• Shubham Kumar
• 07 Sep, 2024
• • Code

Make String a Subsequence Using Cyclic Increments

This is the LeetCode problem number 2825. Cyclic increment This is when you increase an entity by an amount and when you reach the end you circle back to start and continue the count. If a is increased cyclicly by 1, we will get b. If a is increased cyclicly by 2, we will get c. But if z is increased cyclicly by 1, we get a. By 2 we will get b. String increaseCyclic (String str, int index) { char ch = str.charAt(index); char newch = (char) ((ch - 'a' + 1) % 26 + 'a'); return str.substring(0, index) + newChar + str.substring(index + 1); }Subsequence String str1 is said to contain the subsequence of str2 if we can delete some characters from str1 to get str2. During this deletion we are not allowed to disturb the relative order of chars in the str1. A code to check if str1 contains subsequence str2. We can iterate over all the characters of str1 sequencely and check if all the letters are there as in str2. boolean isSubsequence(String str1, String str2) { int p1 = 0; int p2 = 0; while (p1 < str1.length() && p2 < str2.length()) { if (str1.charAt(p1) == str2.charAt(p2)) { p2++; } p1++; } return p2 == str2.length(); }Solution The problem asks us that we are allowed to cyclic increase any number of chars in str1. And check whether we are able to say str1 will contain a subsequence of str2. We can solve this problem by just merging both the problems. Instead of checking just the characters equality, we can add an additional check on character of str1 after increasig it cyclicly. public boolean canMakeSubsequence(String str1, String str2) { int p1 = 0; int p2 = 0; while (p1 < str1.length() && p2 < str2.length()) { char cyclicCh = (char) ((str1.charAt(p1) - 'a' + 1) % 26 + 'a'); if (str1.charAt(p1) == str2.charAt(p2) || cyclicCh == str2.charAt(p2)) { p2++; } p1++; } return p2 == str2.length(); }