Welcome to Subscribe On Youtube

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

• Java
• C++
• Python
• Go

• 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);
      }
  }
  

• 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);
      }
  };
  

• 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
  
  

• 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
  }
  

All Problems

All Solutions