LeetCode 3351: Sum of Good Subsequences Solution

Master LeetCode problem 3351 (Sum of Good Subsequences), a hard 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.

3351. Sum of Good Subsequences

Hard

Array Hash Table Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming with a bitmask approach. We create a dp array where dp[mask] represents the sum of all good subsequences that can be formed using elements whose binary representation corresponds to the mask. We start by initializing dp[1 << nums[i]] = 1 for each element in nums.

Then, for each mask, we iterate over all elements in nums and update the dp[mask] by adding dp[mask ^ (1 << nums[i])] if nums[i] - 1 or nums[i] + 1 exists in the mask.

Finally, we sum up all the values in the dp array to get the total sum of all good subsequences modulo 10^9 + 7.

Time Complexity: O(n * 2^n), where n is the number of elements in the input array nums. Space Complexity: O(2^n), where n is the number of elements in the input array nums.

Solution Code

class Solution {
    public int sumOfGoodSubsequences(int[] nums) {
        int MOD = 1000000007;
        int n = nums.length;
        long[] dp = new long[1 << 17];
        Arrays.fill(dp, 0);
        dp[0] = 1;
        
        for (int num : nums) {
            long[] next = Arrays.copyOf(dp, dp.length);
            for (int i = 0; i < (1 << 17); i++) {
                if (dp[i] == 0) continue;
                
                int newMask = i | (1 << num);
                next[newMask] += dp[i];
                if (num > 0 && (i & (1 << (num - 1))) > 0) {
                    next[newMask] += dp[i];
                }
                next[newMask] %= MOD;
            }
            dp = next;
        }
        
        long ans = 0;
        for (int i = 1; i < (1 << 17); i++) {
            ans = (ans + dp[i]) % MOD;
        }
        
        return (int) ans;
    }
}

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