Change two lines of code of: def solution(A, K): n = len(A) for i in range(n - 1): if (A[i] + 1 < A[i + 1]): return False if (A[0] != 1 and A[n - 1] != K): return False else: return True SO that it solves the question : you are given a implementation of a function def solution (A,K) This function, given a non-empty array A of N integers (sorted in non-decreasing order) and integer K, checks whether A contains numbers 1,2..K (every numbers from 1 to K at least once and no other numbers (for example given array A, K=3 ; A(0)=1, A(1)=1, A(2)=2, A(3)=3, A(4)=3 then the function should return True while for the following array A, K=2; A(0)=1, A(1)=1, A(2)=3 then the function should return False.) Assume that: -N and K are integers within the range (1....300,000) -each element of array A is an integer within the range (0...1,000,000,000) -array A is sorted in non-decreasing order
Change two lines of code of:
def solution(A, K):
n = len(A)
for i in range(n - 1):
if (A[i] + 1 < A[i + 1]):
return False
if (A[0] != 1 and A[n - 1] != K):
return False
else:
return True
SO that it solves the question
: you are given a implementation of a function def solution (A,K) This function, given a non-empty array A of N integers (sorted in non-decreasing order) and integer K, checks whether A contains numbers 1,2..K (every numbers from 1 to K at least once and no other numbers (for example given array A, K=3 ; A(0)=1, A(1)=1, A(2)=2, A(3)=3, A(4)=3 then the function should return True while for the following array A, K=2; A(0)=1, A(1)=1, A(2)=3 then the function should return False.) Assume that: -N and K are integers within the range (1....300,000) -each element of array A is an integer within the range (0...1,000,000,000) -array A is sorted in non-decreasing order
Trending now
This is a popular solution!
Step by step
Solved in 2 steps
There is too much lines changed in this case 5 lines get changed