Sync LeetCode submission Runtime - 87 ms (69.32%), Memory - 24.7 MB (12.98%)

Author: jimit105

1370-count-number-of-nice-subarrays/README.md

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+<p>Given an array of integers <code>nums</code> and an integer <code>k</code>. A continuous subarray is called <strong>nice</strong> if there are <code>k</code> odd numbers on it.</p>
+
+<p>Return <em>the number of <strong>nice</strong> sub-arrays</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,1,2,1,1], k = 3
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> The only sub-arrays with 3 odd numbers are [1,1,2,1] and [1,2,1,1].
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [2,4,6], k = 1
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> There are no odd numbers in the array.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [2,2,2,1,2,2,1,2,2,2], k = 2
+<strong>Output:</strong> 16
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 50000</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10^5</code></li>
+	<li><code>1 &lt;= k &lt;= nums.length</code></li>
+</ul>

1370-count-number-of-nice-subarrays/solution.py

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+# Approach 1: Hashing
+
+# Time: O(n)
+# Space: O(n)
+
+class Solution:
+    def numberOfSubarrays(self, nums: List[int], k: int) -> int:
+        curr_sum = 0
+        subarrays = 0
+        prefix_sum = {curr_sum : 1}
+
+        for num in nums:
+            curr_sum += num % 2
+            if curr_sum - k in prefix_sum:
+                subarrays += prefix_sum[curr_sum - k]
+            prefix_sum[curr_sum] = prefix_sum.get(curr_sum, 0) + 1
+
+        return subarrays
+