"Implemented the Java code to find the largest island. It is similar to count the island. But in this we need to keep track of max island and compute its perimeter."
Techzen I. - "Implemented the Java code to find the largest island. It is similar to count the island. But in this we need to keep track of max island and compute its perimeter."See full answer
"bool isValidBST(TreeNode* root, long min = LONGMIN, long max = LONGMAX){
if (root == NULL)
return true;
if (root->val val >= max)
return false;
return isValidBST(root->left, min, root->val) &&
isValidBST(root->right, root->val, max);
}
`"
Alvaro R. - "bool isValidBST(TreeNode* root, long min = LONGMIN, long max = LONGMAX){
if (root == NULL)
return true;
if (root->val val >= max)
return false;
return isValidBST(root->left, min, root->val) &&
isValidBST(root->right, root->val, max);
}
`"See full answer
"from typing import List
def traprainwater(height: List[int]) -> int:
if not height:
return 0
l, r = 0, len(height) - 1
leftMax, rightMax = height[l], height[r]
res = 0
while l < r:
if leftMax < rightMax:
l += 1
leftMax = max(leftMax, height[l])
res += leftMax - height[l]
else:
r -= 1
rightMax = max(rightMax, height[r])
"
Anonymous Roadrunner - "from typing import List
def traprainwater(height: List[int]) -> int:
if not height:
return 0
l, r = 0, len(height) - 1
leftMax, rightMax = height[l], height[r]
res = 0
while l < r:
if leftMax < rightMax:
l += 1
leftMax = max(leftMax, height[l])
res += leftMax - height[l]
else:
r -= 1
rightMax = max(rightMax, height[r])
"See full answer
"Make current as root.
2 while current is not null,
if p and q are less than current,
go left.
If p and q are greater than current,
go right.
else return current.
return null"
Vaibhav D. - "Make current as root.
2 while current is not null,
if p and q are less than current,
go left.
If p and q are greater than current,
go right.
else return current.
return null"See full answer