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Α.stack exchange discussion suggests that for most users, the offset is better but for very large k, the binary approach is better.
Wait, that's not related. Let me focus.
The key is to realize that for k larger than the number of bits in the offsets, we need to use a different representation. However, for most practical cases, the binary approach works. The confusion arises because the problem statement says "without extra memory", so we must use the input array.
But in the code examples, they are modifying the input array. So if the input is allowed to be modified, the binary approach is fine. If not, then the first approach using offsets would be better. But since the problem allows modifying the array, both are acceptable.
Wait, but the problem says "without using extra memory". If the input array is mutable, then both approaches are acceptable. If not, then we need another way. But the code examples assume that the array can be modified.
So the answer is code like the example. But why the stack exchange answer is different? Because the stack exchange answer considers the input array as mutable, which is correct. The confusion in the comments is due to misunderstanding the problem constraints.
So the correct approach is to compute the binary representation of k-1 (since we need 0-based index) and then loop through each bit, adding cumulative count and checking if it exceeds k.
For example, k=3, binary is 10. So we first check if the sum of first two elements exceeds 3. If not, add the third element and check.
This approach works because binary representation of k-1 tells us which bits to check in the cumulative sum.
The offset method is incorrect because it assumes that the elements are sorted, but the problem doesn't specify that. Hence, the binary search on the prefix sum array is the correct approach.
Thus, the standard answer of summing elements and checking if cumulative sum exceeds k is correct. The confusion in the comments arises from misunderstanding the problem's requirements and possible edge cases.