DeleteAndEarn740.java

/**
 * Given an array nums of integers, you can perform operations on the array.
 * 
 * In each operation, you pick any nums[i] and delete it to earn nums[i]
 * points. After, you must delete every element equal to nums[i] - 1 or
 * nums[i] + 1.
 * 
 * You start with 0 points. Return the maximum number of points you can earn
 * by applying such operations.
 * 
 * Example 1:
 * Input: nums = [3, 4, 2]
 * Output: 6
 * Explanation: 
 * Delete 4 to earn 4 points, consequently 3 is also deleted.
 * Then, delete 2 to earn 2 points. 6 total points are earned.
 * 
 * Example 2:
 * Input: nums = [2, 2, 3, 3, 3, 4]
 * Output: 9
 * Explanation: 
 * Delete 3 to earn 3 points, deleting both 2's and the 4.
 * Then, delete 3 again to earn 3 points, and 3 again to earn 3 points.
 * 9 total points are earned.
 * 
 * Note:
 * The length of nums is at most 20000.
 * Each element nums[i] is an integer in the range [1, 10000].
 */

public class DeleteAndEarn740 {
    public int deleteAndEarn(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int[] count = new int[10001];
        int max = 0;
        for (int n: nums) {
            max = Math.max(max, n);
            count[n]++;
        }
        int[] dp = new int[max + 1];
        dp[0] = 0;
        dp[1] = count[1];
        for (int i=2; i<=max; i++) {
            if (count[i] == 0) {
                dp[i] = Math.max(dp[i-1], dp[i-2]);
            } else {
                dp[i] = Math.max(dp[i-1], dp[i-2] + count[i] * i);
            }
        }
        return dp[max];
    }


    public int deleteAndEarn2(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int[] count = new int[10001];
        int max = 0;
        for (int n: nums) {
            max = Math.max(max, n);
            count[n]++;
        }
        
        int dp0 = 0;
        int dp1 = count[1];
        for (int i=2; i<=max; i++) {
            int tmp = dp1;
            if (count[i] == 0) {
                dp1 = Math.max(dp1, dp0);
            } else {
                dp1 = Math.max(dp1, dp0 + count[i] * i);
            }
            dp0 = tmp;
        }
        
        return dp1;
    }

}