Repeated subsequence of length 2 or more - GeeksforGeeks
Given a string, find if there is any subsequence of length 2 or more that repeats itself such that the two subsequence don't have same character at same position, i.e., any 0'th or 1st character in the two subsequences shouldn't have the same index in the original string.
Read full article from Repeated subsequence of length 2 or more - GeeksforGeeks
Given a string, find if there is any subsequence of length 2 or more that repeats itself such that the two subsequence don't have same character at same position, i.e., any 0'th or 1st character in the two subsequences shouldn't have the same index in the original string.
The problem is classic variation of longest common subsequence problem. We have discussed Dynamic programming solution here. Dynamic programming solution takes O(n2) time and space.
In this post, O(n) time and space approach is discussed.
The idea is to remove all the non-repeated characters from the string and check if the resultant string is palindrome or not. If the remaining string is palindrome then it is not repeated, else there is a repetition. One special case we need to handle for inputs like “AAA”, which are palindrome but their repeated subsequence exists. Repeated subsequence exists for a palindrome string if it is of odd length and its middle letter is same as left(or right) character.
bool isPalindrome(char str[], int l, int h){ // l and h are leftmost and rightmost corners of str // Keep comparing characters while they are same while (h > l) if (str[l++] != str[h--]) return false; return true;}// The main function that checks if repeated// subsequence exists in the stringint check(char str[]){ // Find length of input string int n = strlen(str); // Create an array to store all characters and their // frequencies in str[] int freq[MAX_CHAR] = { 0 }; // Traverse the input string and store frequencies // of all characters in freq[] array. for (int i = 0; i < n; i++) { freq[str[i]]++; // If the character count is more than 3 // we found a repetition if (freq[str[i]] > 3) return true; } // In-place remove non-repeating characters // from the string int k = 0; for (int i = 0; i < n; i++) if (freq[str[i]] > 1) str[k++] = str[i]; str[k] = '\0'; // check if the resultant string is palindrome if (isPalindrome(str, 0, k-1)) { // special case - if length is odd // return true if the middle characer is // same as previous one if (k & 1) return str[k/2] == str[k/2 - 1]; // return false if string is a palindrome return false; } // return true if string is not a palindrome return true;}