Added another array problem named reverse_array_in_groups

Arrays/reverse_array_in_groups.py

@@ -0,0 +1,58 @@
+'''
+Reverse Array in groups
+
+Given an Array of size N. Modify the array in-place by reversing every sub-array of size K.
+
+Note : If at any instance, there are no more subarrays of size greater than or equal to K, 
+        then reverse the last subarray (irrespective of its size).
+        You shouldn't return any array, modify the given array in-place.
+
+Input: [1, 2, 3, 4, 5]
+Output: [3, 2, 1, 5, 4]
+Output explanation: 1, 2, 3 => 3, 2, 1; 4, 5 => 5, 4
+
+=========================================
+Find the start and end of each sublist and reverse it in-place.
+    Time Complexity:    O(N)
+    Space Complexity:   O(1)
+'''
+
+
+############
+# Solution #
+############
+
+def reverse_array_in_groups(arr, k):
+    # Initialize a variable to keep track of sub arrays of length k
+    i = 0
+    n = len(arr)
+    while i<n:
+        # The sub-array can be of length k or less so the right index in chosen accordingly
+        left = i
+        right = min(i+k-1, n-1)
+        while left < right:
+            # After finding the required indices we reverse the array by swapping two elements at a time
+            arr[left], arr[right] = arr[right], arr[left]
+            # Increasing the left index and decreasing the right index.
+            left += 1
+            right -= 1
+        # After succesfully reversing one sub-array move to next by increasing i by a value of k.
+        i += k
+    return arr
+
+
+###########
+# Testing #
+###########
+
+# Test 1
+# Correct result => [3, 2, 1, 5, 4]
+print(reverse_array_in_groups([1, 2, 3, 4, 5], 3))
+
+# Test 2
+# Correct result => [2, 3, 4, 5, 1]
+print(reverse_array_in_groups([5, 4, 3, 2, 1], 4))
+
+# Test 3
+# Correct result => [10, 7, 5, 7, 2, 4, 3, 1, 8]
+print(reverse_array_in_groups([5, 7, 10, 4, 2, 7, 8, 1, 3], 3))