diff --git a/Domains/CompetitiveProgramming/Programs/C++/Jump_Game.cpp b/Domains/CompetitiveProgramming/Programs/C++/Jump_Game.cpp
new file mode 100644
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+++ b/Domains/CompetitiveProgramming/Programs/C++/Jump_Game.cpp
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+/*
+Problem: Jump Game
+Platform: LeetCode
+Problem Code/Number: 55
+Difficulty: Medium
+Link: https://leetcode.com/problems/jump-game/
+
+Problem Statement:
+You are given an integer array nums. You are initially positioned at the array's first index, 
+and each element in the array represents your maximum jump length at that position.
+Return true if you can reach the last index, or false otherwise.
+
+Approach:
+Use a greedy approach to track the farthest index you can reach:
+- Initialize a variable 'farthest' to 0.
+- Traverse the array:
+  - If the current index i is greater than 'farthest', return false (stuck, cannot move forward).
+  - Update 'farthest' as max(farthest, i + nums[i]).
+  - If 'farthest' >= last index, return true.
+- If the loop ends, return true.
+
+Time Complexity: O(n)
+Space Complexity: O(1)
+
+Contributor: alisha1510
+*/
+
+#include <iostream>
+#include <stack>
+#include <vector>
+#include <string>
+using namespace std;
+
+class Solution {
+public:
+    bool canJump(vector<int>& nums) {
+        int farthest = 0;
+        int n = nums.size();
+        
+        for (int i = 0; i < n; i++) {
+            if (i > farthest) return false;
+            farthest = max(farthest, i + nums[i]);
+            if (farthest >= n - 1) return true;
+        }
+        
+        return true;
+    }
+};
diff --git a/Domains/CompetitiveProgramming/Programs/C++/Valid_Parentheses.cpp b/Domains/CompetitiveProgramming/Programs/C++/Valid_Parentheses.cpp
new file mode 100644
index 00000000..98a31fd7
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+++ b/Domains/CompetitiveProgramming/Programs/C++/Valid_Parentheses.cpp
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+/*
+Problem: Valid Parentheses
+Platform: LeetCode
+Problem Code/Number: 20
+Difficulty: Easy
+Link: https://leetcode.com/problems/valid-parentheses/
+
+Problem Statement:
+Given a string s containing just the characters '(', ')', '{', '}', '[' and ']', 
+determine if the input string is valid.
+An input string is valid if:
+1. Open brackets must be closed by the same type of brackets.
+2. Open brackets must be closed in the correct order.
+3. Every close bracket has a corresponding open bracket of the same type.
+
+Approach:
+Use a stack to store opening brackets. Traverse the string:
+- Push opening brackets onto the stack.
+- For closing brackets, check if the top of the stack matches the corresponding opening bracket.
+- If it matches, pop the stack; otherwise, the string is invalid.
+- The string is valid if the stack is empty at the end.
+
+Time Complexity: O(n)
+Space Complexity: O(n)
+
+Contributor: alisha1510
+*/
+
+#include <iostream>
+#include <stack>
+#include <vector>
+#include <string>
+using namespace std;
+
+class Solution {
+public:
+    bool isValid(string s) {
+        stack<char> st;
+        for (char c : s) {
+            if (c == '(' || c == '{' || c == '[') {
+                st.push(c);
+            } else {
+                if (st.empty()) return false;
+                char top = st.top();
+                if ((c == ')' && top != '(') ||
+                    (c == '}' && top != '{') ||
+                    (c == ']' && top != '[')) {
+                    return false;
+                }
+                st.pop();
+            }
+        }
+        return st.empty();
+    }
+};
+
+int main() {
+    Solution sol;
+    vector<string> testCases = {"()", "()[]{}", "(]", "([])", "([)]"};
+    for (auto &s : testCases) {
+        cout << s << " -> " << (sol.isValid(s) ? "Valid" : "Invalid") << endl;
+    }
+    return 0;
+}