Need help with Code Rush X question 2!

#	User	Rating
1	tourist	4009
2	jiangly	3831
3	Radewoosh	3646
4	jqdai0815	3620
4	Benq	3620
6	orzdevinwang	3529
7	ecnerwala	3446
8	Um_nik	3396
9	gamegame	3386
10	ksun48	3373

#	User	Contrib.
1	cry	164
1	maomao90	164
3	Um_nik	163
4	atcoder_official	160
5	-is-this-fft-	158
6	awoo	157
7	adamant	156
8	TheScrasse	154
8	nor	154
10	Dominater069	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))

Rev.	By	When	Δ	Comment
en3	iron_nicko	2023-01-28 23:38:58	873	Changed the code
en2	iron_nicko	2023-01-28 22:34:53	50	Spolier-ed the code
en1	iron_nicko	2023-01-28 22:03:44	1681	Initial revision (published)

Rev.

Lang.

When

Comment

en3