Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 17.6 MB (97.86%)

Author: jimit105

3242-count-elements-with-maximum-frequency/README.md

@@ -0,0 +1,32 @@
+<p>You are given an array <code>nums</code> consisting of <strong>positive</strong> integers.</p>
+
+<p>Return <em>the <strong>total frequencies</strong> of elements in</em><em> </em><code>nums</code>&nbsp;<em>such that those elements all have the <strong>maximum</strong> frequency</em>.</p>
+
+<p>The <strong>frequency</strong> of an element is the number of occurrences of that element in the array.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,2,3,1,4]
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> The elements 1 and 2 have a frequency of 2 which is the maximum frequency in the array.
+So the number of elements in the array with maximum frequency is 4.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,3,4,5]
+<strong>Output:</strong> 5
+<strong>Explanation:</strong> All elements of the array have a frequency of 1 which is the maximum.
+So the number of elements in the array with maximum frequency is 5.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
+</ul>

3242-count-elements-with-maximum-frequency/solution.py

@@ -0,0 +1,24 @@
+# Approach 3: One-Pass Sum Max Frequencies
+
+# Time: O(n)
+# Space: O(n)
+
+class Solution:
+    def maxFrequencyElements(self, nums: List[int]) -> int:
+        freq = {}
+        max_freq = 0
+        total_freq = 0
+
+        for num in nums:
+            freq[num] = freq.get(num, 0) + 1
+            curr_freq = freq[num]
+
+            if curr_freq > max_freq:
+                max_freq = curr_freq
+                total_freq = curr_freq
+
+            elif curr_freq == max_freq:
+                total_freq += curr_freq
+
+        return total_freq
+