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229. Majority Element II

Description

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?

Solutions

Solution 1

Thinking

At most two values can appear more than \(\lfloor n/3\rfloor\) times. Hashing works but uses extra memory. Boyer–Moore can keep two candidates and their offsetting counts.

A match increments a count; otherwise both counts decrement when they are positive. A second pass verifies the true frequencies.

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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]

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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;
    }
}

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class Solution {
public:
    vector<int> majorityElement(vector<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;
            }
        }
        vector<int> ans;
        if (count(nums.begin(), nums.end(), m1) > nums.size() / 3) ans.push_back(m1);
        if (count(nums.begin(), nums.end(), m2) > nums.size() / 3) ans.push_back(m2);
        return ans;
    }
};

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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
}

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public class Solution {
    public IList<int> MajorityElement(int[] nums) {
        int n1 = 0, n2 = 0;
        int m1 = 0, m2 = 1;
        foreach (int m in 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;
            }
        }
        var ans = new List<int>();
        ans.Add(m1);
        ans.Add(m2);
        return ans.Where(m => nums.Count(n => n == m) > nums.Length / 3).ToList();
    }
}

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class Solution {
    /**
     * @param Integer[] $nums
     * @return Integer[]
     */
    function majorityElement($nums) {
        $rs = [];
        $n = count($nums);
        for ($i = 0; $i < $n; $i++) {
            $hashmap[$nums[$i]] += 1;
            if ($hashmap[$nums[$i]] > $n / 3) {
                array_push($rs, $nums[$i]);
            }
        }
        return array_unique($rs);
    }
}

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