[pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

data_structures/arrays/array_leader.py

@@ -2,20 +2,21 @@
 Given an array arr[] of size n, the task is to find all the Leaders in the array.
  An element is a Leader if it is greater than all the elements to its right side.
 
-Note: The rightmost element is always a leader. 
+Note: The rightmost element is always a leader.
 """
 
+
 def array_leader(arr):
     leaders = []
     for i in range(len(arr)):
-        for j in range(i+1, len(arr)):
+        for j in range(i + 1, len(arr)):
             if arr[i] < arr[j]:
                 break
         else:
             leaders.append(arr[i])
-    
+
     return " ".join(map(str, leaders))
 
 
 # Test the function with the provided input
-print(array_leader([16, 17, 4, 3, 5, 2]))  # Expected output: 17 5 2
+print(array_leader([16, 17, 4, 3, 5, 2]))  # Expected output: 17 5 2

data_structures/arrays/max_product_subarray.py

@@ -1,10 +1,11 @@
 """
-Given an integer array, the task is to find the maximum product of any subarray. 
+Given an integer array, the task is to find the maximum product of any subarray.
 Input: arr[] = {-1, -3, -10, 0, 60}
 Output: 60
 Explanation: The subarray with maximum product is {60}.
 """
 
+
 def max_product(arr):
     result = 0
 
@@ -15,5 +16,6 @@ def max_product(arr):
             result = max(result, current_product)
     return result
 
+
 # Test the function with the provided input
-print(max_product([-1, -3, -10, 0, 60]))  # Expected output: 60
+print(max_product([-1, -3, -10, 0, 60]))  # Expected output: 60

data_structures/arrays/max_subarray.py

@@ -12,6 +12,7 @@ Output: 25
 Explanation: The subarray {5, 4, 1, 7, 8} has the largest sum 25.
 """
 
+
 def max_subarray(arr):
     result = arr[0]
 
@@ -22,7 +23,8 @@ def max_subarray(arr):
             result = max(result, current_sum)
     return result
 
+
 # Test the function with the provided inputs
 print(max_subarray([2, 3, -8, 7, -1, 2, 3]))  # Expected output: 11
 print(max_subarray([-2, -4]))  # Expected output: -2
-print(max_subarray([5, 4, 1, 7, 8]))  # Expected output: 25
+print(max_subarray([5, 4, 1, 7, 8]))  # Expected output: 25

data_structures/arrays/missing_number.py

@@ -1,4 +1,4 @@
-'''
+"""
 Given an array arr[] of size n-1 with integers in the range of [1, n], the task is to find the missing number from the first N integers.
 
 Note: There are no duplicates in the list.
@@ -10,16 +10,18 @@ Explanation: Here the size of the array is 8, so the range will be [1, 8]. The m
 Input: arr[] = {1, 2, 3, 5}, n = 5
 Output: 4
 Explanation: Here the size of the array is 4, so the range will be [1, 5]. The missing number between 1 to 5 is 4
-'''
+"""
+
 
 def find_missing_number(arr, n):
-    hash = [0] * (n+1)
+    hash = [0] * (n + 1)
 
     for num in arr:
         hash[num] += 1
-    for i in range(1, (n+1)):
+    for i in range(1, (n + 1)):
         if hash[i] == 0:
             return i
     return -1
 
-print(find_missing_number([1,2,3,4,6,7,8], 8))
+
+print(find_missing_number([1, 2, 3, 4, 6, 7, 8], 8))

data_structures/arrays/rotate_elements.py

@@ -1,6 +1,6 @@
 """
-Given an Array of size N and a value K, 
-around which we need to right rotate the array. 
+Given an Array of size N and a value K,
+around which we need to right rotate the array.
 How do you quickly print the right rotated array?
 
 Input: Array[] = {1, 3, 5, 7, 9}, K = 2.
@@ -9,11 +9,13 @@ Explanation:
 After 1st rotation – {9, 1, 3, 5, 7}After 2nd rotation – {7, 9, 1, 3, 5}
 """
 
+
 def right_rotate(arr, k):
     n = len(arr)
     k = k % n
-    arr = arr[n-k:] + arr[:n-k]
+    arr = arr[n - k :] + arr[: n - k]
     return arr
 
+
 # Test the function with the provided input
 print(right_rotate([1, 3, 5, 7, 9], 2))  # Expected output: [7, 9, 1, 3, 5]