LeetCode #1753 — MEDIUM

Maximum Score From Removing Stones

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

The Problem

Problem Statement

You are playing a solitaire game with three piles of stones of sizes a, b, and c respectively. Each turn you choose two different non-empty piles, take one stone from each, and add 1 point to your score. The game stops when there are fewer than two non-empty piles (meaning there are no more available moves).

Given three integers a, b, and c, return the maximum score you can get.

Example 1:

Input: a = 2, b = 4, c = 6
Output: 6
Explanation: The starting state is (2, 4, 6). One optimal set of moves is:
- Take from 1st and 3rd piles, state is now (1, 4, 5)
- Take from 1st and 3rd piles, state is now (0, 4, 4)
- Take from 2nd and 3rd piles, state is now (0, 3, 3)
- Take from 2nd and 3rd piles, state is now (0, 2, 2)
- Take from 2nd and 3rd piles, state is now (0, 1, 1)
- Take from 2nd and 3rd piles, state is now (0, 0, 0)
There are fewer than two non-empty piles, so the game ends. Total: 6 points.

Example 2:

Input: a = 4, b = 4, c = 6
Output: 7
Explanation: The starting state is (4, 4, 6). One optimal set of moves is:
- Take from 1st and 2nd piles, state is now (3, 3, 6)
- Take from 1st and 3rd piles, state is now (2, 3, 5)
- Take from 1st and 3rd piles, state is now (1, 3, 4)
- Take from 1st and 3rd piles, state is now (0, 3, 3)
- Take from 2nd and 3rd piles, state is now (0, 2, 2)
- Take from 2nd and 3rd piles, state is now (0, 1, 1)
- Take from 2nd and 3rd piles, state is now (0, 0, 0)
There are fewer than two non-empty piles, so the game ends. Total: 7 points.

Example 3:

Input: a = 1, b = 8, c = 8
Output: 8
Explanation: One optimal set of moves is to take from the 2nd and 3rd piles for 8 turns until they are empty.
After that, there are fewer than two non-empty piles, so the game ends.

Constraints:

1 <= a, b, c <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are playing a solitaire game with three piles of stones of sizes a, b, and c respectively. Each turn you choose two different non-empty piles, take one stone from each, and add 1 point to your score. The game stops when there are fewer than two non-empty piles (meaning there are no more available moves). Given three integers a, b, and c, return the maximum score you can get.

Baseline thinking

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

Pattern signal: Math · Greedy

Example 1

2
4
6

Example 2

4
4
6

Example 3

1
8
8

Core Insight

What unlocks the optimal approach

It's optimal to always remove one stone from the biggest 2 piles
Note that the limits are small enough for simulation

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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 #1753: Maximum Score From Removing Stones
class Solution {
    public int maximumScore(int a, int b, int c) {
        int[] s = new int[] {a, b, c};
        Arrays.sort(s);
        int ans = 0;
        while (s[1] > 0) {
            ++ans;
            s[1]--;
            s[2]--;
            Arrays.sort(s);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1753: Maximum Score From Removing Stones
func maximumScore(a int, b int, c int) (ans int) {
	s := []int{a, b, c}
	sort.Ints(s)
	for s[1] > 0 {
		ans++
		s[1]--
		s[2]--
		sort.Ints(s)
	}
	return
}

# Accepted solution for LeetCode #1753: Maximum Score From Removing Stones
class Solution:
    def maximumScore(self, a: int, b: int, c: int) -> int:
        s = sorted([a, b, c])
        ans = 0
        while s[1]:
            ans += 1
            s[1] -= 1
            s[2] -= 1
            s.sort()
        return ans

// Accepted solution for LeetCode #1753: Maximum Score From Removing Stones
struct Solution;

impl Solution {
    fn maximum_score(a: i32, b: i32, c: i32) -> i32 {
        let sum = a + b + c;
        let min = a.max(b).max(c);
        (sum / 2).min(sum - min)
    }
}

#[test]
fn test() {
    let a = 2;
    let b = 4;
    let c = 6;
    let res = 6;
    assert_eq!(Solution::maximum_score(a, b, c), res);
    let a = 4;
    let b = 4;
    let c = 6;
    let res = 7;
    assert_eq!(Solution::maximum_score(a, b, c), res);
    let a = 1;
    let b = 8;
    let c = 8;
    let res = 8;
    assert_eq!(Solution::maximum_score(a, b, c), res);
}

// Accepted solution for LeetCode #1753: Maximum Score From Removing Stones
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1753: Maximum Score From Removing Stones
// class Solution {
//     public int maximumScore(int a, int b, int c) {
//         int[] s = new int[] {a, b, c};
//         Arrays.sort(s);
//         int ans = 0;
//         while (s[1] > 0) {
//             ++ans;
//             s[1]--;
//             s[2]--;
//             Arrays.sort(s);
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

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.