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‎Challenges/2021/January-LeetCoding-Challenge.md‎

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|1646.|[Get Maximum in Generated Array](https://leetcode.com/problems/get-maximum-in-generated-array/)|[Python](/Easy/1646.GetMaximuminGeneratedArray.py)|Easy|
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|215.|[Kth Largest Element in an Array](https://leetcode.com/problems/kth-largest-element-in-an-array/)|[Python](/Medium/215.KthLargestElementinanArray.py)|Medium|
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|1641.|[Count Sorted Vowel Strings](https://leetcode.com/problems/count-sorted-vowel-strings/)|[Python](/Medium/1641.CountSortedVowelStrings.py)|Medium|
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|1679.|[Max Number of K-Sum Pairs](https://leetcode.com/problems/max-number-of-k-sum-pairs/)|[Python](/Medium/1679.MaxNumberofK-SumPairs.py)|Medium|
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‎Challenges/README.md‎

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- [December LeetCoding Challenge](/Challenges/2020/December-LeetCoding-Challenge.md) - 27/31
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2021:
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- [January LeetCoding Challenge](/Challenges/2021/January-LeetCoding-Challenge.md) - 15/31
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- [January LeetCoding Challenge](/Challenges/2021/January-LeetCoding-Challenge.md) - 16/31

‎Easy/1725.NumberOfRectanglesThatCanFormTheLargestSquare.py‎

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'''
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You are given an array rectangles where rectangles[i] = [li, wi]
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represents the ith rectangle of length li and width wi.
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You can cut the ith rectangle to form a square with a
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side length of k if both k <= li and k <= wi. For
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example, if you have a rectangle [4,6], you can cut it
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to get a square with a side length of at most 4.
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Let maxLen be the side length of the largest square you
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can obtain from any of the given rectangles.
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Return the number of rectangles that can make a square
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with a side length of maxLen.
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Example:
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Input: rectangles = [[5,8],[3,9],[5,12],[16,5]]
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Output: 3
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Explanation: The largest squares you can get from each
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rectangle are of lengths [5,3,5,5].
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The largest possible square is of length 5,
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and you can get it out of 3 rectangles.
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Example:
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Input: rectangles = [[2,3],[3,7],[4,3],[3,7]]
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Output: 3
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Constraints:
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-1 <= rectangles.length <= 1000
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-rectangles[i].length == 2
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-1 <= li, wi <= 10^9
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-li != wi
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'''
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#Difficulty: Easy
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#68 / 68 test cases passed.
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#Runtime: 176 ms
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#Memory Usage: 14.8 MB
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#Runtime: 176 ms, faster than 100.00% of Python3 online submissions for Number Of Rectangles That Can Form The Largest Square.
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#Memory Usage: 14.8 MB, less than 66.67% of Python3 online submissions for Number Of Rectangles That Can Form The Largest Square.
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from collections import defaultdict
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class Solution:
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def countGoodRectangles(self, rectangles: List[List[int]]) -> int:
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squares = defaultdict(int)
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for rectangle in rectangles:
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squares[min(rectangle)] += 1
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return squares[max(squares)]

‎Medium/1019.NextGreaterNodeInLinkedList.py‎

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'''
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We are given a linked list with head as the first node.
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Let's number the nodes in the list: node_1, node_2,
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node_3, ... etc.
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Each node may have a next larger value: for node_i,
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next_larger(node_i) is the node_j.val such that j > i,
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node_j.val > node_i.val, and j is the smallest possible
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choice. If such a j does not exist, the next larger
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value is 0.
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Return an array of integers answer, where answer[i] =
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next_larger(node_{i+1}).
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Note that in the example inputs (not outputs) below,
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arrays such as [2,1,5] represent the serialization of
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a linked list with a head node value of 2, second node
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value of 1, and third node value of 5.
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Example:
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Input: [2,1,5]
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Output: [5,5,0]
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Example:
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Input: [2,7,4,3,5]
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Output: [7,0,5,5,0]
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Example:
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Input: [1,7,5,1,9,2,5,1]
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Output: [7,9,9,9,0,5,0,0]
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Note:
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1. 1 <= node.val <= 10^9 for each node in the linked l
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ist.
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2. The given list has length in the range [0, 10000].
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'''
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#Difficulty: Medium
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#76 / 76 test cases passed.
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#Runtime: 320 ms
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#Memory Usage: 18.6 MB
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#Runtime: 320 ms, faster than 66.26% of Python3 online submissions for Next Greater Node In Linked List.
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#Memory Usage: 18.6 MB, less than 66.04% of Python3 online submissions for Next Greater Node In Linked List.
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, val=0, next=None):
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# self.val = val
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# self.next = next
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class Solution:
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def nextLargerNodes(self, head: ListNode) -> List[int]:
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result = []
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stack = []
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length = 0
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vals = []
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while head:
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result.append(0)
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length += 1
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vals.append(head.val)
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head = head.next
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for i in range(length):
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if stack:
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while stack and vals[stack[-1]] < vals[i]:
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result[stack.pop()] = vals[i]
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stack.append(i)
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return result

‎Medium/1448.CountGoodNodesinBinaryTree(optimized).py‎

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'''
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Given a binary tree root, a node X in the tree is named
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good if in the path from root to X there are no nodes
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with a value greater than X.
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Return the number of good nodes in the binary tree.
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Example:
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3
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/ \
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1 4
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/ / \
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3 1 5
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Input: root = [3,1,4,3,null,1,5]
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Output: 4
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Explanation: - Nodes in blue are good.
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- Root Node (3) is always a good node.
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- Node 4 -> (3,4) is the maximum value in
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the path starting from the root.
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- Node 5 -> (3,4,5) is the maximum value
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in the path
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- Node 3 -> (3,1,3) is the maximum value
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in the path.
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Example:
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3
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/
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3
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/ \
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4 2
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Input: root = [3,3,null,4,2]
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Output: 3
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Explanation: - Node 2 -> (3, 3, 2) is not good, because
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"3" is higher than it.
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Example:
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Input: root = [1]
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Output: 1
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Explanation: Root is considered as good.
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Constraints:
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- The number of nodes in the binary tree is in the
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range [1, 10^5].
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- Each node's value is between [-10^4, 10^4].
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'''
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#Difficulty: Medium
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#63 / 63 test cases passed.
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#Runtime: 228 ms
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#Memory Usage: 33.5 MB
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#Runtime: 228 ms, faster than 92.42% of Python3 online submissions for Count Good Nodes in Binary Tree.
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#Memory Usage: 33.5 MB, less than 43.74% of Python3 online submissions for Count Good Nodes in Binary Tree.
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def goodNodes(self, root: TreeNode) -> int:
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self.count = 0
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self.dfs(root, float(-inf))
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return self.count
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def dfs(self, root, max_val):
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if not root:
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return
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if root.val >= max_val:
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self.count += 1
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max_val = root.val
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self.dfs(root.left, max_val)
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self.dfs(root.right, max_val)

‎Medium/1448.CountGoodNodesinBinaryTree.py‎

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'''
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Given a binary tree root, a node X in the tree is named
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good if in the path from root to X there are no nodes
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with a value greater than X.
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Return the number of good nodes in the binary tree.
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Example:
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3
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/ \
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1 4
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/ / \
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3 1 5
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Input: root = [3,1,4,3,null,1,5]
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Output: 4
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Explanation: - Nodes in blue are good.
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- Root Node (3) is always a good node.
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- Node 4 -> (3,4) is the maximum value in
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the path starting from the root.
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- Node 5 -> (3,4,5) is the maximum value
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in the path
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- Node 3 -> (3,1,3) is the maximum value
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in the path.
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Example:
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3
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/
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3
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/ \
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4 2
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Input: root = [3,3,null,4,2]
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Output: 3
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Explanation: - Node 2 -> (3, 3, 2) is not good, because
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"3" is higher than it.
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Example:
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Input: root = [1]
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Output: 1
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Explanation: Root is considered as good.
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Constraints:
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- The number of nodes in the binary tree is in the
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range [1, 10^5].
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- Each node's value is between [-10^4, 10^4].
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'''
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#Difficulty: Medium
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#63 / 63 test cases passed.
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#Runtime: 304 ms
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#Memory Usage: 33 MB
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#Runtime: 304 ms, faster than 16.81% of Python3 online submissions for Count Good Nodes in Binary Tree.
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#Memory Usage: 33 MB, less than 79.45% of Python3 online submissions for Count Good Nodes in Binary Tree.
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def goodNodes(self, root: TreeNode) -> int:
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self.count = 1
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self.dfs(root, float(-inf))
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return self.count
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def dfs(self, root, max_val):
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if not root:
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return
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max_val = max(max_val, root.val)
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if root.left:
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if root.left.val >= max_val:
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self.count += 1
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self.dfs(root.left, max_val)
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if root.right:
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if root.right.val >= max_val:
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self.count += 1
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self.dfs(root.right, max_val)

‎README.md‎

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# Python solutions of LeetCode problems.
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![Language](https://img.shields.io/badge/language-Python-blue.svg)&nbsp;
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![Problems Solved](https://img.shields.io/badge/problems%20solved-491%2F1567-orange)&nbsp;
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![Problems Solved](https://img.shields.io/badge/problems%20solved-494%2F1571-orange)&nbsp;
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[![License](https://img.shields.io/badge/license-MIT-green.svg)](./LICENSE)&nbsp;
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![Update](https://img.shields.io/badge/update-Daily-brightgreen.svg)&nbsp;
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<br><br>
@@ -19,7 +19,7 @@ In this repository provided my Python solutions of LeetCode problems.
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- [December LeetCoding Challenge](/Challenges/2020/December-LeetCoding-Challenge.md) - 27/31
2020

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2021:
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- [January LeetCoding Challenge](/Challenges/2021/January-LeetCoding-Challenge.md) - 15/31
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- [January LeetCoding Challenge](/Challenges/2021/January-LeetCoding-Challenge.md) - 16/31
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<br><br>
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## Solutions
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*P.S. If you like this, please leave me a star.* ★
@@ -355,6 +355,7 @@ In this repository provided my Python solutions of LeetCode problems.
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|1009.|[Complement of Base 10 Integer](https://leetcode.com/problems/complement-of-base-10-integer/)|[Python](/Easy/1009.ComplementofBase10Integer.py)|Easy||
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|1010.|[Pairs of Songs With Total Durations Divisible by 60](https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/)|[Python](/Medium/1010.PairsofSongsWithTotalDurationsDivisibleby60.py)|Medium||
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|1015.|[Smallest Integer Divisible by K](https://leetcode.com/problems/smallest-integer-divisible-by-k/)|[Python](/Medium/1015.SmallestIntegerDivisiblebyK.py)|Medium|counting lenght, Brute Force|
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|1019.|[Next Greater Node In Linked List](https://leetcode.com/problems/next-greater-node-in-linked-list/)|[Python](/Medium/1019.NextGreaterNodeInLinkedList.py)|Medium||
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|1021.|[Remove Outermost Parentheses](https://leetcode.com/problems/remove-outermost-parentheses/)|[Python](/Easy/1021.RemoveOutermostParentheses.py)|Easy|Overcode|
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|1022.|[Sum of Root To Leaf Binary Numbers](https://leetcode.com/problems/sum-of-root-to-leaf-binary-numbers/)|[Python](/Easy/1022.SumofRootToLeafBinaryNumbers.py)|Easy|`DFS`, `Recursion`|
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|1025.|[Divisor Game](https://leetcode.com/problems/divisor-game/)|[Python](/Easy/1025.DivisorGame.py)|Easy||
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|1437.|[Check If All 1's Are at Least Length K Places Away](https://leetcode.com/problems/check-if-all-1s-are-at-least-length-k-places-away/)|[Python](/Medium/1437.CheckIfAll1sAreatLeastLengthKPlacesAway.py)|Medium||
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|1441.|[Build an Array With Stack Operations](https://leetcode.com/problems/build-an-array-with-stack-operations/)|[Python](/Easy/1441.BuildanArrayWithStackOperations.py)|Easy|`List`|
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|1446.|[Consecutive Characters](https://leetcode.com/problems/consecutive-characters/)|[Python](/Easy/1446.ConsecutiveCharacters.py)|Easy|`Two pointers`|
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|1448.|[Count Good Nodes in Binary Tree](https://leetcode.com/problems/count-good-nodes-in-binary-tree/)|[Python](/Medium/1448.CountGoodNodesinBinaryTree.py)|Medium|`DFS`, `Binary Tree`|
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|1448.|[Count Good Nodes in Binary Tree](https://leetcode.com/problems/count-good-nodes-in-binary-tree/)|[Python](/Medium/1448.CountGoodNodesinBinaryTree(optimized).py)|Medium|optimized `DFS`, `Binary Tree`|
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|1450.|[Number of Students Doing Homework at a Given Time](https://leetcode.com/problems/number-of-students-doing-homework-at-a-given-time/)|[Python](/Easy/1450.NumberofStudentsDoingHomeworkataGivenTime.py)|Easy|`Enumerate`|
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|1455.|[Check If a Word Occurs As a Prefix of Any Word in a Sentence](https://leetcode.com/problems/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence/)|[Python](/Easy/1455.CheckIfaWordOccursAsaPrefixofAnyWordinaSentence.py)|Easy|`Enumerate`|
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|1457.|[Pseudo-Palindromic Paths in a Binary Tree](https://leetcode.com/problems/pseudo-palindromic-paths-in-a-binary-tree/)|[Python](/Medium/1457.Pseudo-PalindromicPathsinaBinaryTree.py)|Medium|`DFS`, `recursion`|
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|1716.|[Calculate Money in Leetcode Bank](https://leetcode.com/problems/calculate-money-in-leetcode-bank/)|[Python](/Easy/1716.CalculateMoneyinLeetcodeBank.py)|Easy||
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|1720.|[Decode XORed Array](https://leetcode.com/problems/decode-xored-array/)|[Python](/Easy/1720.DecodeXORedArray.py)|Easy||
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|1721.|[Swapping Nodes in a Linked List](https://leetcode.com/problems/swapping-nodes-in-a-linked-list/)|[Python](/Medium/1721.SwappingNodesinaLinkedList.py)|Medium|`Linked List`, `Queue`|
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|1725.|[Number Of Rectangles That Can Form The Largest Square](https://leetcode.com/problems/number-of-rectangles-that-can-form-the-largest-square/)|[Python](/Easy/1725.NumberOfRectanglesThatCanFormTheLargestSquare.py)|Easy|`collections.defaultdict`|
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<b><a href="#python-solutions-of-leetcode-problems">Back to Top</a></b>

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