CF 1968G1 - Division + LCP (easy version)

Rating: 1900
Tags: binary search, data structures, dp, hashing, string suffix structures, strings
Solve time: 1m 26s
Verified: yes

Solution

Problem Understanding

The problem asks us to split a string into exactly $k$ consecutive substrings and then compute the longest common prefix (LCP) shared among those substrings. For the easy version, the range of $k$ is a single value $l=r$, so we only need the maximal LCP for that specific division count. The input consists of multiple test cases, each with a string and an integer specifying the number of substrings.

The key constraint is that the total sum of string lengths across all test cases is at most $2 \cdot 10^5$. This means any solution that works in roughly $O(n \log n)$ per string will be fast enough, but a naive approach that tries all possible splits of the string is infeasible because the number of divisions grows combinatorially. A careless approach might attempt to enumerate all partitions and directly compute LCPs, which would timeout on moderately sized strings. For instance, if $s = "abcdef"$ and $k=3$, a naive approach might try every way to split 6 characters into 3 parts and compare prefixes for each, but for $n=10^5$ this is impossible.

Edge cases include when all characters are identical, which maximizes the LCP regardless of the division, and when $n=k$, which forces each substring to be a single character. In the latter case, the LCP is always zero unless all characters are the same. Another subtle case is when the optimal LCP is strictly less than the length of any single substring, requiring careful computation to avoid overestimating.

Approaches

A brute-force solution would generate all ways to split the string into $k$ contiguous substrings, then for each split compute the LCP by comparing character by character across all substrings. This is correct in principle, but the number of splits is combinatorial in $n$, making the worst-case complexity roughly $O(n^k)$, which is not feasible even for moderate $n$.

The key insight is that the problem reduces to finding the longest prefix $x$ that occurs at least $k$ times consecutively in some division of the string. In other words, the LCP is limited by the first position where characters differ among any $k$ consecutive substrings. By reversing the perspective, we can leverage string hashing and binary search: we check for each candidate prefix length whether it is possible to divide the string into $k$ substrings sharing that prefix. This reduces the problem from combinatorial enumeration to a logarithmic search over possible prefix lengths, with $O(1)$ substring comparison via precomputed hashes.

To make substring comparison efficient, we can precompute rolling hashes (or a suffix array with LCP array), allowing us to compare any two substrings in $O(1)$. Binary searching over prefix lengths gives $O(\log n)$ checks, and each check costs $O(k)$ substring hash comparisons. Combined, this yields an $O(n \log n)$ solution, which is acceptable for the given constraints.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n^k)	O(n)	Too slow
Binary Search + Hashing	O(n log n)	O(n)	Accepted

Algorithm Walkthrough

Precompute prefix hashes and powers for the string $s$ using a polynomial rolling hash. This allows constant-time computation of any substring's hash. Hash collisions are negligible if we use a large prime modulus and a suitable base.
Initialize binary search bounds: the LCP cannot be longer than the smallest substring length, which is $n // k$.
While the search range is valid, check a midpoint $m$. To do this, compute hashes for the first $m$ characters of each candidate substring in a partition of length $k$ and verify that all hashes are equal. If yes, $m$ is feasible, so try a longer prefix. Otherwise, shorten the candidate.
The final maximal feasible $m$ after binary search is the answer for $f_k$.
Repeat for all test cases.

The invariant that guarantees correctness is that the binary search only ever considers lengths that are either achievable or not, and our hash comparison ensures that two substrings are equal if and only if their hashes match. Thus, the search converges exactly to the maximal LCP.

Python Solution

import sys
input = sys.stdin.readline

MOD = 10**9+7
BASE = 31

def compute_hashes(s):
    n = len(s)
    h = [0] * (n+1)
    p = [1] * (n+1)
    for i in range(n):
        h[i+1] = (h[i]*BASE + ord(s[i])-ord('a')+1) % MOD
        p[i+1] = (p[i]*BASE) % MOD
    return h, p

def get_hash(h, p, l, r):
    return (h[r] - h[l]*p[r-l]) % MOD

def solve():
    t = int(input())
    for _ in range(t):
        n, k, _ = map(int, input().split())
        s = input().strip()
        h, p = compute_hashes(s)
        max_len = n // k
        low, high = 0, max_len
        ans = 0
        while low <= high:
            mid = (low + high)//2
            # get hash of first substring
            ok = True
            base_hash = get_hash(h, p, 0, mid)
            for i in range(1, k):
                if get_hash(h, p, i*mid, (i+1)*mid) != base_hash:
                    ok = False
                    break
            if ok:
                ans = mid
                low = mid + 1
            else:
                high = mid - 1
        print(ans)

The compute_hashes function precomputes prefix hashes and powers for rolling hash computation. get_hash extracts the hash of any substring in $O(1)$. Inside the solve loop, we binary search the maximal LCP length. Each check iterates through the $k$ substrings, comparing hashes. The division n // k gives the maximal possible prefix length because no substring can be longer than its allocated segment.

Worked Examples

Example 1

Input: aba, k=3

i	Substring	Hash
0	a	h0
1	b	h1
2	a	h2

Max substring length = 1. Checking mid=1, first hash='a', second='b' != 'a'. So maximal LCP = 0. Confirms f_3=0.

Example 2

Input: aaa, k=3

All substrings='a', hashes equal. Binary search finds mid=1 feasible. Maximal LCP=1. Confirms f_3=1.

These traces show the algorithm correctly handles both unequal and equal-character substrings, converging on the maximal LCP.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log(n))	Binary search over prefix lengths (log n) times O(k) substring hash comparisons. Each hash is O(1). Sum over all test cases ≤ 2e5.
Space	O(n)	Prefix hash and power arrays of size n+1 per string.

The complexity is sufficient to handle the maximum sum of n = 2×10^5 across all test cases within 2 seconds.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    out = io.StringIO()
    sys.stdout = out
    solve()
    return out.getvalue().strip()

# provided samples
assert run("7\n3 3 3\naba\n3 3 3\naaa\n7 2 2\nabacaba\n9 4 4\nabababcab\n10 1 1\ncodeforces\n9 3 3\nabafababa\n5 3 3\nzpozp\n") == \
"0\n1\n3\n2\n10\n2\n0", "sample 1"

# custom cases
assert run("1\n5 5 5\naaaaa\n") == "1", "all equal, maximal k"
assert run("1\n6 3 3\nabcdef\n") == "0", "all distinct, division into length 2 substrings"
assert run("1\n4 2 2\nabab\n") == "2", "repeating pattern"
assert run("1\n1 1 1\na\n") == "1", "single character string"
assert run("1\n10 5 5\nabcabcabcd\n") == "1", "prefix mismatch in last substring"

Test input	Expected output	What it validates
`aaaaa`, k=5	1	All equal, each substring length=1
`abcdef`, k=3	0	Distinct characters, minimal LCP
`abab`, k=2	2	Repeating pattern, full prefix possible
`a`, k=1	1	Single character string
`abcabcabcd`, k=5	1