LeetCode #2212 — MEDIUM

Maximum Points in an Archery Competition

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

The Problem

Problem Statement

Alice and Bob are opponents in an archery competition. The competition has set the following rules:

Alice first shoots numArrows arrows and then Bob shoots numArrows arrows.
The points are then calculated as follows:
1. The target has integer scoring sections ranging from 0 to 11 inclusive.
2. For each section of the target with score k (in between 0 to 11), say Alice and Bob have shot a_k and b_k arrows on that section respectively. If a_k >= b_k, then Alice takes k points. If a_k < b_k, then Bob takes k points.
3. However, if a_k == b_k == 0, then nobody takes k points.

For example, if Alice and Bob both shot 2 arrows on the section with score 11, then Alice takes 11 points. On the other hand, if Alice shot 0 arrows on the section with score 11 and Bob shot 2 arrows on that same section, then Bob takes 11 points.

You are given the integer numArrows and an integer array aliceArrows of size 12, which represents the number of arrows Alice shot on each scoring section from 0 to 11. Now, Bob wants to maximize the total number of points he can obtain.

Return the array bobArrows which represents the number of arrows Bob shot on each scoring section from 0 to 11. The sum of the values in bobArrows should equal numArrows.

If there are multiple ways for Bob to earn the maximum total points, return any one of them.

Example 1:

Input: numArrows = 9, aliceArrows = [1,1,0,1,0,0,2,1,0,1,2,0]
Output: [0,0,0,0,1,1,0,0,1,2,3,1]
Explanation: The table above shows how the competition is scored. 
Bob earns a total point of 4 + 5 + 8 + 9 + 10 + 11 = 47.
It can be shown that Bob cannot obtain a score higher than 47 points.

Example 2:

Input: numArrows = 3, aliceArrows = [0,0,1,0,0,0,0,0,0,0,0,2]
Output: [0,0,0,0,0,0,0,0,1,1,1,0]
Explanation: The table above shows how the competition is scored.
Bob earns a total point of 8 + 9 + 10 = 27.
It can be shown that Bob cannot obtain a score higher than 27 points.

Constraints:

1 <= numArrows <= 10⁵
aliceArrows.length == bobArrows.length == 12
0 <= aliceArrows[i], bobArrows[i] <= numArrows
sum(aliceArrows[i]) == numArrows

Step 01

Brute Force Baseline

Problem summary: Alice and Bob are opponents in an archery competition. The competition has set the following rules: Alice first shoots numArrows arrows and then Bob shoots numArrows arrows. The points are then calculated as follows: The target has integer scoring sections ranging from 0 to 11 inclusive. For each section of the target with score k (in between 0 to 11), say Alice and Bob have shot ak and bk arrows on that section respectively. If ak >= bk, then Alice takes k points. If ak < bk, then Bob takes k points. However, if ak == bk == 0, then nobody takes k points. For example, if Alice and Bob both shot 2 arrows on the section with score 11, then Alice takes 11 points. On the other hand, if Alice shot 0 arrows on the section with score 11 and Bob shot 2 arrows on that same section, then Bob takes 11 points. You are given the integer numArrows and an integer array aliceArrows of size 12, which

Baseline thinking

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

Pattern signal: Array · Backtracking · Bit Manipulation

Example 1

9
[1,1,0,1,0,0,2,1,0,1,2,0]

Example 2

3
[0,0,1,0,0,0,0,0,0,0,0,2]

Core Insight

What unlocks the optimal approach

To obtain points for some certain section x, what is the minimum number of arrows Bob must shoot?
Given the small number of sections, can we brute force which sections Bob wants to win?
For every set of sections Bob wants to win, check if we have the required amount of arrows. If we do, it is a valid selection.

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 #2212: Maximum Points in an Archery Competition
class Solution {
    public int[] maximumBobPoints(int numArrows, int[] aliceArrows) {
        int st = 0, mx = 0;
        int m = aliceArrows.length;
        for (int mask = 1; mask < 1 << m; ++mask) {
            int cnt = 0, s = 0;
            for (int i = 0; i < m; ++i) {
                if ((mask >> i & 1) == 1) {
                    s += i;
                    cnt += aliceArrows[i] + 1;
                }
            }
            if (cnt <= numArrows && s > mx) {
                mx = s;
                st = mask;
            }
        }
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            if ((st >> i & 1) == 1) {
                ans[i] = aliceArrows[i] + 1;
                numArrows -= ans[i];
            }
        }
        ans[0] += numArrows;
        return ans;
    }
}

// Accepted solution for LeetCode #2212: Maximum Points in an Archery Competition
func maximumBobPoints(numArrows int, aliceArrows []int) []int {
	st, mx := 0, 0
	m := len(aliceArrows)
	for mask := 1; mask < 1<<m; mask++ {
		cnt, s := 0, 0
		for i, x := range aliceArrows {
			if mask>>i&1 == 1 {
				s += i
				cnt += x + 1
			}
		}
		if cnt <= numArrows && s > mx {
			mx = s
			st = mask
		}
	}
	ans := make([]int, m)
	for i, x := range aliceArrows {
		if (st>>i)&1 == 1 {
			ans[i] = x + 1
			numArrows -= ans[i]
		}
	}
	ans[0] += numArrows
	return ans
}

# Accepted solution for LeetCode #2212: Maximum Points in an Archery Competition
class Solution:
    def maximumBobPoints(self, numArrows: int, aliceArrows: List[int]) -> List[int]:
        st = mx = 0
        m = len(aliceArrows)
        for mask in range(1, 1 << m):
            cnt = s = 0
            for i, x in enumerate(aliceArrows):
                if mask >> i & 1:
                    s += i
                    cnt += x + 1
            if cnt <= numArrows and s > mx:
                mx = s
                st = mask
        ans = [0] * m
        for i, x in enumerate(aliceArrows):
            if st >> i & 1:
                ans[i] = x + 1
                numArrows -= ans[i]
        ans[0] += numArrows
        return ans

// Accepted solution for LeetCode #2212: Maximum Points in an Archery Competition
impl Solution {
    pub fn maximum_bob_points(num_arrows: i32, alice_arrows: Vec<i32>) -> Vec<i32> {
        let mut st = 0;
        let mut mx = 0;
        let m = alice_arrows.len();
        for mask in 1..(1 << m) {
            let mut cnt = 0;
            let mut s = 0;
            for i in 0..m {
                if (mask >> i) & 1 == 1 {
                    s += i as i32;
                    cnt += alice_arrows[i] + 1;
                }
            }
            if cnt <= num_arrows && s > mx {
                mx = s;
                st = mask;
            }
        }
        let mut ans = vec![0; m];
        let mut num_arrows = num_arrows;
        for i in 0..m {
            if (st >> i) & 1 == 1 {
                ans[i] = alice_arrows[i] + 1;
                num_arrows -= ans[i];
            }
        }
        ans[0] += num_arrows;
        ans
    }
}

// Accepted solution for LeetCode #2212: Maximum Points in an Archery Competition
function maximumBobPoints(numArrows: number, aliceArrows: number[]): number[] {
    let [st, mx] = [0, 0];
    const m = aliceArrows.length;
    for (let mask = 1; mask < 1 << m; mask++) {
        let [cnt, s] = [0, 0];
        for (let i = 0; i < m; i++) {
            if ((mask >> i) & 1) {
                cnt += aliceArrows[i] + 1;
                s += i;
            }
        }
        if (cnt <= numArrows && s > mx) {
            mx = s;
            st = mask;
        }
    }
    const ans: number[] = Array(m).fill(0);
    for (let i = 0; i < m; i++) {
        if ((st >> i) & 1) {
            ans[i] = aliceArrows[i] + 1;
            numArrows -= ans[i];
        }
    }
    ans[0] += numArrows;
    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!)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.