LeetCode #948 — MEDIUM

Bag of Tokens

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

The Problem

Problem Statement

You start with an initial power of power, an initial score of 0, and a bag of tokens given as an integer array tokens, where each tokens[i] denotes the value of token_i.

Your goal is to maximize the total score by strategically playing these tokens. In one move, you can play an unplayed token in one of the two ways (but not both for the same token):

Face-up: If your current power is at least tokens[i], you may play token_i, losing tokens[i] power and gaining 1 score.
Face-down: If your current score is at least 1, you may play token_i, gaining tokens[i] power and losing 1 score.

Return the maximum possible score you can achieve after playing any number of tokens.

Example 1:

Input: tokens = [100], power = 50

Output: 0

Explanation: Since your score is 0 initially, you cannot play the token face-down. You also cannot play it face-up since your power (50) is less than tokens[0] (100).

Example 2:

Input: tokens = [200,100], power = 150

Output: 1

Explanation: Play token₁ (100) face-up, reducing your power to 50 and increasing your score to 1.

There is no need to play token₀, since you cannot play it face-up to add to your score. The maximum score achievable is 1.

Example 3:

Input: tokens = [100,200,300,400], power = 200

Output: 2

Explanation: Play the tokens in this order to get a score of 2:

Play token₀ (100) face-up, reducing power to 100 and increasing score to 1.
Play token₃ (400) face-down, increasing power to 500 and reducing score to 0.
Play token₁ (200) face-up, reducing power to 300 and increasing score to 1.
Play token₂ (300) face-up, reducing power to 0 and increasing score to 2.

The maximum score achievable is 2.

Constraints:

0 <= tokens.length <= 1000
0 <= tokens[i], power < 10⁴

Step 01

Brute Force Baseline

Problem summary: You start with an initial power of power, an initial score of 0, and a bag of tokens given as an integer array tokens, where each tokens[i] denotes the value of tokeni. Your goal is to maximize the total score by strategically playing these tokens. In one move, you can play an unplayed token in one of the two ways (but not both for the same token): Face-up: If your current power is at least tokens[i], you may play tokeni, losing tokens[i] power and gaining 1 score. Face-down: If your current score is at least 1, you may play tokeni, gaining tokens[i] power and losing 1 score. Return the maximum possible score you can achieve after playing any number of tokens.

Baseline thinking

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

Pattern signal: Array · Two Pointers · Greedy

Example 1

[100]
50

Example 2

[200,100]
150

Example 3

[100,200,300,400]
200

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #948: Bag of Tokens
class Solution {
    public int bagOfTokensScore(int[] tokens, int power) {
        Arrays.sort(tokens);
        int ans = 0, score = 0;
        for (int i = 0, j = tokens.length - 1; i <= j;) {
            if (power >= tokens[i]) {
                power -= tokens[i++];
                ans = Math.max(ans, ++score);
            } else if (score > 0) {
                power += tokens[j--];
                --score;
            } else {
                break;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #948: Bag of Tokens
func bagOfTokensScore(tokens []int, power int) (ans int) {
	sort.Ints(tokens)
	i, j := 0, len(tokens)-1
	score := 0
	for i <= j {
		if power >= tokens[i] {
			power -= tokens[i]
			i++
			score++
			ans = max(ans, score)
		} else if score > 0 {
			power += tokens[j]
			j--
			score--
		} else {
			break
		}
	}
	return
}

# Accepted solution for LeetCode #948: Bag of Tokens
class Solution:
    def bagOfTokensScore(self, tokens: List[int], power: int) -> int:
        tokens.sort()
        ans = score = 0
        i, j = 0, len(tokens) - 1
        while i <= j:
            if power >= tokens[i]:
                power -= tokens[i]
                score, i = score + 1, i + 1
                ans = max(ans, score)
            elif score:
                power += tokens[j]
                score, j = score - 1, j - 1
            else:
                break
        return ans

// Accepted solution for LeetCode #948: Bag of Tokens
struct Solution;

impl Solution {
    fn bag_of_tokens_score(mut tokens: Vec<i32>, mut p: i32) -> i32 {
        let n = tokens.len();
        if n == 0 {
            return 0;
        }
        tokens.sort_unstable();
        let mut l = 0;
        let mut r = n - 1;
        let mut point = 0;
        let mut res = 0;
        while l <= r {
            if tokens[l] <= p {
                p -= tokens[l];
                point += 1;
                res = res.max(point);
                l += 1;
            } else {
                if point > 0 {
                    p += tokens[r];
                    point -= 1;
                    r -= 1;
                } else {
                    break;
                }
            }
        }
        res
    }
}

#[test]
fn test() {
    let tokens = vec![100];
    let p = 50;
    let res = 0;
    assert_eq!(Solution::bag_of_tokens_score(tokens, p), res);
    let tokens = vec![100, 200];
    let p = 150;
    let res = 1;
    assert_eq!(Solution::bag_of_tokens_score(tokens, p), res);
    let tokens = vec![100, 200, 300, 400];
    let p = 200;
    let res = 2;
    assert_eq!(Solution::bag_of_tokens_score(tokens, p), res);
}

// Accepted solution for LeetCode #948: Bag of Tokens
function bagOfTokensScore(tokens: number[], power: number): number {
    tokens.sort((a, b) => a - b);
    let [i, j] = [0, tokens.length - 1];
    let [ans, score] = [0, 0];
    while (i <= j) {
        if (power >= tokens[i]) {
            power -= tokens[i++];
            ans = Math.max(ans, ++score);
        } else if (score) {
            power += tokens[j--];
            score--;
        } else {
            break;
        }
    }
    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(log n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

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.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

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.