LeetCode 740: Delete and Earn Solution

Master LeetCode problem 740 (Delete and Earn), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

740. Delete and Earn

Medium

Array Hash Table Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. We first count the occurrences of each number in the input array. Then, we create two arrays skip and take, where skip[i] represents the maximum points we can get if we skip the i-th element, and take[i] represents the maximum points we can get if we take the i-th element.

We iterate over the values from 0 to the maximum value in the input array, and for each value i, we update the skip and take arrays based on the optimal strategy of taking or skipping the current value.

Finally, the maximum points we can earn is the maximum value between the last elements of skip and take arrays.

Time Complexity: The time complexity of this solution is O(N) where N is the number of elements in the input array.

Space Complexity: The space complexity of this solution is O(N) where N is the number of elements in the input array.

Solution Code

class Solution {
    public int deleteAndEarn(int[] nums) {
        int maxVal = 0;
        for (int num : nums) {
            maxVal = Math.max(maxVal, num);
        }
        
        int[] count = new int[maxVal + 1];
        for (int num : nums) {
            count[num]++;
        }
        
        int[] skip = new int[maxVal + 1];
        int[] take = new int[maxVal + 1];
        
        for (int i = 1; i <= maxVal; i++) {
            skip[i] = Math.max(skip[i - 1], take[i - 1]);
            take[i] = skip[i - 1] + count[i] * i;
        }
        
        return Math.max(skip[maxVal], take[maxVal]);
    }
}

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