Today I have done leetcode's November LeetCoding Challenge with cpp.

November LeetCoding Challenge 18

Description

Defuse the Bomb

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:
  vector<int> decrypt(vector<int>& code, int k) {
    int len = code.size();
    vector<int> prefix(len);
    for(int i = 0; i < len; ++i) {
      prefix[i] = code[i];
      if(i) prefix[i] += prefix[i - 1];
    }
    vector<int> answer(len);
    if(!k) return answer;
    for(int i = 0; i < len; ++i) {
      int result = 0;
      if(k < 0) {
        if(i + k <= 0) {
          answer[i] = prefix[len - 1] - prefix[i + k + len - 1]  + prefix[i] - code[i];
        } else {
          answer[i] = prefix[i - 1] - prefix[i + k - 1];
        }
      } else {
        if(i + k >= len) {
          answer[i] = prefix[len - 1] - prefix[i] + prefix[i + k - len];
        } else {
          answer[i] = prefix[i + k] - prefix[i];
        }
      }
    }
    return answer;
  }
};