October leetcoding challenge, week 2, Q 4

Author: tans105

ds-algorithms-competitive-programming/src/main/java/competitiveProgramming/leetcode/thirtyDaysLeetcodingChallenge/october/week2/RemoveDuplicateLetters.java

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+package competitiveProgramming.leetcode.thirtyDaysLeetcodingChallenge.october.week2;
+
+
+import java.util.Stack;
+
+/*
+https://leetcode.com/explore/challenge/card/october-leetcoding-challenge/560/week-2-october-8th-october-14th/3491/
+
+Remove Duplicate Letters
+Given a string s, remove duplicate letters so that every letter appears once and only once. You must make sure your result is the smallest in lexicographical order among all possible results.
+
+Note: This question is the same as 1081: https://leetcode.com/problems/smallest-subsequence-of-distinct-characters/
+
+
+
+Example 1:
+
+Input: s = "bcabc"
+Output: "abc"
+Example 2:
+
+Input: s = "cbacdcbc"
+Output: "acdb"
+
+
+Constraints:
+
+1 <= s.length <= 104
+s consists of lowercase English letters.
+ */
+public class RemoveDuplicateLetters {
+    public static void main(String[] args) {
+        System.out.println(removeDuplicateLetters("bcabc"));
+        System.out.println(removeDuplicateLetters("cbacdcbc"));
+    }
+
+
+    public static String removeDuplicateLetters(String sr) {
+
+        int[] res = new int[26]; //will contain number of occurences of character (i+'a')
+        boolean[] visited = new boolean[26]; //will contain if character (i+'a') is present in current result Stack
+        char[] ch = sr.toCharArray();
+        for (char c : ch) {  //count number of occurences of character
+            res[c - 'a']++;
+        }
+        Stack<Character> st = new Stack<>(); // answer stack
+        int index;
+        for (char s : ch) {
+            index = s - 'a';
+            res[index]--;   //decrement number of characters remaining in the string to be analysed
+            if (visited[index]) //if character is already present in stack, dont bother
+                continue;
+            //if current character is smaller than last character in stack which occurs later in the string again
+            //it can be removed and  added later e.g stack = bc remaining string abc then a can pop b and then c
+            while (!st.isEmpty() && s < st.peek() && res[st.peek() - 'a'] != 0) {
+                visited[st.pop() - 'a'] = false;
+            }
+            st.push(s); //add current character and mark it as visited
+            visited[index] = true;
+        }
+
+        StringBuilder sb = new StringBuilder();
+        //pop character from stack and build answer string from back
+        while (!st.isEmpty()) {
+            sb.insert(0, st.pop());
+        }
+        return sb.toString();
+    }
+}