diff --git a/algorithms/CPlusPlus/Arrays/.cph/.subarray-product.cpp_7825d78a6905f552e3a1c581ec6241ce.prob b/algorithms/CPlusPlus/Arrays/.cph/.subarray-product.cpp_7825d78a6905f552e3a1c581ec6241ce.prob
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+{"name":"Local: subarray-product","url":"c:\\Users\\HP\\OneDrive\\Desktop\\DSA Contrib\\DSA\\algorithms\\CPlusPlus\\Arrays\\subarray-product.cpp","tests":[{"id":1688797590384,"input":"5","output":""}],"interactive":false,"memoryLimit":1024,"timeLimit":3000,"srcPath":"c:\\Users\\HP\\OneDrive\\Desktop\\DSA Contrib\\DSA\\algorithms\\CPlusPlus\\Arrays\\subarray-product.cpp","group":"local","local":true}
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diff --git a/algorithms/CPlusPlus/Arrays/subarray-product.cpp b/algorithms/CPlusPlus/Arrays/subarray-product.cpp
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+/*
+A C++ (CPP) code to find how many subarrays ( contiguous part of an array) are present in an array whose product is less than a
+ given number 'k'
+
+ e.g.-
+  Input array -  arr[5]= {1, 3, 4 ,5 ,2}
+  Input No.   -   k = 5
+
+  Subarrays of arr having product less than (k=5) are - {1} , {3} , {4} , {2} , {1 , 3}
+*/
+
+/*
+
+Idea - Use Sliding Window Algorithm
+
+Procedure -
+
+Go in a loop ( j=0 to n) and create a sliding window each time starting from current element and increase its size (in second loop)
+till both of these  conditions are true  ( product of all eleemnts in the window < k and iterator < last_index )
+
+Proof -
+ If at any instance , the product of all elements in the current window (subarray) exceeds k , then all further windows's product>k
+ and hence we shift to the element to immediate right and do the same process .
+*/
+
+// Implementation
+
+#include <bits/stdc++.h>
+using namespace std;
+#define ll long long int
+
+// required function
+
+int count_SubArray_Product_Less_Than_K(int a[], int n, int k)
+{
+    // Actual answer
+    int ans = 0;
+
+    for (int j = 0; j < n; j++)
+    {
+        // traverse  each element
+        int i = j;
+
+        // initialize the product as 1 for current window .
+
+        long long int product = 1;
+
+        // making  a window that is expanding continuously till (product < k) AND (counter variable < size of array) .
+        while (i < n && product < k)
+        {
+            product *= a[i];
+            if (product < k)
+            {
+                ans++;
+                i++;
+            }
+        }
+    }
+
+    return ans;
+}
+
+int main()
+{
+
+    // Taking INPUT from user
+    int n;
+
+    cout << "Enter the no. of elements in array : \n";
+    cin >> n;
+
+    int arr[n];
+
+    cout << "Enter the elements : \n";
+
+    for (int j = 0; j < n; j++)
+    {
+        cin >> arr[j];
+    }
+
+    int k;
+
+    cout << "Enter the number k : \n";
+    cin >> k;
+
+    // Calling the required Function
+    int ans = count_SubArray_Product_Less_Than_K(arr, n, k);
+
+    // Giving Output
+    cout << "The Number of subarrays in given array whose product is less than " << k << "are- " << endl;
+    cout << ans << endl;
+
+    return 0;
+}