LeetCode #229 — MEDIUM

Majority Element II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.

Example 1:

Input: nums = [3,2,3]
Output: [3]

Example 2:

Input: nums = [1]
Output: [1]

Example 3:

Input: nums = [1,2]
Output: [1,2]

Constraints:

1 <= nums.length <= 5 * 10⁴
-10⁹ <= nums[i] <= 10⁹

Follow up: Could you solve the problem in linear time and in O(1) space?

Step 01

Brute Force Baseline

Problem summary: Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[3,2,3]

Example 2

[1]

Example 3

[1,2]

Core Insight

What unlocks the optimal approach

Think about the possible number of elements that can appear more than ⌊ n/3 ⌋ times in the array.
It can be at most two. Why?
Consider using Boyer-Moore Voting Algorithm, which is efficient for finding elements that appear more than a certain threshold.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #229: Majority Element II
class Solution {
    public List<Integer> majorityElement(int[] nums) {
        int n1 = 0, n2 = 0;
        int m1 = 0, m2 = 1;
        for (int m : nums) {
            if (m == m1) {
                ++n1;
            } else if (m == m2) {
                ++n2;
            } else if (n1 == 0) {
                m1 = m;
                ++n1;
            } else if (n2 == 0) {
                m2 = m;
                ++n2;
            } else {
                --n1;
                --n2;
            }
        }
        List<Integer> ans = new ArrayList<>();
        n1 = 0;
        n2 = 0;
        for (int m : nums) {
            if (m == m1) {
                ++n1;
            } else if (m == m2) {
                ++n2;
            }
        }
        if (n1 > nums.length / 3) {
            ans.add(m1);
        }
        if (n2 > nums.length / 3) {
            ans.add(m2);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #229: Majority Element II
func majorityElement(nums []int) []int {
	var n1, n2 int
	m1, m2 := 0, 1
	for _, m := range nums {
		if m == m1 {
			n1++
		} else if m == m2 {
			n2++
		} else if n1 == 0 {
			m1, n1 = m, 1
		} else if n2 == 0 {
			m2, n2 = m, 1
		} else {
			n1, n2 = n1-1, n2-1
		}
	}
	n1, n2 = 0, 0
	for _, m := range nums {
		if m == m1 {
			n1++
		} else if m == m2 {
			n2++
		}
	}
	var ans []int
	if n1 > len(nums)/3 {
		ans = append(ans, m1)
	}
	if n2 > len(nums)/3 {
		ans = append(ans, m2)
	}
	return ans
}

# Accepted solution for LeetCode #229: Majority Element II
class Solution:
    def majorityElement(self, nums: List[int]) -> List[int]:
        n1 = n2 = 0
        m1, m2 = 0, 1
        for m in nums:
            if m == m1:
                n1 += 1
            elif m == m2:
                n2 += 1
            elif n1 == 0:
                m1, n1 = m, 1
            elif n2 == 0:
                m2, n2 = m, 1
            else:
                n1, n2 = n1 - 1, n2 - 1
        return [m for m in [m1, m2] if nums.count(m) > len(nums) // 3]

// Accepted solution for LeetCode #229: Majority Element II
struct Solution;

type Pair = (i32, usize);

impl Solution {
    fn majority_element(nums: Vec<i32>) -> Vec<i32> {
        let n = nums.len();
        let mut pairs: Vec<Pair> = vec![];
        for &x in &nums {
            if pairs.iter().any(|p| p.0 == x) {
                pairs = pairs
                    .into_iter()
                    .map(|p| (p.0, if p.0 == x { p.1 + 1 } else { p.1 }))
                    .collect();
            } else {
                if pairs.len() < 2 {
                    pairs.push((x, 1));
                } else {
                    pairs = pairs.into_iter().map(|p| (p.0, p.1 - 1)).collect();
                    pairs = pairs.into_iter().filter(|p| p.1 > 0).collect();
                }
            }
        }
        let mut res = vec![];
        for pair in pairs {
            let sum = nums
                .iter()
                .fold(0, |acc, &x| if x == pair.0 { acc + 1 } else { acc });
            if sum > n / 3 {
                res.push(pair.0);
            }
        }
        res
    }
}

#[test]
fn test() {
    let nums = vec![3, 2, 3];
    let res = vec![3];
    assert_eq!(Solution::majority_element(nums), res);
    let nums = vec![1, 1, 1, 3, 3, 2, 2, 2];
    let res = vec![1, 2];
    assert_eq!(Solution::majority_element(nums), res);
}

// Accepted solution for LeetCode #229: Majority Element II
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #229: Majority Element II
// class Solution {
//     public List<Integer> majorityElement(int[] nums) {
//         int n1 = 0, n2 = 0;
//         int m1 = 0, m2 = 1;
//         for (int m : nums) {
//             if (m == m1) {
//                 ++n1;
//             } else if (m == m2) {
//                 ++n2;
//             } else if (n1 == 0) {
//                 m1 = m;
//                 ++n1;
//             } else if (n2 == 0) {
//                 m2 = m;
//                 ++n2;
//             } else {
//                 --n1;
//                 --n2;
//             }
//         }
//         List<Integer> ans = new ArrayList<>();
//         n1 = 0;
//         n2 = 0;
//         for (int m : nums) {
//             if (m == m1) {
//                 ++n1;
//             } else if (m == m2) {
//                 ++n2;
//             }
//         }
//         if (n1 > nums.length / 3) {
//             ans.add(m1);
//         }
//         if (n2 > nums.length / 3) {
//             ans.add(m2);
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.