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Commit 5b009ad

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Merge pull request #2 from 149ps/main
Main
2 parents 01cf981 + b8c7c0e commit 5b009ad

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3 files changed

+67
-25
lines changed

3 files changed

+67
-25
lines changed

‎1046. Last Stone Weight/1046. Last Stone Weight.py‎

Lines changed: 5 additions & 9 deletions
Original file line numberDiff line numberDiff line change
@@ -31,12 +31,8 @@ def lastStoneWeight(self, stones: List[int]) -> int:
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max_heap = [-val for val in stones]
3232
heapq.heapify(max_heap)
3333
while len(max_heap) > 1:
34-
y = (-1) * heapq.heappop(max_heap)
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x = (-1) * heapq.heappop(max_heap)
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if x == y:
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if not max_heap: # consider a case [1,1]
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return 0
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continue
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else:
41-
heapq.heappush(max_heap,x-y)
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return max_heap[0] if max_heap[0] > 0 else (-1)*max_heap[0]
34+
stone1 = (-1) * heapq.heappop(max_heap)
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stone2 = (-1) * heapq.heappop(max_heap)
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if stone1 != stone2:
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heapq.heappush(max_heap,-(stone1-stone2))
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return -max_heap[0] if max_heap else 0

‎523. Continuous Subarray Sum/523. Continuous Subarray Sum.py‎

Lines changed: 8 additions & 16 deletions
Original file line numberDiff line numberDiff line change
@@ -22,20 +22,12 @@
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"""
2323
class Solution:
2424
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
25-
"""
26-
The idea here is if two sums have the same module and they are atleast two indexes apart then we can return true otherwise false.
27-
For [0,2], k=2 prefix sum would be [0,2] and sum2 = 2 and sum1 = 0. sum2 - sum1 can be divided by k and the distance between those two indexes is atleast 2.
28-
"""
29-
hmap = {0:-1} # initially at index -1 the sum would be zero.
30-
for i in range(1,len(nums)):
31-
nums[i] += nums[i-1] # prefix sum
32-
for i in range(len(nums)):
33-
temp = nums[i]
34-
if k:
35-
temp %= k
36-
if temp in hmap.keys():
37-
if i - hmap[temp] >= 2:
38-
return True
25+
hmap,total = {},0
26+
for i,num in enumerate(nums):
27+
total = (total + num) % k
28+
if total == 0 and i > 0: return True # if total % k =0 that means we have a multiple of k present already and check if the array size is greater than or equal to 2.
29+
if total not in hmap:
30+
hmap[total] = i
3931
else:
40-
hmap[temp] =i
41-
return False
32+
ifi-hmap[total] >=2: returnTrue
33+
return False
Lines changed: 54 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,54 @@
1+
"""
2+
You are given the root of a binary tree where each node has a value in the range [0, 25] representing the letters 'a' to 'z'.
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Return the lexicographically smallest string that starts at a leaf of this tree and ends at the root.
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As a reminder, any shorter prefix of a string is lexicographically smaller.
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For example, "ab" is lexicographically smaller than "aba".
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A leaf of a node is a node that has no children.
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Example 1:
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Input: root = [0,1,2,3,4,3,4]
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Output: "dba"
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Example 2:
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Input: root = [25,1,3,1,3,0,2]
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Output: "adz"
23+
Example 3:
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25+
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Input: root = [2,2,1,null,1,0,null,0]
27+
Output: "abc"
28+
29+
30+
Constraints:
31+
32+
The number of nodes in the tree is in the range [1, 8500].
33+
0 <= Node.val <= 25
34+
"""
35+
# Definition for a binary tree node.
36+
# class TreeNode:
37+
# def __init__(self, val=0, left=None, right=None):
38+
# self.val = val
39+
# self.left = left
40+
# self.right = right
41+
class Solution:
42+
def smallestFromLeaf(self, root: Optional[TreeNode]) -> str:
43+
q,result = collections.deque(),None
44+
q.append((root,chr(root.val+97)))
45+
while q:
46+
node,path = q.popleft()
47+
if not node.left and not node.right:
48+
if not result:
49+
result = path
50+
else:
51+
result = min(result,path)
52+
if node.left: q.append((node.left,chr(node.left.val+97)+path))
53+
if node.right: q.append((node.right,chr(node.right.val+97)+path))
54+
return result

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