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1798. Maximum Number of Consecutive Values You Can Make

Description

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.

 

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: nums = [1,4,10,3,1]
Output: 20

 

Constraints:

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

Solutions

Solution 1: Sorting + Greedy

First, we sort the array. Then we define $ans$ as the current number of consecutive integers that can be constructed, initialized to $1$.

We traverse the array, for the current element $v$, if $v > ans$, it means that we cannot construct $ans+1$ consecutive integers, so we directly break the loop and return $ans$. Otherwise, it means that we can construct $ans+v$ consecutive integers, so we update $ans$ to $ans+v$.

Finally, we return $ans$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array.

  • 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;
    }
}

  • class Solution {
public:
    int getMaximumConsecutive(vector<int>& coins) {
        sort(coins.begin(), coins.end());
        int ans = 1;
        for (int& v : coins) {
            if (v > ans) break;
            ans += v;
        }
        return ans;
    }
};

  • class Solution:
    def getMaximumConsecutive(self, coins: List[int]) -> int:
        ans = 1
        for v in sorted(coins):
            if v > ans:
                break
            ans += v
        return ans


  • func getMaximumConsecutive(coins []int) int {
	sort.Ints(coins)
	ans := 1
	for _, v := range coins {
		if v > ans {
			break
		}
		ans += v
	}
	return ans
}

  • 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;
}

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