1183EasyArrayHash TableStringBit ManipulationCounting

Pairs of Labels Using the Same Letters

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Trains the technique from

LeetCode 2506Count Pairs Of Similar Strings

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 shelf holds the labels labels. Two labels are alike when each uses exactly the same set of letters as the other, however many times each letter turns up in either.

Return how many pairs of positions i < j hold labels that are alike.

Examples

Example 1

Input: labels = ["abc", "cba", "bca", "acb"]
Output: 6

All four labels use exactly the letters a, b and c, so all six pairs are alike.

Example 2

Input: labels = ["a", "aa", "aaa"]
Output: 3

Each label uses only the letter a, however many times, so all three pairs are alike.

Example 3

Input: labels = ["ab", "cd", "ef", "ab", "cd"]
Output: 2

Only the two ab labels are alike, and the two cd labels.

Constraints

1 <= labels.length <= 100
1 <= labels[i].length <= 100
each label holds 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 similar_pairs(labels: list[str]) -> int:

Java

public int similarPairs(String[] labels)

The intended approach

Time O(n * m) to build the patterns for n labels of up to m letters.Space O(n) for the tally.

Only which letters a label uses matters, not how many times or in what order, and there are twenty-six letters. So each label boils down to a pattern of twenty-six yes-or-no facts, which fits in a single whole number with one bit per letter. Two labels are alike exactly when their patterns match, so tally how many labels carry each pattern in one pass. A pattern carried by `c` labels then contributes `c * (c - 1) / 2` pairs, since every two of those labels are alike and each pair is counted once. That avoids comparing every pair of labels, and the halving is what stops a pair being counted from both sides.

Pythonentry point: similar_pairs

def similar_pairs(labels):
    tally = {}
    for word in labels:
        pattern = 0
        for ch in word:
            pattern |= 1 << (ord(ch) - 97)
        tally[pattern] = tally.get(pattern, 0) + 1
    total = 0
    for pattern in tally:
        count = tally[pattern]
        total += count * (count - 1) // 2
    return total

The obvious version

Take every pair of positions in increasing order and compare the two labels' sets of letters outright, counting the pairs that match. Costs the label count squared times their length.

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_similar_pairs(labels):
    n = len(labels)
    total = 0
    for i in range(n):
        for j in range(i + 1, n):
            if set(labels[i]) == set(labels[j]):
                total += 1
    return total

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

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