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1652. Defuse the Bomb

Description

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

Solutions

Solution 1: Simulation

We define an answer array ans of length n, initially all elements are 0. According to the problem, if k is 0, return ans directly.

Otherwise, we traverse each position i:

  • If k is a positive number, then the value at position i is the sum of the values at the k positions after position i, that is:

$$ ans[i] = \sum_{j=i+1}^{i+k} code[j \bmod n] $$

  • If k is a negative number, then the value at position i is the sum of the values at the |k| positions before position i, that is:

$$ ans[i] = \sum_{j=i+k}^{i-1} code[(j+n) \bmod n] $$

The time complexity is $O(n \times |k|)$, ignoring the space consumption of the answer, the space complexity is $O(1)$.

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class Solution:
    def decrypt(self, code: List[int], k: int) -> List[int]:
        n = len(code)
        ans = [0] * n
        if k == 0:
            return ans
        for i in range(n):
            if k > 0:
                for j in range(i + 1, i + k + 1):
                    ans[i] += code[j % n]
            else:
                for j in range(i + k, i):
                    ans[i] += code[(j + n) % n]
        return ans

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

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class Solution {
public:
    vector<int> decrypt(vector<int>& code, int k) {
        int n = code.size();
        vector<int> ans(n);
        if (k == 0) {
            return ans;
        }
        for (int i = 0; i < n; ++i) {
            if (k > 0) {
                for (int j = i + 1; j < i + k + 1; ++j) {
                    ans[i] += code[j % n];
                }
            } else {
                for (int j = i + k; j < i; ++j) {
                    ans[i] += code[(j + n) % n];
                }
            }
        }
        return ans;
    }
};

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func decrypt(code []int, k int) []int {
    n := len(code)