1029MediumStringDynamic Programming

Fewest Switch Throws to Settle the Row

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Trains the technique from

LeetCode 926Flip String to Monotone Increasing

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 row of switches reads row, each character '0' for down and '1' for up. One throw changes a single switch, either way.

The row is settled when no switch that is down comes after a switch that is up: all the downs first, then all the ups, and either group may be empty.

Return the fewest throws that settle the row.

Examples

Example 1

Input: row = "111000"
Output: 3

Three ups then three downs. Throwing all three ups down settles the row, and so does throwing all three downs up, either way three throws. Nothing shorter works, since every up sits before every down.

Example 2

Input: row = "0000"
Output: 0

Every switch is down, which is already settled.

Example 3

Input: row = "10"
Output: 1

One up followed by one down. A single throw on either switch settles it.

Constraints

1 <= row.length <= 10^5
Every character of row is '0' or '1'.

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 min_flips_mono_incr(row: str) -> int:

Java

public int minFlipsMonoIncr(String row)

The intended approach

Time O(n) over the row's length.Space O(1).

A settled row is decided by one thing only: where the downs stop and the ups begin. For a given boundary, the throws needed are the ups sitting before it, each thrown down, plus the downs sitting after it, each thrown up. Sliding the boundary from left to right updates both counts in a step, so trying every boundary costs one pass. Neater still, the boundary need never be named. Carry the fewest throws that settle everything read so far. Reading an up forces nothing more, since an up can always sit at the end of a settled stretch. Reading a down leaves a choice: throw it up, which costs one more than the best so far, or leave it down and throw down every up read before it, which costs the tally of ups. Take the smaller of the two. One pass and two running numbers, and the tally of ups only ever climbs on an up.

Pythonentry point: min_flips_mono_incr

def min_flips_mono_incr(row):
    ups = 0
    throws = 0
    for ch in row:
        if ch == '1':
            ups += 1
        else:
            throws += 1
            if ups < throws:
                throws = ups
    return throws

The obvious version

Try every boundary outright. For each place the downs might stop, count the ups before it and the downs after it, add them, and keep the smallest total. Costs the row's length squared rather than a single pass.

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_min_flips_mono_incr(row):
    n = len(row)
    best = None
    for cut in range(n + 1):
        throws = 0
        for at in range(cut):
            if row[at] == '1':
                throws += 1
        for at in range(cut, n):
            if row[at] == '0':
                throws += 1
        if best is None or throws < best:
            best = throws
    return best

How this problem was checked

Reference solution executed against 15 test cases — 15 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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