Sync LeetCode submission Runtime - 824 ms (60.33%), Memory - 36.1 MB (71.92%)

Author: jimit105

0523-continuous-subarray-sum/README.md

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+<p>Given an integer array nums and an integer k, return <code>true</code> <em>if </em><code>nums</code><em> has a <strong>good subarray</strong> or </em><code>false</code><em> otherwise</em>.</p>
+
+<p>A <strong>good subarray</strong> is a subarray where:</p>
+
+<ul>
+	<li>its length is <strong>at least two</strong>, and</li>
+	<li>the sum of the elements of the subarray is a multiple of <code>k</code>.</li>
+</ul>
+
+<p><strong>Note</strong> that:</p>
+
+<ul>
+	<li>A <strong>subarray</strong> is a contiguous part of the array.</li>
+	<li>An integer <code>x</code> is a multiple of <code>k</code> if there exists an integer <code>n</code> such that <code>x = n * k</code>. <code>0</code> is <strong>always</strong> a multiple of <code>k</code>.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [23,<u>2,4</u>,6,7], k = 6
+<strong>Output:</strong> true
+<strong>Explanation:</strong> [2, 4] is a continuous subarray of size 2 whose elements sum up to 6.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [<u>23,2,6,4,7</u>], k = 6
+<strong>Output:</strong> true
+<strong>Explanation:</strong> [23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42.
+42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [23,2,6,4,7], k = 13
+<strong>Output:</strong> false
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+	<li><code>0 &lt;= sum(nums[i]) &lt;= 2<sup>31</sup> - 1</code></li>
+	<li><code>1 &lt;= k &lt;= 2<sup>31</sup> - 1</code></li>
+</ul>

0523-continuous-subarray-sum/solution.py

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+# Approach 1: Prefix Sum and Hashing
+
+# Time: O(n)
+# Space: O(n)
+
+class Solution:
+    def checkSubarraySum(self, nums: List[int], k: int) -> bool:
+        prefix_mod = 0
+        mod_seen = {0: -1}
+        
+        for i in range(len(nums)):
+            prefix_mod = (prefix_mod + nums[i]) % k
+
+            if prefix_mod in mod_seen:
+                # ensures that the size of subarray is at least 2
+                if i - mod_seen[prefix_mod] > 1:
+                    return True
+            else:
+                # mark the value of prefix_mod with the current index.
+                mod_seen[prefix_mod] = i
+
+        return False