{"id":478,"date":"2025-06-29T20:54:49","date_gmt":"2025-06-29T15:24:49","guid":{"rendered":"https:\/\/codeanddebug.in\/blog\/?p=478"},"modified":"2025-06-29T20:54:50","modified_gmt":"2025-06-29T15:24:50","slug":"majority-element-ii-leetcode-229","status":"publish","type":"post","link":"https:\/\/codeanddebug.in\/blog\/majority-element-ii-leetcode-229\/","title":{"rendered":"Majority Element II | Leetcode 229 | Explained in Python"},"content":{"rendered":"\n

You\u2019re given an integer array nums<\/code> of length n<\/code>.
An element is called a majority element II<\/strong> if it appears strictly more than \u230a n \/ 3 \u230b times<\/strong>.
Because n \/ 3 \u2248 33 %<\/code>, at most two<\/strong> such elements can exist.
Return them in any order<\/strong>.<\/p>\n\n\n\n

Here’s the [Problem Link<\/span><\/mark><\/a><\/strong>] to begin with.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

Input<\/th>Explanation<\/th>Output<\/th><\/tr><\/thead>
[3,2,3]<\/code><\/td>3<\/code> appears 2 > 3\/3 = 1 times<\/td>[3]<\/code><\/td><\/tr>
[1,1,1,3,3,2,2,2]<\/code><\/td>1<\/code> & 2<\/code> each appear 3 > 8\/3 = 2 times<\/td>[1,2]<\/code><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Typical interview follow-ups:<\/p>\n\n\n\n