Created Monotonic _stack.py in Python stack Directory

python/Stack/Monotonic-Stack.py

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+"""
+Next Greater Element
+
+Given an array of integers . Our goal is to  return an answer array which contains 
+Next greater element of every elements in an array. if an element don't have any greater 
+element. we add -1. 
+
+Examples:
+
+numbers = [ 4, 12, 5, 31, 2, 5]
+answer = [ 12, 31, 31, -1, 5, -1]
+
+Explanation:
+
+First element in numbers array is 4  -  for 4 next greater element is 12 . the very next index itself so in answer array 
+                                       we add 12 .
+
+Next element in numbers array is 12 -  for 12 , the next element 5 is not greater than 12 so we move the next index 
+                                       5 is also not greater so we move, 31 is greater.so we add 31 in the answer array.
+
+Next element in numbers array is 5  -  for 5, 31 is next greater element. so add 31 to answer array
+
+
+Next element in numbers array is 31 -  for 31 , there is no greater element , so we add -1
+
+Next element in numbers array is 2  -  for 2 , 5 is a greater element , so we add 5.
+
+Next element in numbers array is 5  -  5 is the last element, so there is no possibility of next greater element so we add -1
+
+
+
+
+How do We approach this problem?
+
+Brute Force approcah:
+      In Brute Force approach we can make 2 for loops. for every element we have to check any other element comes
+      after the element is greater or not . if greater we can add the element and break the loop
+      But this approach takes time complexity of O(N^2) . Can we optimise it?
+
+ Of course we can use Montonic Stack.Don't Worry I wil tell you ,What is Monotonic Stack?
+
+ Montonic Stack is a stack which stores the value either in an increasing order or decreasing order.
+
+ Montonic Stack Approach:
+
+        We are going to use montonic stack of increasing order.
+
+        step 1 : First we are going to create  one empty montonic_stack . we used
+                 python list but we are going to add montonic functionalityto it. Don't be panic, that's very easy we will learn that at the go.
+
+        step 2 : next create an answer array and the size of answer array should be equal to given array . 
+
+        step 3 :  Traverse the given array in a reverse order .
+
+        step 4 :  First we  have to check monotonic stack has elements. if it have, we are going to check the last element in 
+                  monotonic stack is lesser than or equal to the element from given array . if the last value in montonic stack 
+                  is lesser or equal we are going to pop those values. Basically we are trying to remove all lesser values in 
+                  Monotonic stack. Beacause we only want the greater elements
+
+
+        step 5 :  After this operation if a montonic stack is empty .In answer array, in the index of traversed element we are going to put -1. beacuse there is 
+                  no Greater element on the monotonic stack . if its not empty add the last value in monotonic stack to the respective position in answer
+                  array because that is the larger value then our given number.
+
+        step 6 :  add  this traversed  element from given array to monotonic stack.
+
+        step 7 :  Repeat step 4, 5 and 6 until we reach the first the element of given array.
+
+
+
+ Time complexity : O(N)
+ space Complexity: o(N)
+
+
+
+"""
+
+def next_greater(numbers):
+    monotonic_stack = []
+    # intially fill the answer array with any value with  length of an given array
+    answer = [0 for i in range(len(numbers))]
+
+    # Traverse the given array in reverse order
+    for i in range(len(numbers)-1, -1, -1):
+
+        # condition to remove all the lesser values in  montonic stack
+        while monotonic_stack and monotonic_stack[-1] <= numbers[i]:
+            monotonic_stack.pop()
+
+        # after removing all the lesser values if montonic stack is empty add -1 in the position of an element in answer array
+        if monotonic_stack == []:
+            answer[i] = -1
+
+        # else add the last element of monotonic stack to its respective position in answer array
+        else:
+            answer[i] = monotonic_stack[-1]
+
+        # add the traversed element to monotonic stack
+        monotonic_stack.append(numbers[i])
+
+    return answer  # return the answer array
+
+
+if __name__ == "__main__":
+    arr = [4, 12, 5, 31, 2, 5]
+    arr1 = [12, 3, 5, 8, 1, 67]
+    print(next_greater(arr))  # [12, 31, 31, -1, 5, -1]
+    print(next_greater(arr1))  # [67, 5, 8, 67, 67, -1]
+