Find index i such that prefix of S1 and suffix of S2 till i form a palindrome when concatenated

Given two strings A and B of equal lengths, the task is to find an index i such that A[0…i] and B[i+1…n-1] give a palindrome when concatenated together. If it is not possible to find such an index then print -1.

Examples:

Input: S1 = “abcdf”, S2 = “sfgba”
Output: 1
S1[0..1] = “ab”, S2[2..n-1] = “gba”
S1 + S2 = “abgba” which is a palindrome.

Input : S1 = “abcda”, S2 = “bacbs”
Output: -1

Simple Approach:

  • Iterate from 0 to n (length of the string) and copy ith character from S1 to another string let’s say S.
  • Now take another temporary string Temp and copy the characters of S2 from index i +1 to n.
  • Now check whether the string (S + Temp) is palindrome or not.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns true if s is palindrome
bool isPalindrome(string s)
{
    int i = 0;
    int j = s.length() - 1;
  
    while (i < j) {
        if (s[i] != s[j])
            return false;
        i++;
        j--;
    }
  
    return true;
}
  
// Function to return the required index
int getIndex(string S1, string S2, int n)
{
  
    string S = "";
  
    for (int i = 0; i < n; i++) {
  
        // Copy the ith character in S
        S = S + S1[i];
        string Temp = "";
  
        // Copy all the character of string s2 in Temp
        for (int j = i + 1; j < n; j++)
            Temp += S2[j];
  
        // Check whether the string is palindrome
        if (isPalindrome(S + Temp)) {
            return i;
        }
    }
  
    return -1;
}
  
// Driver code
int main()
{
    string S1 = "abcdf", S2 = "sfgba";
    int n = S1.length();
  
    cout << getIndex(S1, S2, n);
    return 0;
}

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Output:

1

Time complexity: O(n2)

Efficient Approach: The efficient approach will be to observe that the concatenated string will be palindrome if the first character of first string matches with the last character of second string as we are considering prefix of first string and suffix of second string.

  • Start iterating the first string from start using a pointer say i and second string from end using a pointer say j until i < j and s1[i] == s2[j].
  • Check if both of the pointers i and j are equal at first mismatch.
  • If yes, then return index i, that is we can concatenates strings s1[0..i] and s2[j..N] to form a palindrome.
  • Otherwise, check if either s1[i..j] or s2[i..j] is a palindrome. If yes, then we can still concatenate either s1[0..j] + s2[j+1, N-1] or s1[0..i-1] + s2[i..N-1] to form a palindrome.
  • Else, return -1.

Below is the implementation of the above approach:

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// C++ program to implement the
// above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns true if the sub-string
// starting from index i and ending at index j
// is a palindrome
bool isPalindrome(string s, int i, int j)
{
    while (i < j) {
        if (s[i] != s[j])
            return false;
        i++;
        j--;
    }
    return true;
}
  
// Function to get the required index
int getIndex(string s1, string s2, int len)
{
    int i = 0, j = len - 1;
  
    // Start comparing the two strings
    // from both ends.
    while (i < j) {
        // Break from the loop at first mismatch
        if (s1[i] != s2[j]) {
            break;
        }
  
        i++;
        j--;
    }
  
    // If it is possible to concatenate
    // the strings to form palindrome,
    // return index
    if (i == j) {
        return i - 1;
    }
  
    // If remaining part for s2
    // is palindrome
    else if (isPalindrome(s2, i, j))
        return i - 1;
  
    // If remaining part for s1
    // is palindrome
    else if (isPalindrome(s1, i, j))
        return j;
  
    // If not possible, return -1
    return -1;
}
  
// Driver Code
int main()
{
    string s1 = "abcdf", s2 = "sfgba";
    int len = s1.length();
  
    cout << getIndex(s1, s2, len);
  
    return 0;
}

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Output:

1

Time Complexity: O(N), where N is the length of the string.



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