0784HardStringDynamic Programming

Two Hands On A Lettered Key Panel

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Trains the technique from

LeetCode 1320Minimum Distance to Type a Word Using Two Fingers

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 dispatcher enters a call sign on a panel of lettered keys using two hands.

The panel carries 26 keys, one per uppercase letter, laid out in a rectangle four columns wide and filled in alphabetical order, left to right and then top to bottom. Write t for a letter's place in the alphabet counted from zero, so A is 0 and Z is 25. That letter's key sits in row t / 4 (integer division) and column t % 4. Row 0 therefore holds A B C D, row 1 holds E F G H, and the bottom row 6 holds only Y and Z.

Moving a hand from one key to another costs the number of rows between them plus the number of columns between them.

The letters of call must be pressed in order. Before each press the dispatcher picks which of the two hands makes it; that hand travels from the key it is resting on to the key being pressed and the travel cost is added to the running total. Both hands begin away from the panel, so the first press made by a hand costs nothing. A hand then rests on the key it pressed until it is used again, and the two hands are allowed to rest on the same key.

Return the smallest total travel cost for entering the whole call sign.

Examples

Example 1

Input: call = "MAP"
Output: 3

One hand presses M at row 3 column 0, the other hand presses A at row 0 column 0 for nothing since it is that hand's first press, and then the first hand slides from M to P at row 3 column 3, three columns across.

Example 2

Input: call = "ZEBRA"
Output: 5

One hand can take Z at row 6 column 1 and later R at row 4 column 1, travelling two rows. The other hand takes E at row 1 column 0 free of charge, then B at row 0 column 1 for two steps, then A at row 0 column 0 for one more, which totals 5.

Example 3

Input: call = "KKKKKK"
Output: 0

Every press is on the same key, so a single hand can stay put and travel nothing.

Constraints

1 <= call.length <= 300
call consists of uppercase 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 min_hand_travel(call: str) -> int:

Java

public int minHandTravel(String call)

The intended approach

Time O(n) for a call sign of n letters: 27 states are swept per letter, plus a fixed 676-entry distance table.Space O(1) beyond the input: 27 running costs and the fixed 26 by 26 distance table.

Work through the call sign one letter at a time. After a press, one hand is certainly resting on the key just pressed, so the only thing worth remembering is where the other hand rests, giving 27 states per letter once an extra value is reserved for a hand still off the panel. From each state the next press is either made by the resting hand, which pays the gap between the two consecutive keys and leaves the other hand alone, or by the other hand, which pays its own gap (nothing if it has not pressed yet) and leaves the previously pressed key occupied. Distances come from a precomputed 26 by 26 table.

Pythonentry point: min_hand_travel

def min_hand_travel(call):
    # State: one hand always rests on the key just pressed, so only the other hand's
    # resting key has to be remembered (26 stands for "still off the panel"). Each press
    # either continues with the resting hand or brings the other hand across, which makes
    # the previously pressed key the other hand's resting place.
    # Time O(n) with 27 states carried per letter, space O(1) for fixed-size tables.
    keys = [ord(ch) - 65 for ch in call]
    n = len(keys)
    if n < 2:
        return 0
    gap = [[0] * 26 for _ in range(26)]
    for a in range(26):
        for b in range(26):
            gap[a][b] = abs(a // 4 - b // 4) + abs(a % 4 - b % 4)

    BIG = 10 ** 9
    dp = [BIG] * 27
    dp[26] = 0
    for i in range(1, n):
        prev = keys[i - 1]
        cur = keys[i]
        step = gap[prev][cur]
        nxt = [BIG] * 27
        for other in range(27):
            cost = dp[other]
            if cost >= BIG:
                continue
            stay = cost + step
            if stay < nxt[other]:
                nxt[other] = stay
            swap = cost + (0 if other == 26 else gap[other][cur])
            if swap < nxt[prev]:
                nxt[prev] = swap
        dp = nxt
    return min(dp)

The obvious version

Enumerate every way of handing each press to one of the two hands as a bit mask, add up the travel each hand does under that assignment, and keep the cheapest total.

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_hand_travel(call):
    keys = [ord(ch) - 65 for ch in call]
    n = len(keys)
    best = None
    for mask in range(1 << n):
        rest = [None, None]
        total = 0
        for i in range(n):
            hand = (mask >> i) & 1
            k = keys[i]
            at = rest[hand]
            if at is not None:
                total += abs(at // 4 - k // 4) + abs(at % 4 - k % 4)
            rest[hand] = k
        if best is None or total < best:
            best = total
    return best

How this problem was checked

Reference solution executed against 13 test cases — 13 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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