From 4cf62a9abdb1ae67b9a83b97d3922a4a7b1c4d9e Mon Sep 17 00:00:00 2001
From: RajSpeed <116724067+RajSpeed@users.noreply.github.com>
Date: Wed, 26 Oct 2022 22:13:41 +0530
Subject: [PATCH 1/2] Create 316_Remove_Duplicate_Letters.cpp

---
 C++/316_Remove_Duplicate_Letters.cpp | 50 ++++++++++++++++++++++++++++
 1 file changed, 50 insertions(+)
 create mode 100644 C++/316_Remove_Duplicate_Letters.cpp

diff --git a/C++/316_Remove_Duplicate_Letters.cpp b/C++/316_Remove_Duplicate_Letters.cpp
new file mode 100644
index 0000000..5466c57
--- /dev/null
+++ b/C++/316_Remove_Duplicate_Letters.cpp
@@ -0,0 +1,50 @@
+// Name: Rajesh Sharma
+// Date: 26/10/2022
+
+/*
+Given a string s, return the lexicographically smallest subsequence of s that contains all the distinct characters of s exactly once.
+*/
+
+// This question is the same as 1081: https://leetcode.com/problems/smallest-subsequence-of-distinct-characters/
+// That is language variation.
+
+/*
+Approach: As the question suggests, We can greedily pick the characters which affect the answer
+and can be picked in future where they may be coming after lower lexicographical characters.
+
+Time Complexity: O(n)
+Space Complexity: O(n)
+*/
+
+class Solution {
+  public:
+    string smallestSubsequence(string s) {
+      vector < bool > vis(26, false); // visited array to mark if char already added to answer
+      vector < int > lastOccurance(26, 1e5); // last occurance of each character is stored in this, useful to check if we can ignore this char now and pick it later
+      for (int i = 0; i < s.length(); i++) lastOccurance[s[i] - 'a'] = i;
+      stack < char > st; // stack to access previous character
+      for (int i = 0; i < s.length(); i++) {
+        char c = s[i];
+        if (vis[c - 'a']) continue;
+        if (st.size() == 0) { // stack size is zero
+          st.push(c);
+          vis[c - 'a'] = true;
+          continue;
+        }
+
+        while (st.size() && st.top() > c && lastOccurance[st.top() - 'a'] > i) { // while our previous character is > us and can be ignored right now, we pop it from the stack
+          vis[st.top() - 'a'] = false;
+          st.pop();
+        }
+        st.push(c);
+        vis[c - 'a'] = true;
+      }
+      string ans = "";
+      while (st.size()) { //fill stack chars in the answer string, note they will have to be reversed.
+        ans.push_back(st.top());
+        st.pop();
+      }
+      reverse(ans.begin(), ans.end()); // reversing
+      return ans;
+    }
+};

From 5ebcc77283f13d3f000c24d5553032a56f287904 Mon Sep 17 00:00:00 2001
From: RajSpeed <116724067+RajSpeed@users.noreply.github.com>
Date: Wed, 26 Oct 2022 22:17:55 +0530
Subject: [PATCH 2/2] Update 316_Remove_Duplicate_Letters.cpp

---
 C++/316_Remove_Duplicate_Letters.cpp | 3 ++-
 1 file changed, 2 insertions(+), 1 deletion(-)

diff --git a/C++/316_Remove_Duplicate_Letters.cpp b/C++/316_Remove_Duplicate_Letters.cpp
index 5466c57..b433b15 100644
--- a/C++/316_Remove_Duplicate_Letters.cpp
+++ b/C++/316_Remove_Duplicate_Letters.cpp
@@ -2,7 +2,8 @@
 // Date: 26/10/2022
 
 /*
-Given a string s, return the lexicographically smallest subsequence of s that contains all the distinct characters of s exactly once.
+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
 */
 
 // This question is the same as 1081: https://leetcode.com/problems/smallest-subsequence-of-distinct-characters/