diff --git a/coding1.java b/coding1.java
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index 00000000..967258f4
--- /dev/null
+++ b/coding1.java
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+// Approach:
+// Use binary search to compare each element with its expected value (index + 1).
+// If the values match, the missing number lies in the right half; otherwise, it lies in the left half.
+// Continue until the search space is exhausted and return the missing number.
+
+// Time Complexity: O(log N)
+// Space Complexity: O(1)
+
+class Solution {
+    int missingNumber(int arr[]) {
+        // code here
+        int n=arr.length;
+        int low=0;
+        int high=n-1;
+        while(low<=high){
+            int mid=low+(high-low)/2;
+            if(arr[mid]==mid+1){
+                low=mid+1;
+            }else{
+                high=mid-1;
+            }
+        }
+        return low+1;
+    }
+}