solution to Minimum String Length After Removing Substrings

october_2024/potd_07_10_2024.cpp

@@ -0,0 +1,55 @@
+// 2696.  Minimum String Length After Removing Substrings
+
+/*
+Problem Description:
+You are given a string s consisting only of uppercase English letters.
+
+You can apply some operations to this string where, in one operation, you can remove any occurrence of one of the substrings "AB" or "CD" from s.
+
+Return the minimum possible length of the resulting string that you can obtain.
+
+Note that the string concatenates after removing the substring and could produce new "AB" or "CD" substrings.
+
+ 
+
+Example 1:
+
+Input: s = "ABFCACDB"
+Output: 2
+Explanation: We can do the following operations:
+- Remove the substring "ABFCACDB", so s = "FCACDB".
+- Remove the substring "FCACDB", so s = "FCAB".
+- Remove the substring "FCAB", so s = "FC".
+So the resulting length of the string is 2.
+It can be shown that it is the minimum length that we can obtain.
+
+Example 2:
+
+Input: s = "ACBBD"
+Output: 5
+Explanation: We cannot do any operations on the string so the length remains the same.
+*/
+
+
+// Solution:
+
+
+class Solution {
+public:
+    int minLength(string s) {
+        stack<char> st;               // Initialize an empty stack to store characters
+        int n = s.size();              // Store the length of the string
+
+        for (int i = 0; i < n; i++) {
+            char curr = s[i];
+            if (!st.empty() && 
+                ((st.top() == 'A' && curr == 'B') || st.top() == 'C' && curr == 'D')) {
+                st.pop();              // Remove the top character if a valid pair is found
+            } else {
+                st.push(curr);         // Push the current character onto the stack if no pair is formed
+            }
+        }
+        return st.size();              // Return the size of the stack, which gives the remaining unpaired characters
+    }
+};
+