0928EasyArrayStringSimulation

Heaviest Label by Letter Weight

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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 weight is given for every letter of the alphabet: weights[0] for 'a', weights[1] for 'b' and so on to weights[25] for 'z'.

A label's weight is the total of its letters' weights, counting a repeated letter every time it appears.

Return the heaviest label in words. When two labels weigh the same, return whichever is listed first.

Examples

Example 1

Input: words = ["fig", "kiwi", "melon", "plum"], weights = [3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4, 3, 3]
Output: "melon"

Adding up each label's letter weights, the heaviest of the four is the one returned.

Example 2

Input: words = ["cab", "bac", "abc"], weights = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
Output: "cab"

Every letter weighs one, so all three labels weigh three, and the first listed wins.

Example 3

Input: words = ["aa", "zz"], weights = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26]
Output: "zz"

With weights rising through the alphabet, two z's weigh 52 against two a's at 2, so the second label wins even though it is listed later.

Constraints

1 <= words.length <= 100
1 <= words[i].length <= 10
weights.length == 26
1 <= weights[i] <= 100
Every entry of words consists of lowercase 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 map_word_weights(words: list[str], weights: list[int]) -> str:

Java

public String mapWordWeights(String[] words, int[] weights)

The intended approach

Time O(total) over the total length of all the labels.Space O(1) beyond the answer.

A letter's weight is found by its place in the alphabet, which the character code gives directly, so a label's weight is one pass over its letters. Walk the labels in the order given, keep the heaviest weight seen along with the label that reached it, and only replace it on a strictly heavier label. Insisting on strictly heavier is what makes an equal weight fall to whichever label came first.

Pythonentry point: map_word_weights

def map_word_weights(words, weights):
    best_word = None
    best_weight = -1
    for word in words:
        total = 0
        for ch in word:
            total += weights[ord(ch) - 97]
        if total > best_weight:
            best_weight = total
            best_word = word
    return best_word

The obvious version

Work out every label's weight by searching the alphabet for each of its letters, collect the weights alongside the labels, and pick the first label reaching the largest weight found. Scans the alphabet per letter and makes two passes over the labels.

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_map_word_weights(words, weights):
    alphabet = "abcdefghijklmnopqrstuvwxyz"
    totals = []
    for word in words:
        total = 0
        for ch in word:
            for at in range(26):
                if alphabet[at] == ch:
                    total += weights[at]
                    break
        totals.append(total)
    top = totals[0]
    for t in totals:
        if t > top:
            top = t
    for i in range(len(words)):
        if totals[i] == top:
            return words[i]
    return words[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

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