Hey, I know the contest was postponed, but I was trying to solve this question:
Question:
Consider the integer sequence A[], where the elements are first N natural numbers in order.
You are now given two integers, L and S. Determine whether there exists a subarray with length L and sum S after removing at most one element from A.
A subarray of an array is a non-empty sequence obtained by removing zero or more elements from the front of the array, and zero or more elements from the back of the array.
1 <= N <= 10^9
1 <= L <= N - 1
I came up with something like this:
My Code
def solve(n, l, s):
lo = 1
hi = n
def helper(i):
one = i * (i+1) >> 1
two = 0
if i - l > 0:
two = (i-l)*(i-l+1) >> 1
return one - two
for _ in range(100):
mid = ((hi - lo) >> 1) + lo
val = helper(mid)
if abs(s - val) < 2:
if lo == 1:
if s - val >= 0:
return "YES"
elif mid == n:
if s - val <= 0:
return "YES"
else:
return "YES"
if s > val:
lo = mid + 1
elif s < val:
hi = mid - 1
return "NO"
for _ in range(int(input())):
n, l, s = map(int, input().split())
print(solve(n, l, s))
Test Cases with the answers:
3
5 3 11 # YES
5 3 5 # NO
5 3 6 # YES
My code seems to pass the initial test cases, but I'd like to know if this will work for all cases. Please let me know if this is correct.
Updated Codedef solve(n, l, s):
lo = l
hi = n
def helper(i):
one = i * (i+1) >> 1
two = 0
if i - l > 0:
two = (i-l)*(i-l+1) >> 1
return one - two
for _ in range(100):
if hi < lo: break
mid = ((hi - lo) >> 1) + lo
val = helper(mid)
if abs(s - val) < 2:
if lo == l:
if s - val >= 0:
return "YES"
elif mid == n:
if s - val <= 0:
return "YES"
else:
return "YES"
if s > val:
lo = mid + 1
elif s < val:
hi = mid - 1
return "NO"
for _ in range(int(input())):
n, l, s = map(int, input().split())
print(solve(n, l, s))
