diff --git a/Subarrays.cpp b/Subarrays.cpp
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+++ b/Subarrays.cpp
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+/*C++ program to find total number of
+even-odd subarrays present in given array*/
+#include <bits/stdc++.h>
+using namespace std;
+
+// function that returns the count of subarrays that
+// contain equal number of odd as well as even numbers
+int countSubarrays(int arr[], int n)
+{
+	// initialize difference and answer with 0
+	int difference = 0;
+	int ans = 0;
+
+	// create two auxiliary hash arrays to count frequency
+	// of difference, one array for non-negative difference
+	// and other array for negative difference. Size of these
+	// two auxiliary arrays is 'n+1' because difference can
+	// reach maximum value 'n' as well as minimum value '-n'
+	int hash_positive[n + 1], hash_negative[n + 1];
+
+	// initialize these auxiliary arrays with 0
+	fill_n(hash_positive, n + 1, 0);
+	fill_n(hash_negative, n + 1, 0);
+
+	// since the difference is initially 0, we have to
+	// initialize hash_positive[0] with 1
+	hash_positive[0] = 1;
+
+	// for loop to iterate through whole
+	// array (zero-based indexing is used)
+	for (int i = 0; i < n ; i++)
+	{
+		// incrementing or decrementing difference based on
+		// arr[i] being even or odd, check if arr[i] is odd
+		if (arr[i] & 1 == 1)
+			difference++;
+		else
+			difference--;
+
+		// adding hash value of 'difference' to our answer
+		// as all the previous occurrences of the same
+		// difference value will make even-odd subarray
+		// ending at index 'i'. After that, we will increment
+		// hash array for that 'difference' value for
+		// its occurrence at index 'i'. if difference is
+		// negative then use hash_negative
+		if (difference < 0)
+		{
+			ans += hash_negative[-difference];
+			hash_negative[-difference]++;
+		}
+		
+		// else use hash_positive
+		else
+		{
+			ans += hash_positive[difference];
+			hash_positive[difference]++;
+		}
+	}
+
+	// return total number of even-odd subarrays
+	return ans;
+}
+
+// Driver code
+int main()
+{
+	int arr[] = {3, 4, 6, 8, 1, 10, 5, 7};
+	int n = sizeof(arr) / sizeof(arr[0]);
+	
+	// Printing total number of even-odd subarrays
+	cout << "Total Number of Even-Odd subarrays"
+		" are " << countSubarrays(arr,n);
+
+	return 0;
+}