1112EasyStringPrefix Sum

Cutting the Strip for the Best Score

Tracked in this browser only

Trains the technique from

LeetCode 1422Maximum Score After Splitting 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 strip marks holds only the characters 0 and 1. Cut it once into a left piece and a right piece, and neither piece may be empty.

A cut scores the number of 0 characters in the left piece plus the number of 1 characters in the right piece.

Return the highest score a cut can reach.

Examples

Example 1

Input: marks = "0011"
Output: 4

Cutting after the two zeros puts both of them on the left and both ones on the right, scoring 2 plus 2.

Example 2

Input: marks = "10"
Output: 0

Only one cut is possible. The left piece is a single 1 and the right piece a single 0, so neither side scores anything.

Example 3

Input: marks = "0000011111"
Output: 10

Cutting where the zeros run out puts all five zeros on the left and all five ones on the right.

Constraints

2 <= marks.length <= 500
marks holds only the characters 0 and 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 max_score(marks: str) -> int:

Java

public int maxScore(String marks)

The intended approach

Time O(n) over the strip.Space O(1).

Sliding the cut one place to the right moves exactly one character from the right piece to the left piece, and the score changes in a way that depends only on that character. So both halves of the score can be carried along in one pass rather than recounted at every cut. Count the `1` characters in the whole strip first: that is the right piece's contribution when the cut sits at the very front. Then walk the cut rightwards. A `0` moving across adds one to the left piece's tally, and a `1` moving across takes one off the right piece's tally. The score after each move is the sum of the two tallies, and the answer is the best of them. The walk stops one place before the end, because the last character has to stay on the right for that piece to be non-empty. Scores can be zero, so the running best starts from the first cut rather than from zero.

Pythonentry point: max_score

def max_score(marks):
    n = len(marks)
    ones = 0
    for ch in marks:
        if ch == "1":
            ones += 1
    zeros = 0
    right = ones
    best = None
    for i in range(n - 1):
        if marks[i] == "0":
            zeros += 1
        else:
            right -= 1
        score = zeros + right
        if best is None or score > best:
            best = score
    return best

The obvious version

Try every legal cut and count both pieces from scratch each time, walking the left piece for its zeros and the right piece for its ones. Costs the strip length squared.

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_max_score(marks):
    n = len(marks)
    best = None
    for cut in range(1, n):
        score = 0
        for i in range(cut):
            if marks[i] == "0":
                score += 1
        for i in range(cut, n):
            if marks[i] == "1":
                score += 1
        if best is None or score > best:
            best = score
    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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in