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# Third Maximum Number LeetCode Solution

## Problem – Third Maximum Number LeetCode Solution

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`

### Third Maximum Number LeetCode Solution in Java

``````    public int thirdMax(int[] nums) {
Integer max1 = null;
Integer max2 = null;
Integer max3 = null;
for (Integer n : nums) {
if (n.equals(max1) || n.equals(max2) || n.equals(max3)) continue;
if (max1 == null || n > max1) {
max3 = max2;
max2 = max1;
max1 = n;
} else if (max2 == null || n > max2) {
max3 = max2;
max2 = n;
} else if (max3 == null || n > max3) {
max3 = n;
}
}
return max3 == null ? max1 : max3;
}
``````

### Third Maximum Number LeetCode Solution in Python

``````class Solution(object):
def thirdMax(self, nums):
v = [float('-inf'), float('-inf'), float('-inf')]
for num in nums:
if num not in v:
if num > v:   v = [num, v, v]
elif num > v: v = [v, num, v]
elif num > v: v = [v, v, num]
return max(nums) if float('-inf') in v else v
``````

### Third Maximum Number LeetCode Solution in C++

``````int thirdMax(vector<int>& nums) {
set<int> top3;
for (int num : nums) {
top3.insert(num);
if (top3.size() > 3)
top3.erase(top3.begin());
}
return top3.size() == 3 ? *top3.begin() : *top3.rbegin();
}
``````
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