LeetCode 2741: Special Permutations Solution

Master LeetCode problem 2741 (Special Permutations), 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.

2741. Special Permutations

Medium

Array Dynamic Programming Bit Manipulation Bitmask

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. We will iterate through the array and keep track of the count of special permutations ending at each index. At each index, we will calculate the count of special permutations by considering the previous elements in the array.

Initialize a list dp to store the count of special permutations ending at each index.
Initialize all values in dp to 1 as each individual element is a special permutation.
For each index i from 1 to n-1:
- For each previous index j from 0 to i-1:
  - If nums[i] % nums[j] == 0 or nums[j] % nums[i] == 0, increment dp[i] by dp[j].

The final answer will be the sum of all elements in the dp list.

Solution Code

class Solution {
    public int countSpecialPermutations(int[] nums) {
        int n = nums.length;
        long[] dp = new long[n];
        Arrays.fill(dp, 1);

        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[i] % nums[j] == 0 || nums[j] % nums[i] == 0) {
                    dp[i] = (dp[i] + dp[j]) % 1000000007;
                }
            }
        }

        long totalSpecialPerms = 0;
        for (int i = 0; i < n; i++) {
            totalSpecialPerms = (totalSpecialPerms + dp[i]) % 1000000007;
        }

        return (int) totalSpecialPerms;
    }
}

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