diff --git a/Sliding Window Maximum b/Sliding Window Maximum
new file mode 100644
index 00000000..c0313b77
--- /dev/null
+++ b/Sliding Window Maximum	
@@ -0,0 +1,50 @@
+CODE:
+class Solution {
+public:
+    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
+        vector<int> ans;
+        int n=nums.size();
+        deque<int>dq;
+        for(int i=0;i<n;i++){
+            while(!dq.empty() && dq.front()<=i-k)dq.pop_front();
+            while(!dq.empty() && nums[dq.back()]<=nums[i])dq.pop_back();
+            dq.push_back(i);
+            if(i>=k-1)ans.push_back(nums[dq.front()]);
+        }
+        return ans;
+    }
+};
+
+
+
+QUESTION:
+You are given an array of integers nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position.
+
+Return the max sliding window.
+
+ 
+
+Example 1:
+
+Input: nums = [1,3,-1,-3,5,3,6,7], k = 3
+Output: [3,3,5,5,6,7]
+Explanation: 
+Window position                Max
+---------------               -----
+[1  3  -1] -3  5  3  6  7       3
+ 1 [3  -1  -3] 5  3  6  7       3
+ 1  3 [-1  -3  5] 3  6  7       5
+ 1  3  -1 [-3  5  3] 6  7       5
+ 1  3  -1  -3 [5  3  6] 7       6
+ 1  3  -1  -3  5 [3  6  7]      7
+Example 2:
+
+Input: nums = [1], k = 1
+Output: [1]
+ 
+
+Constraints:
+
+1 <= nums.length <= 105
+-104 <= nums[i] <= 104
+1 <= k <= nums.length