1755. Defuse The Bomb¶

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1755. Defuse the Bomb

Easy

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

Solution¶

class Solution {
    public int[] decrypt(int[] arr, int k) {
        int n = arr.length;
        int[] res = new int[n];
        if (k == 0) return res;
        else if (k > 0) {
            for (int i = 0; i < n; i++) {
                int s = 0;
                for (int j = 1; j <= k; j++) s += arr[(i + j) % n];
                res[i] = s;
            }
        } 
        else if (k < 0) {
            for (int i = 0; i < n; i++) {
                int s = 0;
                for (int j = 1; j <= -1 * k; j++) s += arr[(i - j + n) % n];
                res[i] = s;
            }
        }
        return res;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here