defuse-the-bomb

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python3

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Algorithms

You have a bomb to defuse, and your time is running out! Your informer will provide you with a circular array <code>code</code>&nbsp;of length of <code>n</code>&nbsp;and a key <code>k</code>.

To decrypt the code, you must replace every number. All the numbers are replaced simultaneously.

<ul>
	<li>If <code>k &gt; 0</code>, replace the <code>ith</code> number with the sum of the next <code>k</code> numbers.</li>
	<li>If <code>k &lt; 0</code>, replace the <code>ith</code> number with the sum of the previous <code>k</code> numbers.</li>
	<li>If <code>k == 0</code>, replace the <code>ith</code> number with <code>0</code>.</li>
</ul>

As <code>code</code> is circular, the next element of <code>code[n-1]</code> is <code>code[0]</code>, and the previous element of <code>code[0]</code> is <code>code[n-1]</code>.

Given the circular array <code>code</code> and an integer key <code>k</code>, return the decrypted code to defuse the bomb!

&nbsp;
Example 1:

<pre>
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.
</pre>

Example 2:

<pre>
Input: code = [1,2,3,4], k = 0
Output: [0,0,0,0]
Explanation: When k is zero, the numbers are replaced by 0. 
</pre>

Example 3:

<pre>
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.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>n == code.length</code></li>
	<li><code>1 &lt;= n&nbsp;&lt;= 100</code></li>
	<li><code>1 &lt;= code[i] &lt;= 100</code></li>
	<li><code>-(n - 1) &lt;= k &lt;= n - 1</code></li>
</ul>

decrypt

Defuse the Bomb

Circular Sentence

circular-sentence

回环句

Shortest Distance to Target String in a Circular Array

shortest-distance-to-target-string-in-a-circular-array

到目标字符串的最短距离

Take K of Each Character From Left and Right

take-k-of-each-character-from-left-and-right

每种字符至少取 K 个

Array

Sliding Window

你有一个炸弹需要拆除，时间紧迫！你的情报员会给你一个长度为 <code>n</code> 的 循环 数组 <code>code</code> 以及一个密钥 <code>k</code> 。

为了获得正确的密码，你需要替换掉每一个数字。所有数字会 同时 被替换。

<ul>
	<li>如果 <code>k > 0</code> ，将第 <code>i</code> 个数字用 接下来 <code>k</code> 个数字之和替换。</li>
	<li>如果 <code>k < 0</code> ，将第 <code>i</code> 个数字用 之前 <code>k</code> 个数字之和替换。</li>
	<li>如果 <code>k == 0</code> ，将第 <code>i</code> 个数字用 <code>0</code> 替换。</li>
</ul>

由于 <code>code</code> 是循环的， <code>code[n-1]</code> 下一个元素是 <code>code[0]</code> ，且 <code>code[0]</code> 前一个元素是 <code>code[n-1]</code> 。

给你 循环 数组 <code>code</code> 和整数密钥 <code>k</code> ，请你返回解密后的结果来拆除炸弹！

 

示例 1：

<pre>
输入：code = [5,7,1,4], k = 3
输出：[12,10,16,13]
解释：每个数字都被接下来 3 个数字之和替换。解密后的密码为 [7+1+4, 1+4+5, 4+5+7, 5+7+1]。注意到数组是循环连接的。
</pre>

示例 2：

<pre>
输入：code = [1,2,3,4], k = 0
输出：[0,0,0,0]
解释：当 k 为 0 时，所有数字都被 0 替换。
</pre>

示例 3：

<pre>
输入：code = [2,4,9,3], k = -2
输出：[12,5,6,13]
解释：解密后的密码为 [3+9, 2+3, 4+2, 9+4] 。注意到数组是循环连接的。如果 k 是负数，那么和为 之前 的数字。
</pre>

 

提示：

<ul>
	<li><code>n == code.length</code></li>
	<li><code>1 <= n <= 100</code></li>
	<li><code>1 <= code[i] <= 100</code></li>
	<li><code>-(n - 1) <= k <= n - 1</code></li>
</ul>