11. Container With Most Water

-

Difficulty: Medium

You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).

Find two lines that together with the x-axis form a container, such that the container contains the most water.

Return the maximum amount of water a container can store.

Notice that you may not slant the container.

 

Example 1:

Input: height = [1,8,6,2,5,4,8,3,7]
Output: 49
Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

Example 2:

Input: height = [1,1]
Output: 1
Python Solution:

class Solution: def maxArea(self, height: List[int]) -> int: i=0 j=len(height)-1 ans=0 while i<j: if height[i]<=height[j]: res=height[i]*(j-i) i=i+1 else: res=height[j]*(j-i) j=j-1 if res>ans: ans=res return ans

If you have any doubt regarding code, just ask me varunkapoor.29085@gmail.com

Comments

Post a Comment

Popular posts from this blog

35. Search Insert Position

-
Difficulty: Easy   Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order. You must write an algorithm with  O(log n)  runtime complexity.   Example 1: Input: nums = [1,3,5,6], target = 5 Output: 2 Example 2: Input: nums = [1,3,5,6], target = 2 Output: 1 Example 3: Input: nums = [1,3,5,6], target = 7 Output: 4 Python Solution: class Solution: def searchInsert(self, nums: List[int], target: int) -> int: if target in nums: return nums.index(target) else: nums.append(target) nums.sort() return nums.index(target)
Read more

118. Pascal's Triangle

-
Image
Difficulty: Easy   Given an integer   numRows , return the first numRows of   Pascal's triangle . In  Pascal's triangle , each number is the sum of the two numbers directly above it as shown:   Example 1: Input: numRows = 5 Output: [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]] Example 2: Input: numRows = 1 Output: [[1]] Python Solution : class Solution: def generate(self, numRows: int) -> List[List[int]]: l=[[1],[1,1]] if numRows==1: return [[1]] else: for _ in range(3,numRows+1): m=[1] for i in range(len(l[-1])-1): m.append(l[-1][i]+l[-1][i+1]) m.append(1) l.append(m) return l
Read more

763. Partition Labels

-
Difficulty: Medium  You are given a string  s . We want to partition the string into as many parts as possible so that each letter appears in at most one part. Note that the partition is done so that after concatenating all the parts in order, the resultant string should be  s . Return  a list of integers representing the size of these parts . Example 1: Input: s = "ababcbacadefegdehijhklij" Output: [9,7,8] Explanation: The partition is "ababcbaca", "defegde", "hijhklij". This is a partition so that each letter appears in at most one part. A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits s into less parts. Example 2: Input: s = "eccbbbbdec" Output: [10] Python SOlution: class Solution: def partitionLabels(self, s: str) -> List[int]: l=[0]*26 for i in range(len(s)): l[ord(s[i])-ord('a')]=i start,end=0,0 lst=[] for i
Read more