1652. Defuse the Bomb

Difficulty: Easy

Topics: Array, Sliding Window

You have a bomb to defuse, and your time is running out! Your informer will provide you with a circular array code of length of n and a key k.

To decrypt the code, you must replace every number. All the numbers are replaced simultaneously.

• If k > 0, replace the ith number with the sum of the next k numbers.
• If k < 0, replace the ith number with the sum of the previous k numbers.
• If k == 0, replace the ith number with 0.

As code is circular, the next element of code[n-1] is code[0], and the previous element of code[0] is code[n-1].

Given the circular array code and an integer key k, return the decrypted code to defuse the bomb!

Example 1:

• Input: code = [5,7,1,4], k = 3
• Output: [12,10,16,13]
• Explanation: Each number is replaced by the sum of the next 3 numbers. The decrypted code is [7+1+4, 1+4+5, 4+5+7, 5+7+1]. Notice that the numbers wrap around.

Example 2:

• Input: code = [1,2,3,4], k = 0
• Output: [0,0,0,0]
• Explanation: When k is zero, the numbers are replaced by 0.

Example 3:

• Input: code = [2,4,9,3], k = -2
• Output: [12,5,6,13]
• Explanation: The decrypted code is [3+9, 2+3, 4+2, 9+4]. Notice that the numbers wrap around again. If k is negative, the sum is of the previous numbers.

Constraints:

• n == code.length
• 1 <= n <= 100
• 1 <= code[i] <= 100
• -(n - 1) <= k <= n - 1

Hint:

1. As the array is circular, use modulo to find the correct index.
2. The constraints are low enough for a brute-force solution.

Solution:

We can implement a function that iterates over the code array and computes the sum of the appropriate numbers based on the value of k.

The general approach will be as follows:

1. If k == 0, replace all elements with 0.
2. If k > 0, replace each element with the sum of the next k elements in the circular array.
3. If k < 0, replace each element with the sum of the previous k elements in the circular array.

The circular nature of the array means that for indices that exceed the bounds of the array, you can use modulo (%) to "wrap around" the array.

Let's implement this solution in PHP: 1652. Defuse the Bomb

<?php
/**
 * @param Integer[] $code
 * @param Integer $k
 * @return Integer[]
 */
function decrypt($code, $k) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Example Usage
$code1 = [5, 7, 1, 4];
$k1 = 3;
print_r(decrypt($code1, $k1)); // Output: [12, 10, 16, 13]

$code2 = [1, 2, 3, 4];
$k2 = 0;
print_r(decrypt($code2, $k2)); // Output: [0, 0, 0, 0]

$code3 = [2, 4, 9, 3];
$k3 = -2;
print_r(decrypt($code3, $k3)); // Output: [12, 5, 6, 13]
?>

Explanation:

1. Initialization:

   • We create a result array initialized with zeros using array_fill.

2. Handling k == 0:

   • If k is zero, the output array is simply filled with zeros, as required by the problem.

3. Iterating Through the Array:

   • For each index i in the array:
     • If k > 0, sum the next k elements using modulo arithmetic to wrap around.
     • If k < 0, sum the previous |k| elements using modulo arithmetic with an offset to handle negative indices.

4. Modulo Arithmetic:

   • We use ($i + $j) % $n to wrap around to the beginning of the array when accessing indices greater than n - 1.
   • Similarly, ($i - $j + $n) % $n handles backward wrapping for negative indices.

5. Complexity:

   • Time Complexity: O(n . |k|), where n is the size of the array and |k| is the absolute value of k.
   • Space Complexity: O(n) for the result array.

Outputs:

The provided examples match the expected results. Let me know if you need further explanation or optimizations!

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