48. Rotate Image

Difficulty: Medium 

You are given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise).

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

 

Example 1:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[7,4,1],[8,5,2],[9,6,3]]

Example 2:

Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]
Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]

Python Solution:

class Solution(object):
 def rotate(self, matrix):
 for i in range(len(matrix)):
 for j in range(i,len(matrix)):
 matrix[i][j],matrix[j][i]=matrix[j][i],matrix[i][j]
 for i in range(len(matrix)):
 matrix[i]=matrix[i][::-1]
 return matrix

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