LeetCode #1798 — MEDIUM

Maximum Number of Consecutive Values You Can Make

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

The Problem

Problem Statement

You are given an integer array coins of length n which represents the n coins that you own. The value of the i^th coin is coins[i]. You can make some value x if you can choose some of your n coins such that their values sum up to x.

Return the maximum number of consecutive integer values that you can make with your coins starting from and including 0.

Note that you may have multiple coins of the same value.

Example 1:

Input: coins = [1,3]
Output: 2
Explanation: You can make the following values:
- 0: take []
- 1: take [1]
You can make 2 consecutive integer values starting from 0.

Example 2:

Input: coins = [1,1,1,4]
Output: 8
Explanation: You can make the following values:
- 0: take []
- 1: take [1]
- 2: take [1,1]
- 3: take [1,1,1]
- 4: take [4]
- 5: take [4,1]
- 6: take [4,1,1]
- 7: take [4,1,1,1]
You can make 8 consecutive integer values starting from 0.

Example 3:

Input: coins = [1,4,10,3,1]
Output: 20

Constraints:

coins.length == n
1 <= n <= 4 * 10⁴
1 <= coins[i] <= 4 * 10⁴

Step 01

Brute Force Baseline

Problem summary: You are given an integer array coins of length n which represents the n coins that you own. The value of the ith coin is coins[i]. You can make some value x if you can choose some of your n coins such that their values sum up to x. Return the maximum number of consecutive integer values that you can make with your coins starting from and including 0. Note that you may have multiple coins of the same value.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[1,3]

Example 2

[1,1,1,4]

Example 3

[1,4,10,3,1]

Core Insight

What unlocks the optimal approach

If you can make the first x values and you have a value v, then you can make all the values <var>≤ v + x</var>
Sort the array of coins. You can always make the value 0 so you can start with x = 0.
Process the values starting from the smallest and stop when there is a value that cannot be achieved with the current x.

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 #1798: Maximum Number of Consecutive Values You Can Make
class Solution {
    public int getMaximumConsecutive(int[] coins) {
        Arrays.sort(coins);
        int ans = 1;
        for (int v : coins) {
            if (v > ans) {
                break;
            }
            ans += v;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1798: Maximum Number of Consecutive Values You Can Make
func getMaximumConsecutive(coins []int) int {
	sort.Ints(coins)
	ans := 1
	for _, v := range coins {
		if v > ans {
			break
		}
		ans += v
	}
	return ans
}

# Accepted solution for LeetCode #1798: Maximum Number of Consecutive Values You Can Make
class Solution:
    def getMaximumConsecutive(self, coins: List[int]) -> int:
        ans = 1
        for v in sorted(coins):
            if v > ans:
                break
            ans += v
        return ans

// Accepted solution for LeetCode #1798: Maximum Number of Consecutive Values You Can Make
struct Solution;

impl Solution {
    fn get_maximum_consecutive(mut coins: Vec<i32>) -> i32 {
        let n = coins.len();
        coins.sort_unstable();
        let mut prefix = 0;
        for i in 0..n {
            if coins[i] > prefix + 1 {
                break;
            } else {
                prefix += coins[i];
            }
        }
        prefix + 1
    }
}

#[test]
fn test() {
    let coins = vec![1, 3];
    let res = 2;
    assert_eq!(Solution::get_maximum_consecutive(coins), res);
    let coins = vec![1, 1, 1, 4];
    let res = 8;
    assert_eq!(Solution::get_maximum_consecutive(coins), res);
    let coins = vec![1, 4, 10, 3, 1];
    let res = 20;
    assert_eq!(Solution::get_maximum_consecutive(coins), res);
}

// Accepted solution for LeetCode #1798: Maximum Number of Consecutive Values You Can Make
function getMaximumConsecutive(coins: number[]): number {
    coins.sort((a, b) => a - b);
    let ans = 1;
    for (const v of coins) {
        if (v > ans) {
            break;
        }
        ans += v;
    }
    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

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.

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.

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.