0142MediumStringStackGreedyMonotonic Stack

Trim the Freight Code

Tracked in this browser only

Trains the technique from

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 freight terminal stamps every pallet with a routing code, handed to you as the digit string num. Before the pallet is re-stamped, the terminal is allowed to erase exactly k of the digit positions. Whatever survives keeps the left-to-right arrangement it already had; surviving digits are never rearranged.

Pick the positions to erase so the re-stamped code reads as the lowest whole number that any legal erasure can produce, and hand that code back as a string. Zeros sitting in front of the first non-zero survivor are dropped when the code is stamped. Should the erasure consume every position, the terminal stamps "0".

Examples

Example 1

Input: num = "4297", k = 2
Output: "27"

Erasing the 4 and the 9 leaves 2 followed by 7. No other pair of survivors reads lower than 27.

Example 2

Input: num = "1122", k = 1
Output: "112"

The digits never step down, so the cheapest erasure is the rightmost position.

Example 3

Input: num = "70081", k = 3
Output: "0"

Holding on to the two zeros stamps a value of zero once the front zero is dropped.

Constraints

1 <= k <= num.length <= 10^5
num is made up of digit characters only.
num carries no zero in front unless it is the lone digit 0.

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 remove_kdigits(num: str, k: int) -> str:

Java

public String removeKdigits(String num, int k)

The intended approach

Time O(n)Space O(n)

Scan the code once while holding the survivors in a stack that never steps down. Whenever the digit on top of the stack is larger than the incoming digit and erasures remain, popping it lowers the code, so pop and spend one erasure. Any erasures still unspent at the end come off the tail, then the front zeros are dropped.

Pythonentry point: remove_kdigits

def remove_kdigits(num, k):
    keep = len(num) - k
    stack = []
    budget = k
    for ch in num:
        while stack and budget > 0 and stack[-1] > ch:
            stack.pop()
            budget -= 1
        stack.append(ch)
    trimmed = ''.join(stack[:keep]).lstrip('0')
    return trimmed or '0'

The obvious version

Enumerate every set of surviving positions of the required size, read each one as an integer and keep the minimum. Exponential, but it needs no greedy insight at all.

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_smallest_code(num, k):
    from itertools import combinations
    n = len(num)
    keep = n - k
    if keep <= 0:
        return '0'
    best = None
    for combo in combinations(range(n), keep):
        value = int(''.join(num[i] for i in combo))
        if best is None or value < best:
            best = value
    return str(best)

How this problem was checked

Reference solution executed against 11 test cases — 11 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