LeetCode 0414. Third Maximum Number Solution in Java, Python, C++, JavaScript, Go & Rust | Explanation + Code

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: Single Pass

We can use three variables m_1, m_2, and m_3 to represent the first, second, and third largest numbers in the array respectively. Initially, we set these three variables to negative infinity.

Then, we iterate through each number in the array. For each number:

• If it equals any of m_1, m_2, or m_3, we skip this number.
• If it is greater than m_1, we update the values of m_1, m_2, and m_3 to m_2, m_3, and this number respectively.
• If it is greater than m_2, we update the values of m_2 and m_3 to m_3 and this number respectively.
• If it is greater than m_3, we update the value of m_3 to this number.

Finally, if the value of m_3 has not been updated, it means that there is no third largest number in the array, so we return m_1. Otherwise, we return m_3.

The time complexity is O(n), where n is the length of the array nums. The space complexity is O(1).

PythonJavaC++Go

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(code-box)

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); } }(code-box)

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); } };(code-box)

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 }(code-box)