Leetcode Shortest Palindrome problem solution YASH PAL, 31 July 2024 In this Leetcode Shortest Palindrome problem solution, You are given a string s. You can convert s to a palindrome by adding characters in front of it. Return the shortest palindrome you can find by performing this transformation. Topics we are covering Toggle Problem solution in Python.Problem solution in Java.Problem solution in C++.Problem solution in C. Problem solution in Python. class Solution(object): def shortestPalindrome(self, s): prefix_idx = 0 for i in range(len(s)-1, -1, -1): if s[i] == s[prefix_idx ]: prefix_idx += 1 if prefix_idx == len(s): return s suffix = s[prefix_idx:] return suffix[::-1] + self.shortestPalindrome(s[:prefix_idx]) + suffix Problem solution in Java. class Solution { public String shortestPalindrome(String s) { String temp = s + "#" + new StringBuilder(s).reverse().toString(); int[] next = getTable(temp); return new StringBuilder(s.substring(next[next.length - 1] + 1)).reverse().toString() + s; } public int[] getTable(String s) { int[] next = new int[s.length()]; next[0] = -1; int k = -1; int j = 0; while (j < s.length() - 1) { if (k == -1 || s.charAt(j) == s.charAt(k)) { j++; k++; next[j] = k; } else { k = next[k]; } } return next; } } Problem solution in C++. class Solution { public: string shortestPalindrome(string s) { string rev_s = s; reverse(rev_s.begin(), rev_s.end()); string l = s + "#" + rev_s; vector<int> p(l.size(), 0); for (int i = 1; i < l.size(); i++) { int j = p[i - 1]; while (j > 0 && l[i] != l[j]) j = p[j - 1]; p[i] = (j += l[i] == l[j]); } return rev_s.substr(0, s.size() - p[l.size() - 1]) + s; } }; Problem solution in C. bool findLongest(char *s, int len, int index1, int index2, int *remain) { while (index1 >= 0 && index2 <= len - 1 && s[index1] == s[index2]) { index1--; index2++; } if (index1 == -1 && *remain > len - index2) { *remain = len - index2; return true; } return false; } char * shortestPalindrome(char * s){ int index = 0, len = strlen(s); int remain = len - 1, new_len = 2 * len - 1; if (len == 0) return calloc(1, sizeof(char)); for (int i=len/2; i>=0; --i) { if (findLongest(s, len, i - 1, i + 1, &remain)) break; if (findLongest(s, len, i - 1, i , &remain)) break; } new_len = len + remain; char *result = (char*)malloc(sizeof(char) * (new_len + 1)); for (int i=0; i<remain; i++) { result[i] = s[len - i - 1]; } for (int i=0; i<len; i++) { result[remain + i] = s[i]; } result[new_len] = 0; return result; } coding problems solutions