Sync LeetCode submission Runtime - 87 ms (38.71%), Memory - 21.9 MB (66.67%)

Author: jimit105

1631-number-of-sub-arrays-with-odd-sum/README.md

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+<p>Given an array of integers <code>arr</code>, return <em>the number of subarrays with an <strong>odd</strong> sum</em>.</p>
+
+<p>Since the answer can be very large, return it modulo <code>10<sup>9</sup> + 7</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> arr = [1,3,5]
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> All subarrays are [[1],[1,3],[1,3,5],[3],[3,5],[5]]
+All sub-arrays sum are [1,4,9,3,8,5].
+Odd sums are [1,9,3,5] so the answer is 4.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> arr = [2,4,6]
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> All subarrays are [[2],[2,4],[2,4,6],[4],[4,6],[6]]
+All sub-arrays sum are [2,6,12,4,10,6].
+All sub-arrays have even sum and the answer is 0.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> arr = [1,2,3,4,5,6,7]
+<strong>Output:</strong> 16
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= arr.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= arr[i] &lt;= 100</code></li>
+</ul>

1631-number-of-sub-arrays-with-odd-sum/solution.py

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+# Approach 3: Prefix Sum with Odd-Even Counting
+
+# Time: O(n)
+# Space: O(1)
+
+class Solution:
+    def numOfSubarrays(self, arr: List[int]) -> int:
+        MOD = 10**9 + 7
+        count = prefix_sum = 0
+
+        odd_count = 0
+        even_count = 1 # even_count starts as 1 since the initial sum (0) is even
+
+        for num in arr:
+            prefix_sum += num
+            # If current prefix sum is even, add the number of odd subarrays
+            if prefix_sum % 2 == 0:
+                count += odd_count
+                even_count += 1
+            
+            # If current prefix sum is odd, add the number of even subarrays
+            else:
+                count += even_count
+                odd_count += 1
+
+            count %= MOD
+
+        return count
+
+