0864MediumString

Fewest Lifts to Order the Bead String

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Trains the technique from

LeetCode 3863Minimum Operations to Sort a String

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A string of beads is given as s, each bead marked with a lowercase letter.

One lift takes a single bead off the string and threads it back on at any position, including either end. The other beads keep their order.

The string is ordered when its letters read in non-decreasing order from one end to the other. Return the fewest lifts that leave it ordered.

Examples

Example 1

Input: s = "cba"
Output: 2

Leaving the bead marked "a" alone and lifting the other two, threading "b" after it and then "c" after that, orders the string in two lifts.

Example 2

Input: s = "abab"
Output: 1

The beads reading "a", "a" and "b" can be left alone as a non-decreasing run, so only the remaining bead has to be lifted.

Example 3

Input: s = "aabbcc"
Output: 0

The string already reads in non-decreasing order, so no bead has to move.

Constraints

1 <= s.length <= 10^5
s consists of lowercase English letters only

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def fewest_lifts(s: str) -> int:

Java

public int fewestLifts(String s)

The intended approach

Time O(n log n) for n beads: one binary search per bead.Space O(n) for the list of tails.

The beads never lifted keep their relative order, so whatever they read must already be non-decreasing, and every lifted bead can be threaded exactly where it belongs. So the fewest lifts is the bead count minus the largest number that can be left alone, which is the longest non-decreasing run that can be picked out of the string without reordering. Find that length by keeping the smallest possible tail letter for each achievable run length: for each bead, replace the first tail strictly greater than it, or extend the list when none is.

Pythonentry point: fewest_lifts

def fewest_lifts(s):
    import bisect
    tails = []
    for ch in s:
        # a non-decreasing run may repeat a letter, so look past equal tails
        at = bisect.bisect_right(tails, ch)
        if at == len(tails):
            tails.append(ch)
        else:
            tails[at] = ch
    return len(s) - len(tails)

The obvious version

Work out the longest non-decreasing run that can be picked out by filling a table: for each bead, the best run ending there is one more than the best run ending at any earlier bead whose letter is no larger. Quadratic, and it never uses a tails list or a binary search.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_fewest_lifts(s):
    n = len(s)
    best = [1] * n
    for i in range(n):
        for j in range(i):
            if s[j] <= s[i] and best[j] + 1 > best[i]:
                best[i] = best[j] + 1
    longest = 0
    for v in best:
        if v > longest:
            longest = v
    return n - longest

How this problem was checked

Reference solution executed against 12 test cases — 12 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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