LeetCode #3320 — HARD

Count The Number of Winning Sequences

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

Alice and Bob are playing a fantasy battle game consisting of n rounds where they summon one of three magical creatures each round: a Fire Dragon, a Water Serpent, or an Earth Golem. In each round, players simultaneously summon their creature and are awarded points as follows:

  • If one player summons a Fire Dragon and the other summons an Earth Golem, the player who summoned the Fire Dragon is awarded a point.
  • If one player summons a Water Serpent and the other summons a Fire Dragon, the player who summoned the Water Serpent is awarded a point.
  • If one player summons an Earth Golem and the other summons a Water Serpent, the player who summoned the Earth Golem is awarded a point.
  • If both players summon the same creature, no player is awarded a point.

You are given a string s consisting of n characters 'F', 'W', and 'E', representing the sequence of creatures Alice will summon in each round:

  • If s[i] == 'F', Alice summons a Fire Dragon.
  • If s[i] == 'W', Alice summons a Water Serpent.
  • If s[i] == 'E', Alice summons an Earth Golem.

Bob’s sequence of moves is unknown, but it is guaranteed that Bob will never summon the same creature in two consecutive rounds. Bob beats Alice if the total number of points awarded to Bob after n rounds is strictly greater than the points awarded to Alice.

Return the number of distinct sequences Bob can use to beat Alice.

Since the answer may be very large, return it modulo 109 + 7.

Example 1:

Input: s = "FFF"

Output: 3

Explanation:

Bob can beat Alice by making one of the following sequences of moves: "WFW", "FWF", or "WEW". Note that other winning sequences like "WWE" or "EWW" are invalid since Bob cannot make the same move twice in a row.

Example 2:

Input: s = "FWEFW"

Output: 18

Explanation:

Bob can beat Alice by making one of the following sequences of moves: "FWFWF", "FWFWE", "FWEFE", "FWEWE", "FEFWF", "FEFWE", "FEFEW", "FEWFE", "WFEFE", "WFEWE", "WEFWF", "WEFWE", "WEFEF", "WEFEW", "WEWFW", "WEWFE", "EWFWE", or "EWEWE".

Constraints:

  • 1 <= s.length <= 1000
  • s[i] is one of 'F', 'W', or 'E'.
Patterns Used
📊Dynamic Programming→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Alice and Bob are playing a fantasy battle game consisting of n rounds where they summon one of three magical creatures each round: a Fire Dragon, a Water Serpent, or an Earth Golem. In each round, players simultaneously summon their creature and are awarded points as follows: If one player summons a Fire Dragon and the other summons an Earth Golem, the player who summoned the Fire Dragon is awarded a point. If one player summons a Water Serpent and the other summons a Fire Dragon, the player who summoned the Water Serpent is awarded a point. If one player summons an Earth Golem and the other summons a Water Serpent, the player who summoned the Earth Golem is awarded a point. If both players summon the same creature, no player is awarded a point. You are given a string s consisting of n characters 'F', 'W', and 'E', representing the sequence of creatures Alice will summon in each round:

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming

Example 1

"FFF"

Example 2

"FWEFW"

Related Problems

  • Predict the Winner (predict-the-winner)
Step 02

Core Insight

What unlocks the optimal approach

  • Use dynamic programming.
  • For <code>0 < i < n - 1</code>, <code>-n < j < n</code>, and <code>k</code> in <code>{’F’, ‘W’, ‘E’}</code>, let <code>dp[i][j][k]</code> be the number of sequences consisting of the first <code>i</code> moves such that the difference between bob’s points and alice’s point is equal to <code>j</code> and the <code>i<sup>th</sup></code> move that Bob played is <code>k</code>.
  • The answer is the sum of <code>dp[n - 1][j][k]</code>over all <code>j > 0</code> and over all <code>k</code>.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3320: Count The Number of Winning Sequences
class Solution {
    private int n;
    private char[] s;
    private int[] d = new int[26];
    private Integer[][][] f;
    private final int mod = (int) 1e9 + 7;

    public int countWinningSequences(String s) {
        d['W' - 'A'] = 1;
        d['E' - 'A'] = 2;
        this.s = s.toCharArray();
        n = this.s.length;
        f = new Integer[n][n + n + 1][4];
        return dfs(0, n, 3);
    }

    private int dfs(int i, int j, int k) {
        if (n - i <= j - n) {
            return 0;
        }
        if (i >= n) {
            return j - n < 0 ? 1 : 0;
        }
        if (f[i][j][k] != null) {
            return f[i][j][k];
        }

        int ans = 0;
        for (int l = 0; l < 3; ++l) {
            if (l == k) {
                continue;
            }
            ans = (ans + dfs(i + 1, j + calc(d[s[i] - 'A'], l), l)) % mod;
        }
        return f[i][j][k] = ans;
    }

    private int calc(int x, int y) {
        if (x == y) {
            return 0;
        }
        if (x < y) {
            return x == 0 && y == 2 ? 1 : -1;
        }
        return x == 2 && y == 0 ? -1 : 1;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × m)
Space
O(n × m)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

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