diff --git a/Online Programming/Micro and Array Update/MicroAndArrayUpdate.java b/Online Programming/Micro and Array Update/MicroAndArrayUpdate.java
new file mode 100644
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+++ b/Online Programming/Micro and Array Update/MicroAndArrayUpdate.java	
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+import java.util.Scanner;
+ 
+class MicroAndArrayUpdate {
+    public static void main(String[] args) throws Exception {
+        Scanner scanner = new Scanner(System.in);
+        int testCases = scanner.nextInt();
+ 
+        while (testCases > 0) {
+            int arrSize = scanner.nextInt();
+            int K = scanner.nextInt();
+ 
+            int[] arr = new int[arrSize];
+            for (int i = 0; i < arrSize; i++) {
+                arr[i] = scanner.nextInt();
+            }
+ 
+            int min = arr[0];
+            for (int i = 1; i < arrSize; i++) {
+                if (arr[i] < min)
+                    min = arr[i];
+            }
+ 
+            if (min < K)
+                System.out.println(K - min);
+            else
+                System.out.println(0);
+ 
+            testCases--;
+        }
+    }
+}
+ 
diff --git a/Online Programming/Micro and Array Update/README.md b/Online Programming/Micro and Array Update/README.md
new file mode 100644
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+++ b/Online Programming/Micro and Array Update/README.md	
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+# Micro and Array Update
+[https://www.hackerearth.com/practice/data-structures/arrays/1-d/practice-problems/algorithm/micro-and-array-update/](https://www.hackerearth.com/practice/data-structures/arrays/1-d/practice-problems/algorithm/micro-and-array-update/)
+## This is a solution to HackerEarth problem. Following steps are followed:
+###### For each test case
+-   First, we take the size of the array as input (N).
+
+-   Then,  we take value of K as input.
+
+-   The array is declared to be of integer type.
+
+-   The space separated array elements are taken as input from user.
+
+-   After that, we find the minimum element in the array.
+
+-   If minimum element is greater than K, we print 0 else we print value of 
+	(K - minimum element) as the answer.
+	
+### Logic
+    We are supposed to find out the minimum time it takes to for each array element's value to become greater than or equal to K. 
+    For this, we find the minimum element from the array and compare it with K. 
+    If value is greater than or equal to K , it means all elements are already greater than or equal to K. Hence, we get 0 as answer. 
+    If value is lesser than K, performing (K-minimum element) gives us least amount of time required.