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PalindromeCheck.java

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public class PalindromeCheck {

// BruteForce Approach

public static boolean isPalindrome(String str) {

String rev = "";

for (int i = str.length() - 1; i >= 0; i--) {

rev = rev + str.charAt(i);

}

return rev.toLowerCase().equals(str);

// The time complexity of concatenating strings in this manner is O(n2) because

// for each character appended, the entire current value of rev is copied.

// T.C=0(n2) S.C=O(n)

}

// Two Pointer Technique

public static boolean isPalindrome1(String str) {

for (int i = 0; i < str.length() - 1; i++) {

if (str.charAt(i) != str.charAt(str.length() - 1 - i)) {

return false;

}

}

return true;

// T.C=O(n) S.C=O(1)

}

// Recursion

public static boolean isPalindrome2(String str, int start, int end) {

if (start > end) {

return true;

}

if (str.charAt(start) == str.charAt(end)) {

return isPalindrome2(str, start + 1, end - 1);

}

return false;

// Time Complexity: O(n)

// Space Complexity: O(n) due to the call stack

}

public static void main(String[] args) {

System.out.println(isPalindrome("pratham"));

System.out.println(isPalindrome("roor"));

System.out.println(isPalindrome1("pratham"));

System.out.println(isPalindrome1("roor"));

String str = "pratham";

String str1 = "opopo";

System.out.println(isPalindrome2(str, 0, str.length() - 1));

System.out.println(isPalindrome2(str1, 0, str1.length() - 1));

}

}