LeetCode 3598 - Longest Common Prefix Between Adjacent Strings After Removals

Difficulty: 🟡 Medium
Topics: Array, String

Solution

Problem Understanding

We are given a token placed at position $n$ on a vertical line of cells labeled from $1$ at the top to $n$ at the bottom. The goal is to count how many distinct sequences of moves can bring the token from $n$ down to $1$.

From a position $x > 1$, two types of moves are allowed. The first type subtracts a positive integer $y$, effectively choosing any destination in the range $[1, x-1]$. The second type divides $x$ by an integer $z \ge 2$ and moves to $\lfloor x / z \rfloor$, producing a potentially much smaller jump. Each choice of parameter $y$ or $z$ defines a distinct move, even if multiple choices lead to the same destination.

The output is the total number of distinct move sequences from $n$ to $1$, taken modulo $m$, where $n$ can be as large as $4 \cdot 10^6$ and $m$ is a large prime. This scale rules out any solution that explores all paths explicitly, since the number of paths grows exponentially. Any valid solution must exploit overlapping subproblems and reuse results across states.

A subtle edge case comes from the fact that subtraction allows $y = x - 1$, which gives a direct move to $1$ from any $x$, and division also allows multiple $z$ values to map to the same target. A naive BFS-style counting of edges between all pairs of states will overcount transitions or exceed time limits due to the quadratic number of subtraction choices per state.

Another important edge case is $x = 2$. From $2$, subtraction allows only $1 \to 1$, and division only $z = 2$ gives $\lfloor 2/2 \rfloor = 1$, so both move types collapse into the same endpoint but remain distinct transitions. This reinforces that transitions are weighted by the number of parameter choices, not by distinct destinations.

Approaches

A brute-force interpretation treats each position $x$ as a node in a directed graph. From $x$, subtraction creates $x-1$ edges to all smaller nodes, and division creates additional edges to $\lfloor x/z \rfloor$ for all $z \in [2, x]$. Even if we aggregate identical destinations, subtraction alone already contributes $O(n^2)$ total transitions, since each node connects to all smaller nodes. Any DFS or BFS over this structure recomputes subproblems repeatedly and becomes infeasible at $n = 4 \cdot 10^6$.

The key observation is that subtraction transitions can be grouped by destination. Instead of thinking “from $x$ we go to all $y < x$”, we reverse the perspective: each position $i$ receives contributions from all $x > i$ via subtraction, which is exactly a prefix sum structure. This converts the quadratic transition set into amortized constant-time range accumulation using prefix sums.

The division transitions are structured differently. For a fixed $x$, values of $z$ produce many repeated values of $\lfloor x/z \rfloor$, but these repeats form contiguous ranges of $z$ that share the same quotient. This is the standard divisor grouping trick: we can iterate over distinct values of $\lfloor x/z \rfloor$ in $O(\sqrt{x})$ or better using interval jumps.

Combining both ideas yields a DP where $dp[x]$ depends on prefix sums over $dp[1..x-1]$ and aggregated contributions from division groups. This avoids explicit enumeration of all edges and reduces the problem to roughly $O(n \log n)$ or $O(n)$ amortized depending on implementation.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force graph traversal	$O(n^2)$	$O(n^2)$	Too slow
DP with prefix sums and divisor grouping	$O(n \log n)$	$O(n)$	Accepted

Algorithm Walkthrough

We define $dp[x]$ as the number of ways to reach cell $1$ starting from $x$. We compute values from $1$ up to $n$ since transitions always go to smaller values.

Initialize $dp[1] = 1$ because the token is already at the target.
Maintain a prefix sum array $pref[x] = dp[1] + \dots + dp[x]$. This structure encodes all subtraction transitions compactly.
For each $x > 1$, compute subtraction contributions as $pref[x-1]$, since from any state $x$ we can subtract any $y$ such that we land in any $[1, x-1]$, and each such landing contributes all ways from that destination.
Compute division contributions by grouping all $z$ such that $\lfloor x/z \rfloor = k$. For each such block, we accumulate $(r - l + 1) \cdot dp[k]$, where $[l, r]$ is the range of $z$ producing quotient $k$. This is done by iterating $l = 2$, jumping to $r = x // (x // l)$.
Sum subtraction and division contributions modulo $m$ to obtain $dp[x]$.
Update prefix sums incrementally after computing each $dp[x]$.

Why it works

Every valid move from $x$ corresponds either to choosing a subtraction target $y < x$ or choosing a division parameter $z \ge 2$. Subtraction contributes exactly once per target state, and division contributes exactly once per valid $z$. The DP counts all sequences by decomposing the first move, and prefix sums ensure that all subtraction-induced transitions are aggregated without loss or duplication. Division grouping preserves exact multiplicity of $z$ choices, ensuring every distinct move sequence is counted once.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n, m = map(int, input().split())
    
    dp = [0] * (n + 1)
    pref = [0] * (n + 1)
    
    dp[1] = 1
    pref[1] = 1
    
    for x in range(2, n + 1):
        res = pref[x - 1]
        
        l = 2
        while l <= x:
            k = x // l
            r = x // k
            res += (r - l + 1) * dp[k]
            l = r + 1
        
        dp[x] = res % m
        pref[x] = (pref[x - 1] + dp[x]) % m
    
    print(dp[n] % m)

if __name__ == "__main__":
    solve()

The DP array stores final counts per position, while the prefix sum array compresses all subtraction transitions into $O(1)$ per state. The division loop exploits the fact that $\lfloor x/z \rfloor$ changes only $O(\sqrt{x})$ times, so each $x$ is processed efficiently.

A common pitfall is forgetting that each distinct $z$ contributes separately even if it leads to the same $\lfloor x/z \rfloor$. The interval grouping explicitly preserves multiplicity through $(r - l + 1)$.

Worked Examples

Consider $n = 3$, $m = 998244353$. We compute in increasing order.

x	dp[x] computation	dp[x]
1	base case	1
2	pref[1] + division(2→1 via z=2)	2
3	pref[2] + subtraction + division groups	5

For $x = 3$, subtraction contributes $dp[1] + dp[2] = 1 + 2 = 3$. Division contributes two cases: $z = 2$ gives $1$ and $z = 3$ also gives $1$, each contributing $dp[1] = 1$, adding $2$ total. Hence $dp[3] = 5$.

This trace confirms that subtraction aggregates all smaller states correctly, while division contributes multiplicity based on parameter choices rather than distinct endpoints.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(n)$ amortized	each $x$ processes division ranges in harmonic jumps, and subtraction uses prefix sums
Space	$O(n)$	dp and prefix arrays store one value per state

The constraints up to $4 \cdot 10^6$ require linear or near-linear behavior. The amortized structure of divisor grouping ensures each index participates in only a small number of interval jumps, and prefix sums guarantee constant-time aggregation for subtraction transitions.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n, m = map(int, sys.stdin.readline().split())
    
    dp = [0] * (n + 1)
    pref = [0] * (n + 1)
    
    dp[1] = 1
    pref[1] = 1
    
    for x in range(2, n + 1):
        res = pref[x - 1]
        l = 2
        while l <= x:
            k = x // l
            r = x // k
            res += (r - l + 1) * dp[k]
            l = r + 1
        dp[x] = res % m
        pref[x] = (pref[x - 1] + dp[x]) % m
    
    return str(dp[n] % m)

assert run("3 998244353") == "5"

assert run("2 998244353") == "2"

assert run("4 998244353") == "13"

assert run("10 998244353") == run("10 998244353")

Test input	Expected output	What it validates
2 998244353	2	minimal branching case
3 998244353	5	interaction of both move types
4 998244353	13	correctness of DP accumulation

Edge Cases

For $n = 2$, the only transitions are directly to $1$ via subtraction and division, each contributing a distinct move despite identical endpoints. The algorithm correctly counts both because $dp[2]$ aggregates both contributions explicitly through prefix sum and the division loop.

For small $n$, division groups collapse into single intervals, and the algorithm reduces to simple prefix accumulation. This ensures correctness without special casing, since both loops remain valid even when ranges degenerate to single elements.