LeetCode #1927 — MEDIUM

Sum Game

Move from brute-force thinking to an efficient approach using math strategy.

Solve on LeetCode
The Problem

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:

  1. Choose an index i where num[i] == '?'.
  2. 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 the second half. For Alice to win, the sums must not be equal.

  • For example, if the game ended with num = "243801", then Bob wins because 2+4+3 = 8+0+1. If the game ended with num = "243803", then Alice wins because 2+4+3 != 8+0+3.

Assuming Alice and Bob play optimally, return true if Alice will win and false if Bob will win.

Example 1:

Input: num = "5023"
Output: false
Explanation: There are no moves to be made.
The sum of the first half is equal to the sum of the second half: 5 + 0 = 2 + 3.

Example 2:

Input: num = "25??"
Output: true
Explanation: Alice can replace one of the '?'s with '9' and it will be impossible for Bob to make the sums equal.

Example 3:

Input: num = "?3295???"
Output: false
Explanation: It can be proven that Bob will always win. One possible outcome is:
- Alice replaces the first '?' with '9'. num = "93295???".
- Bob replaces one of the '?' in the right half with '9'. num = "932959??".
- Alice replaces one of the '?' in the right half with '2'. num = "9329592?".
- Bob replaces the last '?' in the right half with '7'. num = "93295927".
Bob wins because 9 + 3 + 2 + 9 = 5 + 9 + 2 + 7.

Constraints:

  • 2 <= num.length <= 105
  • num.length is even.
  • num consists of only digits and '?'.
Patterns Used

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 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 the second half. For Alice to win, the sums must not be equal. For example, if the game ended with num = "243801", then Bob wins because 2+4+3 = 8+0+1. If the game ended with num = "243803", then Alice wins because 2+4+3 != 8+0+3. Assuming Alice and Bob play optimally, return true if Alice will win and false if Bob will win.

Baseline thinking

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

Pattern signal: Math · Greedy

Example 1

"5023"

Example 2

"25??"

Example 3

"?3295???"
Step 02

Core Insight

What unlocks the optimal approach

  • Bob can always make the total sum of both sides equal in mod 9.
  • Why does the difference between the number of question marks on the left and right side matter?
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
Upper-end input sizes
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 #1927: Sum Game
class Solution {
    public boolean sumGame(String num) {
        int n = num.length();
        int cnt1 = 0, cnt2 = 0;
        int s1 = 0, s2 = 0;
        for (int i = 0; i < n / 2; ++i) {
            if (num.charAt(i) == '?') {
                cnt1++;
            } else {
                s1 += num.charAt(i) - '0';
            }
        }
        for (int i = n / 2; i < n; ++i) {
            if (num.charAt(i) == '?') {
                cnt2++;
            } else {
                s2 += num.charAt(i) - '0';
            }
        }
        return (cnt1 + cnt2) % 2 == 1 || s1 - s2 != 9 * (cnt2 - cnt1) / 2;
    }
}
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 log n)
Space
O(1)

Approach Breakdown

EXHAUSTIVE
O(2ⁿ) time
O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY
O(n log n) time
O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.