Longest Substring that can be made a palindrome by swapping of characters
Last Updated :
25 Sep, 2023
Given a numeric string S, the task is to find the longest non-empty substring that can be made palindrome.
Examples:
Input: S = "3242415"
Output: 5
Explanation: "24241" is the longest such substring which can be converted to the palindromic string "24142".
Input: S = "213123"
Output: 6
Explanation: "213123" such substring which can be converted to the palindromic string "231132".
Naive Approach: The simplest approach to solve this problem is to generate all possible substrings and for each substring check if it can be made a palindrome or not by counting the characters in each substring and check if only one or no odd frequent character is present or not.
Time Complexity: O(N3)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea to solve this problem is to use Bitmasking and Dynamic Programming. A palindrome can be formed if the count of each included number (except maybe one) is even. Follow the steps below to solve the problem:
- Initialize an integer variable, say mask. A bit in our mask is 0 if the count for the corresponding number is even, and 1 if it's odd.
- Traverse through the string and while traversing the string, track odd/even counts in the variable mask.
- If the same mask is encountered again, the subarray between the first position (exclusive) and the current position (inclusive) with the same mask has all numbers with the even count.
- Let the size of the substring be stored in a variable, say res.
- Initialize an array, say dp[], to track the smallest (first) position of each mask of size 1024, since the input only contains 10 digits ("0123456789"), and there can be only 2^10 or 1024 variations of bit masks.
- The size of the substring can be calculated by subtracting it from the current position. Note that the position for zero mask is -1, as the first character is to be included.
- Also, check all masks that are different from the current one by one bit. In other words, if two masks are different by one bit, that means that there is one odd count in the substring.
- Print res as the length of longest such substring.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the Longest
// substring that can be made a
// palindrome by swapping of characters
int longestSubstring(string s)
{
// Initialize dp array of size 1024
int dp[1024];
// Initializeing dp array with length of s
fill(dp, dp + 1024, s.size());
// Initializing mask and res
int res = 0, mask = 0;
dp[0] = -1;
// Traverse the string
for (int i = 0; i < s.size(); ++i)
{
// Find the mask of the current character
mask ^= 1 << (s[i] - 48);
// Finding the length of the longest
// substring in s which is a
// palindrome for even count
res = max(res, i - dp[mask]);
// Finding the length of the longest
// substring in s which is a
// palindrome for one odd count
for (int j = 0; j <= 9; ++j)
// Finding maximum length of
// substring having one odd count
res = max(res, i - dp[mask ^ (1 << j)]);
// dp[mask] is minimum of
// current i and dp[mask]
dp[mask] = min(dp[mask], i);
}
// Return longest length of the substring
// which forms a palindrome with swaps
return res;
}
// Driver Code
int main()
{
// Input String
string s = "3242415";
// Function Call
cout << longestSubstring(s);
return 0;
}
// This code is contributed by subhammahato348.
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to find the Longest
// substring that can be made a
// palindrome by swapping of characters
public static int longestSubstring(String s)
{
// Initialize dp array of size 1024
int dp[] = new int[1024];
// Initializeing dp array with length of s
Arrays.fill(dp, s.length());
// Initializing mask and res
int res = 0, mask = 0;
dp[0] = -1;
// Traverse the string
for (int i = 0; i < s.length(); ++i) {
// Find the mask of the current character
mask ^= 1 << (s.charAt(i) - '0');
// Finding the length of the longest
// substring in s which is a
// palindrome for even count
res = Math.max(res, i - dp[mask]);
// Finding the length of the longest
// substring in s which is a
// palindrome for one odd count
for (int j = 0; j <= 9; ++j)
// Finding maximum length of
// substring having one odd count
res = Math.max(res,
i - dp[mask ^ (1 << j)]);
// dp[mask] is minimum of
// current i and dp[mask]
dp[mask] = Math.min(dp[mask], i);
}
// Return longest length of the substring
// which forms a palindrome with swaps
return res;
}
// Driver Code
public static void main(String[] args)
{
// Input String
String s = "3242415";
// Function Call
System.out.println(longestSubstring(s));
}
}
# Python3 program for the above approach
# Function to find the Longest
# substring that can be made a
# palindrome by swapping of characters
def longestSubstring(s):
# Initialize dp array of size 1024
dp = [1024 for i in range(1024)]
# Initializeing dp array with length of s
# Arrays.fill(dp, s.length());
# Initializing mask and res
res, mask = 0, 0
dp[0] = -1
# Traverse the string
for i in range(len(s)):
# Find the mask of the current character
mask ^= 1 << (ord(s[i]) - ord('0'))
# Finding the length of the longest
# substring in s which is a
# palindrome for even count
res = max(res, i - dp[mask])
# Finding the length of the longest
# substring in s which is a
# palindrome for one odd count
for j in range(10):
# Finding maximum length of
# substring having one odd count
res = max(res, i - dp[mask ^ (1 << j)])
# dp[mask] is minimum of
# current i and dp[mask]
dp[mask] = min(dp[mask], i)
# Return longest length of the substring
# which forms a palindrome with swaps
return res
# Driver Code
if __name__ == '__main__':
# Input String
s = "3242415"
# Function Call
print(longestSubstring(s))
# This code is contributed by mohit kumar 29
// C# program for the above approach
using System;
class GFG
{
// Function to find the Longest
// substring that can be made a
// palindrome by swapping of characters
public static int longestSubstring(string s)
{
// Initialize dp array of size 1024
int []dp = new int[1024];
// Initializeing dp array with length of s
for (int i = 0; i < 1024; ++i)
{
dp[i] = s.Length;
}
// Initializing mask and res
int res = 0, mask = 0;
dp[0] = -1;
// Traverse the string
for (int i = 0; i < s.Length; i++)
{
// Find the mask of the current character
mask = mask ^ (1 << (s[i] - '0'));
// Finding the length of the longest
// substring in s which is a
// palindrome for even count
res = Math.Max(res, i - dp[mask]);
// Finding the length of the longest
// substring in s which is a
// palindrome for one odd count
for (int j = 0; j < 10; j++)
{
// Finding maximum length of
// substring having one odd count
res = Math.Max(res,i - dp[mask ^ (1 << j)]);
}
// dp[mask] is minimum of
// current i and dp[mask]
dp[mask] = Math.Min(dp[mask], i);
}
// Return longest length of the substring
// which forms a palindrome with swaps
return res;
}
// Driver Code
public static void Main(string[] args)
{
// Input String
string s = "3242415";
// Function Call
Console.WriteLine(longestSubstring(s));
}
}
// This code is contributed by AnkThon
<script>
// Javascript program for the above approach
// Function to find the Longest
// substring that can be made a
// palindrome by swapping of characters
function longestSubstring(s)
{
// Initialize dp array of size 1024
let dp = new Array(1024).fill(1);
// Initializing mask and res
let res = 0, mask = 0;
dp[0] = -1;
// Traverse the string
for(let i = 0; i < s.length; ++i)
{
// Find the mask of the current character
mask ^= 1 << (s[i] - '0');
// Finding the length of the longest
// substring in s which is a
// palindrome for even count
res = Math.max(res, i - dp[mask]);
// Finding the length of the longest
// substring in s which is a
// palindrome for one odd count
for(let j = 0; j <= 9; ++j)
// Finding maximum length of
// substring having one odd count
res = Math.max(res,
i - dp[mask ^ (1 << j)]);
// dp[mask] is minimum of
// current i and dp[mask]
dp[mask] = Math.min(dp[mask], i);
}
// Return longest length of the substring
// which forms a palindrome with swaps
return res;
}
// Driver Code
// Input String
let s = "3242415";
// Function Call
document.write(longestSubstring(s));
// This code is contributed by splevel62
</script>
Time Complexity: O(10*N)
Auxiliary Space: O(1024)
New Approach:- Here is another approach to solve this problem
The approach counts the frequency of each digit in the string, adds the counts of even digits to the `maxPalindromeLength`, and considers the counts of odd digits. If there is at least one odd count digit, it subtracts 1 from the count and adds it to `maxPalindromeLength`, considering that one odd digit can be placed at the center of the palindrome. The result is the length of the longest substring that can be made a palindrome by swapping characters in the given string.
Below is the implementation of the above approach:
C++
// C++ program for above approach
#include <iostream>
#include <vector>
using namespace std;
int longestSubstring(string s) {
vector<int> counts(10, 0); // Store the count of each digit
// Count the frequency of each digit in the string
for (char c : s) {
int digit = c - '0';
counts[digit] += 1;
}
// Check for the longest palindrome length
int maxPalindromeLength = 0;
bool oddCount = false; // Flag to track if odd count digit is encountered
for (int count : counts) {
if (count % 2 == 0) {
maxPalindromeLength += count;
} else {
maxPalindromeLength += count - 1;
oddCount = true;
}
}
if (oddCount) {
maxPalindromeLength += 1;
}
return maxPalindromeLength;
}
//Driver code
int main() {
string s = "3242415";
cout << longestSubstring(s) << endl;
return 0;
}
public class Main {
public static int longestSubstring(String s) {
int[] counts = new int[10]; // Store the count of each digit
// Count the frequency of each digit in the string
for (char c : s.toCharArray()) {
int digit = Character.getNumericValue(c);
counts[digit] += 1;
}
// Check for the longest palindrome length
int maxPalindromeLength = 0;
boolean oddCount = false; // Flag to track if odd count digit is encountered
for (int count : counts) {
if (count % 2 == 0) {
maxPalindromeLength += count;
} else {
maxPalindromeLength += count - 1;
oddCount = true;
}
}
if (oddCount) {
maxPalindromeLength += 1;
}
return maxPalindromeLength;
}
public static void main(String[] args) {
String s = "3242415";
System.out.println(longestSubstring(s));
}
}
def longestSubstring(s):
counts = [0] * 10 # Store the count of each digit
# Count the frequency of each digit in the string
for char in s:
digit = int(char)
counts[digit] += 1
# Check for the longest palindrome length
maxPalindromeLength = 0
oddCount = False # Flag to track if odd count digit is encountered
for count in counts:
if count % 2 == 0:
maxPalindromeLength += count
else:
maxPalindromeLength += count - 1
oddCount = True
if oddCount:
maxPalindromeLength += 1
return maxPalindromeLength
# Driver code
s = "3242415"
print(longestSubstring(s))
// C# program for above approach
using System;
using System.Collections.Generic;
class GFG
{
static int LongestSubstring(string s)
{
// Create a dictionary to store the count of each digit
Dictionary<int, int> counts = new Dictionary<int, int>();
// Count the frequency of each digit in the string
foreach (char c in s)
{
int digit = c - '0';
if (counts.ContainsKey(digit))
counts[digit]++;
else
counts[digit] = 1;
}
// Check for the longest palindrome length
int maxPalindromeLength = 0;
bool oddCount = false; // Flag to track if odd count digit is encountered
foreach (int count in counts.Values)
{
if (count % 2 == 0)
{
maxPalindromeLength += count;
}
else
{
maxPalindromeLength += count - 1;
oddCount = true;
}
}
if (oddCount)
{
maxPalindromeLength += 1;
}
return maxPalindromeLength;
}
static void Main()
{
string s = "3242415";
Console.WriteLine(LongestSubstring(s));
}
}
function longestSubstring(s) {
const counts = new Array(10).fill(0); // Store the count of each digit
// Count the frequency of each digit in the string
for (let i = 0; i < s.length; i++) {
const digit = parseInt(s[i]);
counts[digit] += 1;
}
// Check for the longest palindrome length
let maxPalindromeLength = 0;
let oddCount = false; // Flag to track if odd count digit is encountered
for (const count of counts) {
if (count % 2 === 0) {
maxPalindromeLength += count;
} else {
maxPalindromeLength += count - 1;
oddCount = true;
}
}
if (oddCount) {
maxPalindromeLength += 1;
}
return maxPalindromeLength;
}
// Driver code
const s = "3242415";
console.log(longestSubstring(s));
Time complexity: O(n)
Auxiliary Space : O(1)