LeetCode //C - 60. Permutation Sequence

60. Permutation Sequence

The set [1, 2, 3, …, n] contains a total of n! unique permutations.

By listing and labeling all of the permutations in order, we get the following sequence for n = 3:

1. “123”
2. “132”
3. “213”
4. “231”
5. “312”
6. “321”

Given n and k, return the k t h k^{th} kth permutation sequence.
 

Example 1:

Example 2:

Example 3:

Constraints:

• 1 <= n <= 9
• 1 <= k <= n!

From: LeetCode
Link: 60. Permutation Sequence

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Solution:

Ideas:

1. Factorial Calculation: The function starts by calculating factorials, which helps in determining which block or set of permutations the desired permutation falls into.
2. Position Calculation: By dividing k by the factorial of n−1, the function determines the index of the number to place in each position of the resultant string.
3. Update and Shift: After determining the position, the selected number is removed from the available list, effectively reducing the problem size for the next iteration.
4. Memory Management: The function dynamically allocates memory for the result string and a temporary array to hold available numbers, ensuring to free the temporary memory before returning.

Code:

// Helper function to calculate factorial
int factorial(int x) {
    int result = 1;
    for (int i = 2; i <= x; i++) {
        result *= i;
    }
    return result;
}

// Function to get the k-th permutation sequence
char* getPermutation(int n, int k) {
    int i, j, f;
    int len = n;
    k--; // Convert k to zero-indexed for easier calculations

    // Allocate memory for the result
    char *result = malloc((n + 1) * sizeof(char));
    result[n] = '\0'; // Null terminate the string

    // Create an array to hold numbers 1, 2, 3, ..., n
    int *numbers = malloc(n * sizeof(int));
    for (i = 0; i < n; i++) {
        numbers[i] = i + 1;
    }

    for (i = 0; i < len; i++) {
        f = factorial(n - 1);
        j = k / f; // Determine the index of the current digit
        result[i] = numbers[j] + '0'; // Set the current position in result
        k %= f; // Reduce k

        // Remove used number from the array by shifting elements
        for (int m = j; m < n - 1; m++) {
            numbers[m] = numbers[m + 1];
        }
        n--;
    }

    // Clean up and return result
    free(numbers);
    return result;
}