Need help with Code Rush X question 2!

#	User	Rating
1	tourist	4009
2	jiangly	3823
3	Benq	3738
4	Radewoosh	3633
5	jqdai0815	3620
6	orzdevinwang	3529
7	ecnerwala	3446
8	Um_nik	3396
9	ksun48	3390
10	gamegame	3386

#	User	Contrib.
1	cry	167
2	Um_nik	163
3	maomao90	162
3	atcoder_official	162
5	adamant	159
6	-is-this-fft-	158
7	awoo	157
8	TheScrasse	154
9	Dominater069	153
9	nor	153

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 Code

def 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))

iron_nicko's blog