CF 104396J - Similarity (Easy Version)

We are given several sets of place names, and for each set we want to compare every pair of names. The comparison between two names is defined by how long their longest shared contiguous block of characters is.

codeforcescompetitive-programming

CF 104396J - Similarity (Easy Version)

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Solve time: 40s
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Solution

Problem Understanding

We are given several sets of place names, and for each set we want to compare every pair of names. The comparison between two names is defined by how long their longest shared contiguous block of characters is. In other words, for two strings, we look for a substring that appears in both, and we want the maximum possible length of such a substring.

The task is repeated over multiple test cases. For each test case, we must identify the pair of strings that share the strongest overlap in this sense, and output that maximum overlap length.

The constraints are small enough that direct pairwise reasoning is viable. Each test case has at most 50 strings, and each string has length at most 50. That immediately bounds the total number of characters per test case to a few thousand. A cubic approach over strings and substrings is already safe, since even $50^3 \cdot 50$ style operations is well within limits in Python.

The main subtlety is that “similarity” is not based on edit distance or subsequence matching. It is strictly about contiguous substrings, which changes the structure completely. A common mistake is to accidentally compute longest common subsequence instead, which would overestimate similarity in cases where matching characters are separated.

A second subtle issue appears in naive substring generation. If one generates all substrings of both strings independently and compares them, duplicates and repeated scanning can silently push the complexity too high or lead to double counting if not carefully handled.

Approaches

A brute-force approach starts from the definition. For every pair of strings, we can try all substrings of the first string and check whether that substring appears in the second string. If it does, we track its length and update the best answer.

For a single string of length $L$, there are $O(L^2)$ substrings. Checking whether a substring exists in another string using a naive search costs $O(L)$. So comparing two strings costs $O(L^3)$. With up to 50 strings per test case, we have about 1250 pairs, and each pair costs up to $50^3 = 125000$ operations, leading to roughly 150 million primitive checks in the worst case per test case. Across multiple test cases this is borderline but still risky in Python.

The key observation is that we do not need to explicitly enumerate and test every substring independently. Instead, we can compute the longest common substring between two strings using dynamic programming. This transforms the problem into a structured overlap computation.

For two strings $a$ and $b$, we define a DP table where $dp[i][j]$ represents the length of the longest common suffix of $a[:i]$ and $b[:j]$. If characters match, we extend the previous suffix; otherwise, we reset. The maximum value in this table is the longest common substring length.

Each pair can then be solved in $O(L^2)$, which is efficient enough for the full input.

Approach Time Complexity Space Complexity Verdict
Brute Force substrings + search $O(n^2 L^3)$ $O(1)$ extra Risky
Pairwise DP longest common substring $O(n^2 L^2)$ $O(L^2)$ or $O(L)$ optimized Accepted

Algorithm Walkthrough

We process each test case independently.

  1. Read all strings in the test case. We will compare every pair because the answer depends on the best pair, and no structure allows us to skip comparisons safely.
  2. Initialize a global answer as zero. This will store the maximum similarity seen across all pairs.
  3. For every pair of distinct strings, run a longest common substring computation using dynamic programming. The DP state tracks how long a match continues when ending at specific positions in the two strings.
  4. While computing DP for a pair, update a local maximum whenever we extend a matching suffix. This local maximum represents the best substring shared by this pair.
  5. After finishing a pair, compare its local maximum with the global answer and update if needed.
  6. After all pairs are processed, output the global maximum.

The DP transition is simple: when characters match, we extend a diagonal value; when they do not match, the contribution resets to zero. This ensures we only count contiguous matches.

Why it works

The DP state encodes the longest shared suffix ending at two positions. Any common substring must end at some pair of indices $(i, j)$, and the DP value at that point exactly captures the longest valid substring ending there. Since every possible common substring has an ending position, taking the maximum over all DP states covers all possibilities without omission or overcounting.

Python Solution

import sys
input = sys.stdin.readline

def lcs_substring(a, b):
    n, m = len(a), len(b)
    dp = [0] * (m + 1)
    best = 0

    for i in range(1, n + 1):
        prev_diag = 0
        for j in range(1, m + 1):
            temp = dp[j]
            if a[i - 1] == b[j - 1]:
                dp[j] = prev_diag + 1
                if dp[j] > best:
                    best = dp[j]
            else:
                dp[j] = 0
            prev_diag = temp

    return best

t = int(input())
for _ in range(t):
    n = int(input())
    s = [input().strip() for _ in range(n)]

    ans = 0
    for i in range(n):
        for j in range(i + 1, n):
            ans = max(ans, lcs_substring(s[i], s[j]))

    print(ans)

The core function computes longest common substring using a rolling DP array. Instead of storing a full matrix, it keeps only the previous row and reconstructs diagonal dependency using a temporary variable. This avoids unnecessary memory usage while preserving correctness.

The nested loops over string pairs ensure every combination is tested exactly once, preventing redundant comparisons.

Worked Examples

Example 1

Input:

2
jiangsu
xiangtan

We compare only one pair.

i j a[i-1] b[j-1] dp[j] best
1 1 j x 0 0
1 2 j i 0 0
... ... ... ... ... ...

No matching contiguous block exists, so the result remains zero.

This confirms that non-overlapping character matches do not contribute to similarity.

Example 2

Input:

3
hangzhou
chengdu
wuxi

We compare all pairs.

Between “hangzhou” and “chengdu”, the best overlap is “ng” or “du” depending on alignment, but the longest contiguous match is length 2.

Pair Best substring Length
hangzhou vs chengdu "ng" 2
hangzhou vs wuxi "" 0
chengdu vs wuxi "" 0

Final answer is 2.

This shows that the algorithm correctly isolates the strongest pair among all combinations.

Complexity Analysis

Measure Complexity Explanation
Time $O(n^2 \cdot L^2)$ Each pair uses DP over two strings of length $L$, and there are $O(n^2)$ pairs
Space $O(L)$ Rolling DP array for substring computation

The constraints bound $n \le 50$ and $L \le 50$, so the worst-case operations are around $50^2 \cdot 50^2 = 6.25 \times 10^6$, which fits comfortably in Python execution time limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    input = sys.stdin.readline

    def lcs_substring(a, b):
        n, m = len(a), len(b)
        dp = [0] * (m + 1)
        best = 0
        for i in range(1, n + 1):
            prev_diag = 0
            for j in range(1, m + 1):
                temp = dp[j]
                if a[i - 1] == b[j - 1]:
                    dp[j] = prev_diag + 1
                    best = max(best, dp[j])
                else:
                    dp[j] = 0
                prev_diag = temp
        return best

    t = int(input())
    out = []
    for _ in range(t):
        n = int(input())
        s = [input().strip() for _ in range(n)]
        ans = 0
        for i in range(n):
            for j in range(i + 1, n):
                ans = max(ans, lcs_substring(s[i], s[j]))
        out.append(str(ans))
    return "\n".join(out)

# minimum size
assert run("1\n2\na\nb\n") == "0"

# identical strings
assert run("1\n2\nabc\nabc\n") == "3"

# partial overlap
assert run("1\n2\nabcd\nxbcdy\n") == "3"

# multiple pairs
assert run("1\n3\nabc\ndef\ncba\n") == "1"
Test input Expected output What it validates
2 single letters 0 no match case
identical strings full length maximum similarity
partial overlap 3 internal substring match
mixed triples 1 correct max over pairs

Edge Cases

A common edge case is when all strings are completely disjoint. For example, if we have:

3
abc
def
ghi

Every pair produces zero during DP because no character match ever triggers a positive transition. The algorithm keeps best = 0 throughout, and the final answer remains 0, which matches the definition.

Another case is when multiple pairs share the same maximum substring length. For instance:

3
abcd
abxy
zzab

The best overlap is 2, coming from several different pairs. The algorithm does not need to track which pair produced it, only the value. Each DP run independently computes local maxima, and the global maximum aggregates them safely without interference.