What is the time complexity of Sum Game?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Sum Game?

This problem uses Math, String, Greedy, Game Theory. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Sum Game?

Insight 1: The optimal strategy involves recognizing that if the number of '?' is odd, Alice always wins because she can ensure the two halves are unequal. Insight 2: When the number of '?' is even, the crucial factor is the difference between the sums and the difference between the number of '?' in each half. Bob aims to equalize the sums, while Alice aims to create a difference.

Which companies ask LeetCode 1927. Sum Game in interviews?

This problem is frequently asked at ByteDance, DE Shaw.

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Sum Game - LeetCode 1927 Solution

Sum Game - Complete Solution Guide

Sum Game is LeetCode problem 1927, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Alice and Bob take turns playing a game, with Alice starting first . You are given a string num of even length consisting of digits and '?' characters. On each turn, a player will do the following if there is still at least one '?' in num : Choose an index i where num[i] == '?' . Replace num[i] with any digit between '0' and '9' . The game ends when there are no more '?' characters in num . For Bob to win, the sum of the digits in the first half of num must be equal to the sum of the digits in t

Detailed Explanation

The problem describes a game between Alice and Bob played with a string `num` of even length, consisting of digits and question marks ('?'). Alice goes first and players take turns replacing a '?' with a digit from '0' to '9'. The game ends when there are no more question marks. Bob wins if the sum of digits in the first half of the string equals the sum of digits in the second half. Alice wins if the sums are unequal. We need to determine if Alice can win, assuming both players play optimally.

Solution Approach

The solution calculates the sum of digits and the number of question marks in each half of the string. If the total number of question marks is odd, Alice wins immediately. If the total is even, we calculate the difference in sums and the difference in the number of question marks between the two halves. Bob wins if the initial difference in the sums can be equalized to zero by filling the question marks appropriately. Specifically, since each '?' that Alice fills can be balanced by Bob in a way that the combined impact of their moves is 9 (e.g., Alice puts 9, Bob puts 0) the difference is handled as `(q_diff / 2) * 9` and should negate the initial `sum_diff`.

Step-by-Step Algorithm

Step 1: Calculate the length of the input string `num` and find the midpoint `half`.
Step 2: Iterate through the first half of the string, calculating the sum of digits (`s1`) and the number of question marks (`q1`).
Step 3: Iterate through the second half of the string, calculating the sum of digits (`s2`) and the number of question marks (`q2`).
Step 4: Calculate the total number of question marks (`total_q = q1 + q2`).
Step 5: If `total_q` is odd, return `true` (Alice wins).
Step 6: Calculate the difference in sums (`sum_diff = s1 - s2`) and the difference in the number of question marks (`q_diff = q1 - q2`).
Step 7: Check if `sum_diff + (q_diff / 2) * 9 == 0`. If it is, return `false` (Bob wins). Otherwise, return `true` (Alice wins).

Key Insights

Insight 1: The optimal strategy involves recognizing that if the number of '?' is odd, Alice always wins because she can ensure the two halves are unequal.
Insight 2: When the number of '?' is even, the crucial factor is the difference between the sums and the difference between the number of '?' in each half. Bob aims to equalize the sums, while Alice aims to create a difference.
Insight 3: For even numbers of '?', Bob will win if the initial difference between the two halves can be offset by the optimal placement of digits in the question marks.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Math, String, Greedy, Game Theory.

Companies

Asked at: ByteDance, DE Shaw.

Sum Game | LeetCode 1927

Medium

Math String Greedy Game Theory

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean sumGame(String num) {
        int n = num.length();
        int half = n / 2;

        int s1 = 0, q1 = 0;
        for (int i = 0; i < half; i++) {
            if (num.charAt(i) == '?') {
                q1++;
            } else {
                s1 += num.charAt(i) - '0';
            }
        }

        int s2 = 0, q2 = 0;
        for (int i = half; i < n; i++) {
            if (num.charAt(i) == '?') {
                q2++;
            } else {
                s2 += num.charAt(i) - '0';
            }
        }

        int total_q = q1 + q2;
        if (total_q % 2 != 0) {
            return true;
        }

        int sum_diff = s1 - s2;
        int q_diff = q1 - q2;

        if (sum_diff + (q_diff / 2.0) * 9 == 0) {
            return false;
        } else {
            return true;
        }
    }
}

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Problem Info

Description

Problem 1927: Sum Game - Detailed explanation and solution approaches.

Topics

Math String Greedy Game Theory

Companies

ByteDanceDE Shaw

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1927

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Calculate the length of the input string `num` and find the midpoint `half`.
2
Step 2: Iterate through the first half of the string, calculating the sum of digits (`s1`) and the number of question marks (`q1`).
3
Step 3: Iterate through the second half of the string, calculating the sum of digits (`s2`) and the number of question marks (`q2`).
4
Step 4: Calculate the total number of question marks (`total_q = q1 + q2`).
5
Step 5: If `total_q` is odd, return `true` (Alice wins).
6
Step 6: Calculate the difference in sums (`sum_diff = s1 - s2`) and the difference in the number of question marks (`q_diff = q1 - q2`).
7
Step 7: Check if `sum_diff + (q_diff / 2) * 9 == 0`. If it is, return `false` (Bob wins). Otherwise, return `true` (Alice wins).

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because the code iterates through the input string a fixed number of times. The space complexity is constant because the amount of memory used does not depend on the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Simulating the Game

One could theoretically simulate the game by exploring possible moves for Alice and Bob. This approach would involve a search tree, representing each possible state of the game.

Time: Exponential, as the number of possible game states grows exponentially with the number of question marks.Space: Exponential, due to storing the game states in memory.

Trade-offs: This approach is significantly less efficient than the given solution due to its exponential complexity. It's impractical for larger input strings and is mainly useful for smaller examples or for verifying the correctness of the given solution.

✨

Key Insights

1Insight 1: The optimal strategy involves recognizing that if the number of '?' is odd, Alice always wins because she can ensure the two halves are unequal.
2Insight 2: When the number of '?' is even, the crucial factor is the difference between the sums and the difference between the number of '?' in each half. Bob aims to equalize the sums, while Alice aims to create a difference.
3Insight 3: For even numbers of '?', Bob will win if the initial difference between the two halves can be offset by the optimal placement of digits in the question marks.

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: `num` contains no question marks. In this case, `q1` and `q2` are 0, so the condition `sum_diff + (q_diff / 2) * 9 == 0` simplifies to `sum_diff == 0`. The code correctly determines if Alice or Bob wins.
Edge case 2: `num` contains only question marks. In this case, `s1` and `s2` are 0. If `q1 == q2`, `q_diff` is zero and `sum_diff + (q_diff / 2) * 9` is also zero and Bob wins. If the total amount of ? marks is odd Alice wins. Otherwise Bob can equalize by setting 9s and 0s alternately.
Edge case 3: Very large number of '?' marks. The integer arithmetic remains valid, and the condition `sum_diff + (q_diff / 2) * 9 == 0` correctly determines the winner.
Edge case 4: When q_diff is negative: the formula (q_diff / 2) * 9 handles the negative values of q_diff correctly, which correctly adjusts for the imbalance of '?' marks.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to consider the optimal play by both players. Many may attempt to calculate simple sums without incorporating the fact that Alice and Bob will pick the optimal digit.

✗Common mistake 2: Incorrectly calculating the impact of the question marks when the total number of question marks is even. The formula `(q_diff / 2) * 9` correctly represents the offset, but it's easy to miss the division by 2 or the multiplication by 9.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Game theory problems in resource allocation. The core logic of strategic decision-making under constraints can be applied to problems where entities compete for limited resources.

🏦Real-world scenario 2: Competitive strategy in network routing. Determining optimal routes in a network, considering the actions of other entities seeking the same goal, can be modeled using similar game-theoretic concepts.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 6, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Sum Game"?

For this medium-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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