Create Daily 06-04-24.md

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+## Today's 06*04=24 [Problem Link](https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/description/?envType=daily-question&envId=2024-04-06)
+## 1249. Minimum Remove to Make Valid Parentheses
+
+# Intuition
+<!-- Describe your first thoughts on how to solve this problem. -->
+To address this problem, I utilize a stack to manage the indices of unpaired '(' characters. I traverse the string, marking unpaired ')' characters as '#' and popping indices from the stack for each encountered ')', effectively pairing '(' and ')'. Any remaining unpaired '(' characters are marked as '#' after processing. Finally, I remove all '#' characters to obtain the valid string.
+
+# Approach
+<!-- Describe your approach to solving the problem. -->
+**Initialized** an empty stack (`parenthesesStack`) to store indices of unpaired '(' characters.
+
+**Traversed** the input string character by character:
+   - If the current character is '(', push its index onto `parenthesesStack`.
+   - If the current character is ')':
+     - If `parenthesesStack` is empty, marked the ')' character as '#'.
+     - If `parenthesesStack` is not empty, popped the top index to pair '(' and ')'.
+
+**Marked** any remaining unpaired '(' characters as '#' by popping from `parenthesesStack`.
+
+**Removed** all '#' characters from the modified string.
+
+**Returned** the resulting string, which now contains the minimum removed parentheses to make it valid.
+
+--- 
+Have a look at the code , still have any confusion then please let me know in the comments
+Keep Solving.:)
+
+# Complexity
+- Time complexity : $O(s)$
+<!-- Add your time complexity here, e.g. $$O(n)$$ -->
+$s$ :  length of the input string
+- Space complexity : $O(s)$
+<!-- Add your space complexity here, e.g. $$O(n)$$ -->
+
+# Code
+```
+class Solution {
+    public String minRemoveToMakeValid(String s) {
+
+        Deque<Integer> parenthesesStack = new ArrayDeque<>(); // Stores indices of unpaired '(' characters
+        StringBuilder modifiedString = new StringBuilder(s);
+
+        for (int i = 0; i < s.length(); ++i) {
+            if (modifiedString.charAt(i) == '(') {
+                parenthesesStack.push(i); // Record the unpaired '(' index.
+            } 
+            else if (modifiedString.charAt(i) == ')') {
+                if (parenthesesStack.isEmpty()) {
+                    modifiedString.setCharAt(i, '#'); // Mark the unpaired ')' as '#'.
+                } 
+                else {
+                    parenthesesStack.pop(); // Find a pair!
+                }
+            }
+        }
+
+        // Marking the unpaired '(' as '#'.
+        while (!parenthesesStack.isEmpty()) {
+            modifiedString.setCharAt(parenthesesStack.pop(), '#');
+        }
+
+        return modifiedString.toString().replaceAll("#", "");
+    }
+}
+
+```
