Majority Element

L2 Easy Arrays & Hashing

Concept

Once you have a frequency dictionary, "which value is common enough" becomes a direct dictionary scan.

Return the integer that appears more than nums.count / 2 times. If no value is that frequent, return -1.

Examples

▸ nums = [1, 2, 3, 2, 2]

→ 2

▸ nums = [1, 2]

→ -1

▸ nums = [1, 1, 1]

→ 1

Progressive Hints

Hint 1 · Nudge

Find out who shows up more than half the time. You can only know that once you know everyone's tally.

Hint 2 · Plan

Tally every value into a frequency map, then return the first key whose count is greater than half the array size. If none qualifies, return -1.

Hint 3 · Approach

counts = frequency map of nums. For each key: if counts[key] > nums.length / 2 return that key. Otherwise return -1.

Output

// Run your code to see the output here.