by
// the current index
hash[s[i].charCodeAt(0) - "a".charCodeAt(0)] = i;
}
}
// If character appeared
// at least twice
if (ans === 2147483648) return -1;
return ans;
}
// Driver code
var str = "abcdeba";
document.write(count_min_length(str));
</script>
Output:
3
Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time
Auxiliary Space: O(26), as we are using extra space for hash.
