Create Strings.md

Strings.md

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+
+**Q1- First Unique Character in a String** 
+
+**Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.**
+
+```
+s = "InterviewBit"
+return 1.
+```
+
+_Approach:_
+
+The best possible solution here could be of a linear time because to ensure that the character is unique you have to check the whole string anyway.
+
+The idea is to go through the string and save in a hash map the number of times each character appears in the string. That would take O(N) time, where N is a number of characters in the string.
+
+And then we go through the string the second time, this time we use the hash map as a reference to check if a character is unique or not.
+If the character is unique, one could just return its index. The complexity of the second iteration is O(N) as well.
+
+```
+public int firstUniqChar(String s) {
+        HashMap<Character, Integer> count = new HashMap<Character, Integer>();
+        int n = s.length();
+        // build hash map : character and how often it appears
+        for (int i = 0; i < n; i++) {
+            char c = s.charAt(i);
+            count.put(c, count.getOrDefault(c, 0) + 1);
+        }
+        
+        // find the index
+        for (int i = 0; i < n; i++) {
+            if (count.get(s.charAt(i)) == 1) 
+                return i;
+        }
+        return -1;
+}
+```
+
+Time complexity : O(N) since we go through the string of length N two times.
+
+Space complexity : O(N) since we have to keep a hash map with N elements.
+
+**Q2- Isomorphic Strings**
+
+**Given two strings s and t, determine if they are isomorphic.Two strings are isomorphic if the characters in s can be replaced to get t. All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.**
+
+```
+Input: s = "egg", t = "add"
+Output: true
+
+Input: s = "foo", t = "bar"
+Output: false
+
+Input: s = "paper", t = "title"
+Output: true
+```
+
+Solution: 
+
+The idea is that we need to map a char to another one, for example, "egg" and "add", we need to constract the mapping 'e' -> 'a' and 'g' -> 'd'. Instead of directly mapping 'e' to 'a', another way is to mark them with same value, for example, 'e' -> 1, 'a'-> 1, and 'g' -> 2, 'd' -> 2, this works same.
+
+So we use two arrays here m1 and m2, initialized space is 256 (Since the whole ASCII size is 256, 128 also works here). Traverse the character of both s and t on the same position, if their mapping values in m1 and m2 are different, means they are not mapping correctly, returen false; else we construct the mapping, since m1 and m2 are both initialized as 0, we want to use a new value when i == 0, so i + 1 works here.
+
+https://leetcode.com/problems/isomorphic-strings/discuss/57796/My-6-lines-solution
+
+https://leetcode.com/problems/isomorphic-strings/discuss/57802/Java-solution-using-HashMap
+
+**Q3- Valid Anagram**
+
+**Given two strings s and t , write a function to determine if t is an anagram of s.**
+
+```
+Input: s = "anagram", t = "nagaram"
+Output: true
+
+Input: s = "rat", t = "car"
+Output: false
+```
+
+https://leetcode.com/problems/valid-anagram/solution/
+
+**Q4- Check if a string can be obtained by rotating another string 2 places**
+
+```
+Input: string1 = “amazon”, string2 = “azonam”
+Output: Yes
+// rotated anti-clockwise
+
+Input: string1 = “amazon”, string2 = “onamaz”
+Output: Yes
+// rotated clockwise
+```
+
+1- There can be only two cases:
+    a) Clockwise rotated
+    b) Anti-clockwise rotated
+
+2- If clockwise rotated that means elements
+   are shifted in right.
+   So, check if a substring[2.... len-1] of 
+   string2 when concatenated with substring[0,1]
+   of string2 is equal to string1. Then, return true.
+
+3- Else, check if it is rotated anti-clockwise 
+   that means elements are shifted to left.
+   So, check if concatenation of substring[len-2, len-1]
+   with substring[0....len-3] makes it equals to
+   string1. Then return true.
+
+4- Else, return false.
+
+```
+static boolean isRotated(String str1, String str2) 
+    { 
+        if (str1.length() != str2.length()) 
+            return false; 
+       
+        String clock_rot = ""; 
+        String anticlock_rot = ""; 
+        int len = str2.length(); 
+       
+        // Initialize string as anti-clockwise rotation 
+        anticlock_rot = anticlock_rot + 
+                        str2.substring(len-2, len) + 
+                        str2.substring(0, len-2) ; 
+       
+        // Initialize string as clock wise rotation 
+        clock_rot = clock_rot + 
+                    str2.substring(2) + 
+                    str2.substring(0, 2) ; 
+       
+        // check if any of them is equal to string1 
+        return (str1.equals(clock_rot) || 
+                str1.equals(anticlock_rot)); 
+    } 
+    
+    
+```
+Time Complexity: O(n)
+
+https://www.geeksforgeeks.org/check-string-can-obtained-rotating-another-string-2-places/
+
+Followup Question: 
+
+https://www.geeksforgeeks.org/check-if-a-string-can-be-obtained-by-rotating-another-string-d-places/
+
+https://www.geeksforgeeks.org/a-program-to-check-if-strings-are-rotations-of-each-other/
+
+**Q5- Given an input string, reverse the string word by word.**
+
+```
+Input: "the sky is blue"
+Output: "blue is sky the"
+
+Input: "  hello world!  "
+Output: "world! hello"
+Explanation: Your reversed string should not contain leading or trailing spaces.
+```
+
+- A word is defined as a sequence of non-space characters.
+
+- Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces.
+
+- You need to reduce multiple spaces between two words to a single space in the reversed string.
+
+https://leetcode.com/problems/reverse-words-in-a-string/solution/
+
+**Q6- String to Integer (atoi)**
+
+Question: 
+https://leetcode.com/problems/string-to-integer-atoi/
+
+Answer: 
+https://leetcode.com/problems/string-to-integer-atoi/discuss/4643/Java-Solution-with-4-steps-explanations
+
+**Q7- Longest Substring Without Repeating Characters**
+
+**Given a string, find the length of the longest substring without repeating characters.**
+```
+Input: "abcabcbb"
+Output: 3 
+Explanation: The answer is "abc", with the length of 3. 
+
+Input: "pwwkew"
+Output: 3
+Explanation: The answer is "wke", with the length of 3. 
+             Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
+```
+
+https://leetcode.com/problems/longest-substring-without-repeating-characters/solution/
+
+**Q8- Longest Duplicate Substring**
+
+https://leetcode.com/problems/longest-duplicate-substring/
+
+**Homework Problem**
+
+https://codeforces.com/problemset/problem/676/C
+