0091MediumStringDynamic Programming

Command Correction Cost

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 shell offers a suggestion whenever a command is mistyped, and it ranks candidate suggestions by how much repair work each one needs.

You are given the string typed, which is what the user entered, and the string intended, which is the candidate command. A repair is exactly one of these three keystroke-level operations:

drop one character from anywhere in the current text
add one character at any position in the current text
overwrite one character of the current text with a different one

Return the least number of repairs that turns typed into intended. Either string may be empty, and both are made of lowercase letters.

Examples

Example 1

Input: typed = "grep", intended = "grape"
Output: 2

Overwrite the third character with a and add e at the tail; no single repair can close the gap.

Example 2

Input: typed = "flaot", intended = "float"
Output: 2

The two transposed characters each need an overwrite, since swapping a pair is not one of the allowed repairs.

Example 3

Input: typed = "", intended = "cat"
Output: 3

Starting from nothing, every character of the candidate has to be added.

Constraints

0 <= typed.length <= 500
0 <= intended.length <= 500
typed and intended contain only lowercase English letters

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 correction_cost(typed: str, intended: str) -> int:

Java

public int correctionCost(String typed, String intended)

The intended approach

Time O(m * n)Space O(min(m, n))

Let cost(i, j) be the repairs needed to reconcile the first i characters of one string with the first j of the other. Equal characters cost nothing and carry cost(i-1, j-1) forward, otherwise one repair is added on top of the cheapest of dropping, adding or overwriting. Because a row only reads the row above it, keeping a single rolling row over the shorter string is enough.

Pythonentry point: correction_cost

def correction_cost(typed, intended):
    long_side = typed
    short_side = intended
    if len(long_side) < len(short_side):
        long_side, short_side = short_side, long_side
    n = len(short_side)
    prev = list(range(n + 1))
    for i in range(1, len(long_side) + 1):
        ch = long_side[i - 1]
        cur = [i] + [0] * n
        for j in range(1, n + 1):
            if ch == short_side[j - 1]:
                cur[j] = prev[j - 1]
            else:
                drop = prev[j]
                add = cur[j - 1]
                swap = prev[j - 1]
                cur[j] = 1 + min(drop, add, swap)
        prev = cur
    return prev[n]

The obvious version

Treat every (i, j) alignment position as a graph node and run a cheapest-first search from (0, 0) to the far corner, where matching characters are a free edge and each repair is an edge of weight one.

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_correction_cost(typed, intended):
    import heapq
    m = len(typed)
    n = len(intended)
    settled = set()
    queue = [(0, 0, 0)]
    while queue:
        cost, i, j = heapq.heappop(queue)
        if (i, j) in settled:
            continue
        settled.add((i, j))
        if i == m and j == n:
            return cost
        if i < m and j < n and typed[i] == intended[j]:
            heapq.heappush(queue, (cost, i + 1, j + 1))
        if i < m and j < n:
            heapq.heappush(queue, (cost + 1, i + 1, j + 1))
        if i < m:
            heapq.heappush(queue, (cost + 1, i + 1, j))
        if j < n:
            heapq.heappush(queue, (cost + 1, i, j + 1))
    return 0

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