Baozi Leetcode solution 229: Major Element II

Problem Statement 

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

Note: The algorithm should run in linear time and in O(1) space.

Example 1:

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

Example 2:

Input: [1,1,1,3,3,2,2,2]
Output: [1,2]

 Problem link

Video Tutorial

You can find the detailed video tutorial here

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Thought Process

It's a follow up question on Majority Element, given the hint, there should only be at most 2 elements that can satisfy the requirement (You cannot have 3 elements that all appear more than n/3 times). We can perform the voting algorithm on two potential candidates and later do an actual sum to eliminate the false positive (e.g., [1, 2, 1] would have both 1 and 2 as potential candidate, but 2 should not be included)

 

You can refer to Majority Element for a detailed explanation on the voting algorithm. 

 

An extension could be changing n/3 to n/k. We just need extra arrays to store votes and candidates. It will take O(Nk) time complexity

Solutions

public List<Integer> majorityElement(int[] nums) {
    List<Integer> res = new ArrayList<>();

    if (nums == null || nums.length == 0) {
        return res;
    }

    int candidate1 = 1;
    int candidate2 = 1;
    int vote1 = 0;
    int vote2 = 0;

    for (int i = 0; i < nums.length; i++) {
        if (nums[i] == candidate1) {
            vote1++;
        } else if (nums[i] == candidate2) {
            vote2++;
        } else if (vote1 == 0) {  // the order of this if/else actually matters, you cannot put the 0 check first since [8, 8, 7, 7, 7]
            candidate1 = nums[i];
            vote1++;
        } else if (vote2 == 0) {
            candidate2 = nums[i];
            vote2++;
        } else {
            vote1--;
            vote2--;
        }
    }

    /* This is what you should do if we want to put vote1 == 0 check at the beginning 
    for (int i = 0; i < nums.length; i++) {
        if (vote1 == 0 && candidate2 != nums[i]) {
            candidate1 = nums[i];
        } else if (vote2 == 0 && candidate1 != nums[i]) {
            candidate2 = nums[i];
        }
        if (nums[i] == candidate1) {
            vote1++;
        } else if (nums[i] == candidate2) {
            vote2++;
        } else {
            vote1--;
            vote2--;
        }
    }
    */

    int count1 = 0;
    int count2 = 0;

    // need another pass to filter between the 2 candidates, e.g., [3, 2, 3], 2 is candidate but not more than 1/3
    for (int num : nums) {
        if (num == candidate1) {
            count1++;
        } else if (num == candidate2) {
            count2++;
        }
    }

    if (count1 > nums.length / 3) {
        res.add(candidate1);
    }
    if (count2 > nums.length / 3) {
        res.add(candidate2);
    }

    return res;
}

 

 

Voting implementation

Time Complexity: O(N) where N is the array size

Space Complexity: O(1) Constant space

 

References

• Leetcode discussion solution