Palindrome Check

Category: Strings

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Problem Statement

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Approach & Solution

Solution 1

``` Solution 1 Solution 2 Solution 3 Solution 4 class Program { // O(n) time | O(1) space public static boolean isPalindrome(String str) { int leftIdx = 0; int rightIdx = str.length() - 1; while (leftIdx < rightIdx) { if (str.charAt(leftIdx) != str.charAt(rightIdx)) { return false; } leftIdx++; rightIdx--; } return true; } }

Solution 2

``` Solution 1 Solution 2 Solution 3 Solution 4 2 3 class Program { 4 // O(n) time | O(n) space 5 public static boolean isPalindrome(String str) { 6 return isPalindrome(str, 0); 7 } 8 9 public static boolean isPalindrome(String str, int i) { 10 int j = str.length() - 1 - i; 11 return i >= j ? true : str.charAt(i) == str.charAt(j) && isPalindrome(str, i + 1); 12 } 13 } 14

Solution 3

``` Solution 1 Solution 2 Solution 3 Solution 4 class Program { // O(n) time | O(n) space public static boolean isPalindrome(String str) { StringBuilder reversedString = new StringBuilder(); for (int i = str.length() - 1; i >= 0; i--) { reversedString.append(str.charAt(i)); } return str.equals(reversedString.toString()); } }

Solution 4

``` Solution 1 Solution 2 Solution 3 Solution 4 2 3 class Program { 4 // O(n^2) time | O(n) space 5 public static boolean isPalindrome(String str) { 6 String reversedString = ""; 7 for (int i = str.length() - 1; i >= 0; i--) { 8 reversedString += str.charAt(i); 9 } 10 return str.equals(reversedString); 11 } 12 } 13

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Test Cases

``` Test Case 1 {"string": "abcdcba"} Test Case 2 {"string": "a"} Test Case 3 {"string": "ab"} Test Case 4 {"string": "aba"} Test Case 5 {"string": "abb"} Test Case 6 {"string": "abba"} Test Case 7 {"string": "abcdefghhgfedcba"} Test Case 8 {"string": "abcdefghihgfedcba"} Test Case 9 {"string": "abcdefghihgfeddcba"}