700355. Sum of permutations
Here is the detailed Markdown blog post for the "Sum of permutations" problem:
Sum of permutations
Summary
The problem is to calculate the sum of all possible permutations of a given array of integers. The key concept involved is modular arithmetic, as we will be performing calculations with large numbers and taking their remainder when divided by a certain number.
Detailed Explanation
Given an array arr of size n, our task is to find the sum of all possible permutations of this array modulo some prime number p. This problem can be solved using dynamic programming. We first initialize a 2D table dp of size (1 << n) x p, where each cell dp[i][j] will store the cumulative sum of all permutations ending at index i with remainder j.
Here's a step-by-step breakdown of the solution:
- Initialize the table
dpwith zeros. - Iterate over the array
arr. For each elementx, calculate the new cumulative sums by addingxto the previous cumulative sums and taking their remainder modulop. - Update the table
dpaccordingly. - Finally, return the sum of all permutations, which is stored in the last cell of the table
dp.
Here's a diagram illustrating this approach:
+---------------+
| arr = [1, 2, 3] |
+---------------+
|
v
+---------------+
| dp[0][0] = 0 |
| dp[0][1] = 1 |
| ... |
+---------------+
|
v
+---------------+
| dp[1][0] = 1 |
| dp[1][1] = 2 |
| ... |
+---------------+
|
v
+---------------+
| dp[2][0] = 3 |
| dp[2][1] = 4 |
| ... |
+---------------+
|
v
+---------------+
| Return sum of all |
| permutations: dp[(1 << n) - 1][p-1] |
+---------------+
The time complexity of this solution is O(n * p), where n is the size of the array and p is the prime number. The space complexity is O(p).
Optimized Solutions
Java
public int sumOfPermutations(int[] arr, int p) {
int n = arr.length;
int MOD = 1000000007;
long[][] dp = new long[1 << n][p];
for (int i = 0; i < (1 << n); i++) {
for (int j = 0; j < p; j++) {
dp[i][j] = 0;
}
}
for (int x : arr) {
int carry = 0;
for (int i = (1 << n) - 1; i >= 0; i--) {
long sum = dp[i][x % p];
sum += carry;
sum %= MOD;
if ((i & 1) == 1) {
sum = (sum * 2 + x) % MOD;
}
dp[i][x % p] = sum;
}
}
return (int) ((dp[(1 << n) - 1][p-1]) % MOD);
}