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: 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 $O(1)$.
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