Problem Statement
Majority Element
You are given a list of numbers called nums. One value shows up more than half the time. That value is called the majority element. Your job is to find it and return it. You can assume this value always exists, so you never have to handle a list where no value passes the halfway mark.
Signals to notice
Brute force first
Count every element with a hash map. That instinct is useful because it follows the prompt literally, but it usually keeps revisiting work the problem is begging you to organize.
The key insight
Boyer-Moore voting: maintain a candidate and counter. Once you hold onto the right piece of information from moment to moment, the problem feels less like trial and error and more like following a shape that was there all along.
Trace it on nums=[2,2,1,1,1,2,2]
init: candidate=2, count=0 num=2: count==0 -> candidate=2; match -> count=1 num=2: match -> count=2 num=1: no match -> count=1 num=1: no match -> count=0 (candidate vote canceled) num=1: count==0 -> candidate=1; match -> count=1 num=2: no match -> count=0 num=2: count==0 -> candidate=2; match -> count=1; return candidate=2
What must stay true
The majority element survives cancellation because it appears more than all others combined. When you keep that truth intact, each local choice supports the larger solution instead of fighting it.
Shape of the loop
candidate = nums[0]; count = 0
for num in nums:
if count == 0: candidate = num
count += 1 if num == candidate else -1
return candidatePseudocode only — the full worked solution lives in the Solution tab.
Easy way to go wrong
Not knowing Boyer-Moore — the trick is that the majority element always wins the vote. The fix is usually to return to the meaning of each move, not just the steps themselves.