Merge pull request #822 from dsrao711/issue_815

Python solution for Next greater element in DSA450

Author: Saviour1001

DSA 450 GFG/next_greater_element.py

@@ -0,0 +1,48 @@
+#https://leetcode.com/problems/next-greater-element-i/submissions/
+
+# Approach : 
+
+# We need to find the next greater element than the current element which is in the right side of the array
+
+# Since Nums1 is a subset of Nums2 , we will be finding the next greater element in Nums2 and keep storing it in a Hashmap 
+# And return only those values which are are present in Nums1 
+
+# since we need the immediate next element greater than the current element 
+# Create a stack to store an element from the nums , compare the top of the stack with the elements from nums2
+# IF the element from nums is greater than the top of the stack , then store this pair in a hashmap 
+# If the element from the nums is lesser than the top of the stack or if the stack is empty , append the element in the stack 
+# After the nums2 is iterated , check if there is any element remaining in the stack 
+# If there is any element remaining , it means that , these elements dont have any other element greater than it (to the right side of the array)
+# So append these elements to the hashmap as keys with values -1
+
+# Lastly , return the list of elements from the hashmap which are present in nums1 i.e the required output
+
+
+class Solution(object):
+    def nextGreaterElement(self, nums1, nums2):
+        """
+        :type nums1: List[int]
+        :type nums2: List[int]
+        :rtype: List[int]
+        """
+        
+        stack = []
+        d = {}
+        
+        for num in nums2 :
+          while stack and stack[-1] < num:
+            top = stack.pop()
+            d[top] = num
+          
+          if not stack or stack[-1] > num:
+            stack.append(num)
+            
+        while(stack):
+          rest = stack.pop()
+          d[rest] = -1
+        
+        return [d[i] for i in nums1]
+    
+    
+# Time Complexity: O(N) 
+# Auxiliary Space: O(N)