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.

Solution

This is an easy challenge, we only need to iterate the array and calculate the previous(or latter) k-sum.
Note for the negative index, we need to add code.size() and then mod it into range [0, code.size()-1].

Time complexity is O(N*k)

class Solution {
public:
    int try_to_get(vector<int>& code, int cur, int k) {
        int step = k > 0 ? 1 : -1;
        int pos = cur;
        int ret = 0;
        for(int i=0; i<abs(k); i++) {
            pos += step;
            pos = (pos + code.size()) % code.size();
            ret += code[pos];
        }
        return ret;
    }
    vector<int> decrypt(vector<int>& code, int k) {
        vector<int> res;
        for(int i=0; i<code.size(); i++) {
            res.push_back(try_to_get(code, i, k));
        }
        return res;
    }
};

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