diff --git a/coding_freshmen/C++/Aniket_Tripathy/PRODUCT_FINDER.cpp b/coding_freshmen/C++/Aniket_Tripathy/PRODUCT_FINDER.cpp
new file mode 100644
index 0000000..da18cb2
--- /dev/null
+++ b/coding_freshmen/C++/Aniket_Tripathy/PRODUCT_FINDER.cpp
@@ -0,0 +1,35 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+
+int main(){
+  int n;
+  cin>>n;
+  int arr[n];
+  for(int i=0;i<n;i++){
+    cin >> arr[i];
+  }
+  int pivot;
+  cin >> pivot;
+  int prod = 1;
+
+  //edge case
+  if(pivot==0){
+    for (int i = 0; i < n;i++){
+      if(arr[i]==0){
+        continue;
+      }else{
+        prod *= arr[i];
+      }
+    }
+    cout << prod << endl;
+    exit(0);
+  }
+
+  //normal case
+  for(int i=0;i<n;i++){
+    prod *= arr[i];
+  }
+  cout << prod / pivot << endl;
+  return 0;
+}
diff --git a/coding_freshmen/C++/Aniket_Tripathy/README.md b/coding_freshmen/C++/Aniket_Tripathy/README.md
new file mode 100644
index 0000000..eeadb8f
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+++ b/coding_freshmen/C++/Aniket_Tripathy/README.md
@@ -0,0 +1,12 @@
+# Problem Title: PRODUCT_FINDER
+  Find the product of the elements of the array except the given pivot element
+
+# Problem Explanation 🚀
+We need to find the product of all elements of the array while skipping the pivot element 
+
+# Intuition and Logic🧠
+*  Assuming that all the elements in the array are unique we can simply calculate the products of all the elements in a prod variable, and when we encounter the pivot element we can simply use the continue; statememt to skip that element.
+
+# Time Complexity and Space Complexity
+* Time Complexity: O(n)
+* Space Complexity: O(1)