Construct the Largest Palindromic String which lies between two other Strings
Last Updated :
16 Nov, 2023
Given two strings A and B of length N which contains only lowercase characters, find the lexicographically largest string C of length N which contains only lowercase characters such that the following conditions are satisfied.
- C should be lexicographically greater than the string A.
- C should be lexicographically smaller than the string B.
- C should be palindrome.
Examples:
Input: N = 3, A = "aba", B = "ade"
Output: ada
Explanation: "ada" is lexicographically greater than the string A and lexicographically smaller than the string B and it is a palindrome.
Input: N = 1, A = "z", B = "z"
Output: -1
Approach: The idea behind the following approach is:
This problem can be solved with the some observations. Our goal is to make palindrome and it should be lexicographically larger than A but smaller than B. We will perform operations on string B. Divide B to half string and add first half to C, add a middle character (when N is odd) and again add first half to C but in reverse order. Now, to make it smaller than B and greater than A, we will change characters in C accordingly.
Below are the steps involved in the implementation of the code:
- Check A should be smaller than B otherwise return -1.
- Create a empty string ans representing C.
- Split B into two halves and add first half to ans, add a middle character (when N is odd) and again first half of B in backward direction.
- Now, to make it greater than A and smaller than B.
- Iterate in it and try changing character (on both sides) one by one through i and j variables.
- If character present at ans[i] and ans[j] is 'a' skip it.
- If not reduce it by one both the ans[i] and ans[j] and make the characters between i and j index of ans as 'z'. As our goal is to make lexicographically largest.
- If ans > A and ans < B, return ans.
- Otherwise "-1".
Below is the implementation of the code:
C++
// C++ Implementation of the code:
#include <bits/stdc++.h>
using namespace std;
// Function to find String C
string constructString(int N, string A, string B)
{
// If A is greater than or equal to B
if (A >= B) {
return "-1";
}
int mid = N / 2;
string ans = "";
// Form ans upto half of mid of B
for (int i = 0; i < mid; i++)
ans += B[i];
string copy = ans;
reverse(copy.begin(), copy.end());
if (N & 1) {
ans += B[mid];
}
// Making ans string palindrome
ans += copy;
int i = mid, j = mid;
// If length of A is even
if (N % 2 == 0) {
i--;
}
while (i >= 0 && j < A.length()) {
if (ans[i] == 'a') {
i--;
j++;
}
else {
// Check whether ans is smaller
// than B and greater than A
if (ans < B && ans > A) {
return ans;
}
ans[i] = ans[i] - 1;
ans[j] = ans[i];
int l = i + 1;
int r = j - 1;
// Converting other characters to z
// to form biggest lexicographically
// string
while (l <= r) {
ans[l++] = 'z';
ans[r--] = 'z';
}
// If ans is less than A
// lexicographically
if (ans > A)
return ans;
return "-1";
}
}
return "-1";
}
// Driver code
int main()
{
string A = "aaaaa";
string B = "axaae";
int N = A.length();
// Function call
string res = constructString(N, A, B);
cout << res << endl;
return 0;
}
// Java Implementation of the code:
import java.util.*;
public class Main {
// Function to find String C
public static String constructString(int N, String A, String B) {
// If A is greater than or equal to B
if (A.compareTo(B) >= 0) {
return "-1";
}
int mid = N / 2;
StringBuilder ans = new StringBuilder();
// Form ans up to half of mid of B
for (int i = 0; i < mid; i++) {
ans.append(B.charAt(i));
}
StringBuilder copy = new StringBuilder(ans);
copy.reverse();
if (N % 2 == 1) {
ans.append(B.charAt(mid));
}
// Making ans string palindrome
ans.append(copy);
int i = mid, j = mid;
// If length of A is even
if (N % 2 == 0) {
i--;
}
while (i >= 0 && j < A.length()) {
if (ans.charAt(i) == 'a') {
i--;
j++;
} else {
// Check whether ans is smaller
// than B and greater than A
if (ans.toString().compareTo(B) < 0 &&
ans.toString().compareTo(A) > 0) {
return ans.toString();
}
ans.setCharAt(i, (char)(ans.charAt(i) - 1));
ans.setCharAt(j, ans.charAt(i));
int l = i + 1;
int r = j - 1;
// Converting other characters to z
// to form the biggest lexicographically
// string
while (l <= r) {
ans.setCharAt(l, 'z');
ans.setCharAt(r, 'z');
l++;
r--;
}
// If ans is less than A
// lexicographically
if (ans.toString().compareTo(A) > 0) {
return ans.toString();
}
return "-1";
}
}
return "-1";
}
// Driver code
public static void main(String[] args) {
String A = "aaaaa";
String B = "axaae";
int N = A.length();
// Function call
String res = constructString(N, A, B);
System.out.println(res);
}
}
// this code is contributed by uttamdp_10
def constructString(N, A, B):
# If A is greater than or equal to B
if A >= B:
return "-1"
mid = N // 2
ans = []
# Form ans up to half of mid of B
for i in range(mid):
ans.append(B[i])
copy = ans[::-1]
if N % 2 == 1:
ans.append(B[mid])
# Making ans string palindrome
ans.extend(copy)
i, j = mid, mid
# If the length of A is even
if N % 2 == 0:
i -= 1
while i >= 0 and j < len(A):
if ans[i] == 'a':
i -= 1
j += 1
else:
# Check whether ans is smaller
# than B and greater than A
if ''.join(ans) < B and ''.join(ans) > A:
return ''.join(ans)
ans[i] = chr(ord(ans[i]) - 1)
ans[j] = ans[i]
l, r = i + 1, j - 1
# Converting other characters to 'z'
# to form the biggest lexicographically
# string
while l <= r:
ans[l] = 'z'
ans[r] = 'z'
l += 1
r -= 1
# If ans is less than A
# lexicographically
if ''.join(ans) > A:
return ''.join(ans)
return "-1"
return "-1"
# Driver code
A = "aaaaa"
B = "axaae"
N = len(A)
# Function call
res = constructString(N, A, B)
print(res)
# by phasing17
using System;
class GFG
{
// Function to find String C
public static string ConstructString(int N, string A, string B)
{
// If A is greater than or equal to B
if (A.CompareTo(B) >= 0)
return "-1";
int mid = N / 2;
string ans = "";
// Form ans upto half of mid of B
for (int x = 0; x < mid; x++)
ans += B[x];
string copy = new string(ans.ToCharArray());
char[] charArray = copy.ToCharArray();
Array.Reverse(charArray);
copy = new string(charArray);
if (N % 2 == 1)
ans += B[mid];
// Making ans string palindrome
ans += copy;
int i = mid, j = mid;
// If length of A is even
if (N % 2 == 0)
i--;
while (i >= 0 && j < A.Length)
{
if (ans[i] == 'a')
{
i--;
j++;
}
else
{
// Check whether ans is smaller than B and greater than A
if (string.Compare(ans, B) < 0 && string.Compare(ans, A) > 0)
{
return ans;
}
ans = ans.Remove(i, 1).Insert(i, Convert.ToString((char)(ans[i] - 1)));
ans = ans.Remove(j, 1).Insert(j, Convert.ToString(ans[i]));
int l = i + 1;
int r = j - 1;
// Converting other characters to z to form biggest lexicographically string
while (l <= r)
{
ans = ans.Remove(l, 1).Insert(l, "z");
ans = ans.Remove(r, 1).Insert(r, "z");
l++;
r--;
}
// If ans is less than A lexicographically
if (string.Compare(ans, A) > 0)
return ans;
return "-1";
}
}
return "-1";
}
// Driver code
public static void Main(string[] args)
{
string A = "aaaaa";
string B = "axaae";
int N = A.Length;
// Function call
string res = ConstructString(N, A, B);
Console.WriteLine(res);
}
}
function constructString(N, A, B) {
//If A is greater than or equal to B
if (A >= B) {
return "-1";
}
var mid = Math.floor(N / 2);
var ans = [];
//Form ans up to half of mid of B
for (var i = 0; i < mid; i++) {
ans.push(B[i]);
}
var copy = ans.slice().reverse();
if (N % 2 === 1) {
ans.push(B[mid]);
}
//Making ans string palindrome
ans = ans.concat(copy);
var i = mid;
var j = mid;
//If the length of A is even
if (N % 2 === 0) {
i -= 1;
}
while (i >= 0 && j < A.length) {
if (ans[i] === 'a') {
i -= 1;
j += 1;
} else {
//Check whether ans is smaller
//than B and greater than A
if (ans.join('') < B && ans.join('') > A) {
return ans.join('');
}
ans[i] = String.fromCharCode(ans[i].charCodeAt(0) - 1);
ans[j] = ans[i];
var l = i + 1;
var r = j - 1;
//Converting other characters to 'z'
//to form the biggest lexicographically
//string
while (l <= r) {
ans[l] = 'z';
ans[r] = 'z';
l += 1;
r -= 1;
}
//If ans is less than A
//lexicographically
if (ans.join('') > A) {
return ans.join('');
}
return "-1";
}
}
return "-1";
}
//Driver code
var A = "aaaaa";
var B = "axaae";
var N = A.length;
//Function call
var res = constructString(N, A, B);
console.log(res);
Time Complexity: O(N), where N is the length of strings A and B
Auxiliary Space: O(N)
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