1019 Next Greater Node In Linked List.py

Author: rajeevranjancom

1019 Next Greater Node In Linked List.py

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+#!/usr/bin/python3
+"""
+We are given a linked list with head as the first node.  Let's number the nodes
+in the list: node_1, node_2, node_3, ... etc.
+
+Each node may have a next larger value: for node_i, next_larger(node_i) is the
+node_j.val such that j > i, node_j.val > node_i.val, and j is the smallest
+possible choice.  If such a j does not exist, the next larger value is 0.
+
+Return an array of integers answer, where answer[i] = next_larger(node_{i+1}).
+
+Note that in the example inputs (not outputs) below, arrays such as [2,1,5]
+represent the serialization of a linked list with a head node value of 2, second
+node value of 1, and third node value of 5.
+
+Example 1:
+Input: [2,1,5]
+Output: [5,5,0]
+
+Example 2:
+Input: [2,7,4,3,5]
+Output: [7,0,5,5,0]
+
+Example 3:
+Input: [1,7,5,1,9,2,5,1]
+Output: [7,9,9,9,0,5,0,0]
+
+Note:
+1 <= node.val <= 10^9 for each node in the linked list.
+The given list has length in the range [0, 10000].
+"""
+
+# Definition for singly-linked list.
+class ListNode:
+    def __init__(self, x):
+        self.val = x
+        self.next = None
+
+
+from typing import List
+
+
+class Solution:
+    def nextLargerNodes(self, head: ListNode) -> List[int]:
+        """
+        If input is an array, use stack from right to left. Maintain a decreasing stack
+
+        How to make it one-pass? Maintain a stack from left to right for the element
+        waiting for the next larger node
+        """
+        ret = []
+        stk = []  # [[index, value]]
+        i = 0
+        cur = head
+        while cur:
+            while stk and stk[-1][1] < cur.val:
+                idx, _ = stk.pop()
+                ret[idx] = cur.val
+
+            stk.append([i, cur.val])
+            ret.append(0)
+            cur = cur.next
+            i += 1
+
+        return ret