Defuse the Bomb - Array - Easy - LeetCode
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Defuse the Bomb - Array - Easy - LeetCode

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  • 1To decrypt the circular array code, replace each number based on the integer key k, which determines the sum of surrounding elements.
  • 2If k is positive, each number is replaced by the sum of the next k numbers; if k is negative, by the previous k numbers.
  • 3When k is zero, all numbers in the array are replaced with zero, resulting in an array of zeros.

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"To decrypt the circular array code, replace each number based on the integer key k, which determines the sum of surrounding elements."

— 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.

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sumitc91

Published on 15 November 2020 · 2 min read · 346 words

Part of AskGif Blog · coding

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