No. For each achievable length, only the smallest ending value matters for future extensions.

Many subsequences of equal length compete. Keep only the one with the smallest tail because it is never harder to extend. Tails across lengths are ordered, enabling binary search.

What the function promises: Return the length of a strictly increasing subsequence. Empty input returns zero.

def lis_length(values):
    tails = []
    for value in values:
        left, right = 0, len(tails)
        while left < right:
            middle = (left+right)//2
            if tails[middle] >= value: right = middle
            else: left = middle+1
        if left == len(tails): tails.append(value)
        else: tails[left] = value
    return len(tails)

assert lis_length([10,9,2,5,3,7,101,18]) == 4
assert lis_length([2,2,2]) == 1

Work and storage: O(n log n) time and O(n) space.

When to recognize it: For each achievable length, only the smallest ending value matters for future extensions.

Check before applying it: Using upper bound instead of lower bound changes strict increasing into nondecreasing behavior.

Store predecessor indices and which input index currently realizes each tails length.