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by Sumit Chourasia | Nov 15, 2020 | Category :coding | Tags : algorithm array data-structure easy leetcode

Defuse the Bomb - Array - Easy - LeetCode

Defuse the Bomb - Array - Easy - LeetCode

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

 

public class Solution {
    public int[] Decrypt(int[] code, int k) {
        var result = new int[code.Length];
        if(k==0){
            return result;
        }        
        else if(k > 0){
            for(int i=0;i<code.Length;i++){
                int sum = 0;
                for(int j=0;j<k;j++){
                    var index = (i+j+1) % code.Length;
                    sum += code[index];
                }
                result[i]=sum;
            }   
        }
        else{
            for(int i=0;i<code.Length;i++){
                var sum = 0;
                for(int j=0;j<Math.Abs(k);j++){
                    var index = i-j-1;
                    if(index <0){
                        index = code.Length + index;
                    }
                    sum += code[index];
                }
                result[i]=sum;
            }
        }
        
        return result;
    }
}

Time Complexity: O(n*k)

Space Complexity: O(1)

Where n is the length of array and k is the count of number whose sum is required.