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740. Delete and Earn

Description

You are given an integer array nums. You want to maximize the number of points you get by performing the following operation any number of times:

  • Pick any nums[i] and delete it to earn nums[i] points. Afterwards, you must delete every element equal to nums[i] - 1 and every element equal to nums[i] + 1.

Return the maximum number of points you can earn by applying the above operation some number of times.

 

Example 1:

Input: nums = [3,4,2]
Output: 6
Explanation: You can perform the following operations:
- Delete 4 to earn 4 points. Consequently, 3 is also deleted. nums = [2].
- Delete 2 to earn 2 points. nums = [].
You earn a total of 6 points.

Example 2:

Input: nums = [2,2,3,3,3,4]
Output: 9
Explanation: You can perform the following operations:
- Delete a 3 to earn 3 points. All 2's and 4's are also deleted. nums = [3,3].
- Delete a 3 again to earn 3 points. nums = [3].
- Delete a 3 once more to earn 3 points. nums = [].
You earn a total of 9 points.

 

Constraints:

  • 1 <= nums.length <= 2 * 104
  • 1 <= nums[i] <= 104

Solutions

Solution 1

Thinking

Taking \(x\) earns every copy of \(x\) and forbids \(x-1\) and \(x+1\). \(n\le 2\times 10^4\) and values \(\le 10^4\), so searching each choice repeats work.

Each value is all-or-nothing and neighbors conflict—the house-robber recurrence. Bucket scores into \(total[x]\), then walk the value line.

Two rolling scalars are the best through \(i-2\) and \(i-1\); \(i\) takes \(\max(first+total[i], second)\). Time follows the max value.

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class Solution:
    def deleteAndEarn(self, nums: List[int]) -> int:
        mx = -inf
        for num in nums:
            mx = max(mx, num)
        total = [0] * (mx + 1)
        for num in nums:
            total[num] += num
        first = total[0]
        second = max(total[0], total[1])
        for i in range(2, mx + 1):
            cur = max(first + total[i], second)
            first = second
            second = cur
        return second

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class Solution {
    public int deleteAndEarn(int[] nums) {
        if (nums.length == 0) {
            return 0;
        }

        int[] sums = new int[10010];
        int[] select = new int[10010];
        int[] nonSelect = new int[10010];

        int maxV = 0;
        for (int x : nums) {
            sums[x] += x;
            maxV = Math.max(maxV, x);
        }

        for (int i = 1; i <= maxV; i++) {
            select[i] = nonSelect[i - 1] + sums[i];
            nonSelect[i] = Math.max(select[i - 1], nonSelect[i - 1]);
        }
        return Math.max(select[maxV], nonSelect[maxV]);
    }
}

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class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        vector<int> vals(10010);
        for (int& num : nums) {
            vals[num] += num;
        }
        return rob(vals);
    }

    int rob(vector<int>& nums) {
        int a = 0, b = nums[0];
        for (int i = 1; i < nums.size(); ++i) {
            int c = max(nums[i] + a, b);
            a = b;
            b = c;
        }
        return b;
    }
};

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func deleteAndEarn(nums []int) int {

    max := func(x, y int) int {
        if x > y {
            return x
        }
        return y
    }

    mx := math.MinInt32
    for _, num := range nums {
        mx = max(mx, num)
    }
    total := make([]int, mx+1)
    for _, num := range nums {
        total[num] += num
    }
    first := total[0]
    second := max(total[0], total[1])
    for i := 2; i <= mx; i++ {
        cur := max(first+total[i], second)
        first = second
        second = cur
    }
    return second
}

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