414. Third Maximum Number

Description

Given an integer array nums, return the third distinct maximum number in this array. If the third maximum does not exist, return the maximum number.

 

Example 1:

Input: nums = [3,2,1]
Output: 1
Explanation:
The first distinct maximum is 3.
The second distinct maximum is 2.
The third distinct maximum is 1.

Example 2:

Input: nums = [1,2]
Output: 2
Explanation:
The first distinct maximum is 2.
The second distinct maximum is 1.
The third distinct maximum does not exist, so the maximum (2) is returned instead.

Example 3:

Input: nums = [2,2,3,1]
Output: 1
Explanation:
The first distinct maximum is 3.
The second distinct maximum is 2 (both 2's are counted together since they have the same value).
The third distinct maximum is 1.

 

Constraints:

  • 1 <= nums.length <= 104
  • -231 <= nums[i] <= 231 - 1

 

Follow up: Can you find an O(n) solution?

Solutions

Solution 1

Python Code
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class Solution:
    def thirdMax(self, nums: List[int]) -> int:
        m1 = m2 = m3 = -inf
        for num in nums:
            if num in [m1, m2, m3]:
                continue
            if num > m1:
                m3, m2, m1 = m2, m1, num
            elif num > m2:
                m3, m2 = m2, num
            elif num > m3:
                m3 = num
        return m3 if m3 != -inf else m1

Java Code
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class Solution {
    public int thirdMax(int[] nums) {
        long m1 = Long.MIN_VALUE;
        long m2 = Long.MIN_VALUE;
        long m3 = Long.MIN_VALUE;
        for (int num : nums) {
            if (num == m1 || num == m2 || num == m3) {
                continue;
            }
            if (num > m1) {
                m3 = m2;
                m2 = m1;
                m1 = num;
            } else if (num > m2) {
                m3 = m2;
                m2 = num;
            } else if (num > m3) {
                m3 = num;
            }
        }
        return (int) (m3 != Long.MIN_VALUE ? m3 : m1);
    }
}

C++ Code
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class Solution {
public:
    int thirdMax(vector<int>& nums) {
        long m1 = LONG_MIN, m2 = LONG_MIN, m3 = LONG_MIN;
        for (int num : nums) {
            if (num == m1 || num == m2 || num == m3) continue;
            if (num > m1) {
                m3 = m2;
                m2 = m1;
                m1 = num;
            } else if (num > m2) {
                m3 = m2;
                m2 = num;
            } else if (num > m3) {
                m3 = num;
            }
        }
        return (int) (m3 != LONG_MIN ? m3 : m1);
    }
};

Go Code
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func thirdMax(nums []int) int {
	m1, m2, m3 := math.MinInt64, math.MinInt64, math.MinInt64
	for _, num := range nums {
		if num == m1 || num == m2 || num == m3 {
			continue
		}
		if num > m1 {
			m3, m2, m1 = m2, m1, num
		} else if num > m2 {
			m3, m2 = m2, num
		} else if num > m3 {
			m3 = num
		}
	}
	if m3 != math.MinInt64 {
		return m3
	}
	return m1
}