Google_imp_easy

Question 1

Flipping an image(reverse a row and invert 0’s to 1’s)

Answer 1

class Solution(object):
    def flipAndInvertImage(self, A):
        """
        :type A: List[List[int]]
        :rtype: List[List[int]]
        """
        return [[1-i for i in row[::-1]] for row in A]

Question 2

Island Perimeter

Answer 2

# p+=4 in finding island and in for loop of neighbiurs per-=1
class Solution(object):
    def islandPerimeter(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        per=0

    r=len(grid)
    c=len(grid[0])
    x=[0,0,-1,1]
    y=[-1,1,0,0]

    for i in range(r):
        for j in range(c):
            if grid[i][j]==1:
                per+=4

                for k in range(4):
                    nx=i+x[k]
                    ny=j+y[k]

                    if nx>=0 and nx=0 and ny

Question 3

Disappeared in array

Answer 3

return [ i for i in range(1,n+1) if i not in set(k) ]

Question 4

Power of 2

Power of 3

Answer 4

# if n==0 return False
while n if n==1 return true,elif n%2!=0 return false else n=n/2

same instead of 2 use 3

Question 5

Reverse vowels of a string

(make a list and append string which are vowels with indexes in tuples and then loop through this list to half and interchange values while inserting to list )

Answer 5

s=list(s)
vo=[]
for i,j in enumerate(s):
if j.lower() in [‘a’,’e’,’i’,’o’,’u’]:
vo.append((i,j))
for i in range(len(vo)//2):
s[vo[i][0]],s[vo[~i][0]]=vo[~i][1],vo[i][1]

    return ''.join(s)

Question 6

Plus one

Answer 6

for i in range(len(digits)-1,-1,-1):
            if digits[i]<9:
                digits[i]+=1
                break
            else:
                digits[i]=0

    if sum(digits)==0:

        return [1]+digits

    return digits

Question 7

Valid Parenthesis (if multiple parentheses uses stack)

Answer 7

stack=[]
        for i in s:
            if len(stack)==0:
                stack.append(i)
            else:
                if (stack[-1]=='(' and i==')') or (stack[-1]=='[' and i==']') or (stack[-1]=='{' and i=='}'):
                    stack.pop()
                else:
                    stack.append(i)
        if len(stack)==0:
            return True
        else:
            return False