LeetCode #414 — EASY

Third Maximum Number

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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?

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[3,2,1]

Example 2

[1,2]

Example 3

[2,2,3,1]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #414: Third Maximum Number
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);
    }
}

// Accepted solution for LeetCode #414: Third Maximum Number
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
}

# Accepted solution for LeetCode #414: Third Maximum Number
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

// Accepted solution for LeetCode #414: Third Maximum Number
struct Solution;
use std::cmp::Ordering::*;

impl Solution {
    fn third_max(nums: Vec<i32>) -> i32 {
        let mut max1 = nums[0];
        let mut m2: Option<i32> = None;
        let mut m3: Option<i32> = None;
        for x in nums {
            match x.cmp(&max1) {
                Greater => {
                    m3 = m2;
                    m2 = Some(max1);
                    max1 = x;
                }
                Less => {
                    if let Some(max2) = m2 {
                        match x.cmp(&max2) {
                            Greater => {
                                m3 = m2;
                                m2 = Some(x);
                            }
                            Less => {
                                if let Some(max3) = m3 {
                                    if x > max3 {
                                        m3 = Some(x);
                                    }
                                } else {
                                    m3 = Some(x);
                                }
                            }
                            Equal => {}
                        }
                    } else {
                        m2 = Some(x);
                    }
                }
                Equal => {}
            }
        }
        if let Some(max3) = m3 {
            max3
        } else {
            max1
        }
    }
}

#[test]
fn test() {
    let nums = vec![3, 2, 1];
    assert_eq!(Solution::third_max(nums), 1);
    let nums = vec![1, 2];
    assert_eq!(Solution::third_max(nums), 2);
    let nums = vec![2, 2, 3, 1];
    assert_eq!(Solution::third_max(nums), 1);
    let nums = vec![1, 2, 2, 5, 3, 5];
    assert_eq!(Solution::third_max(nums), 2);
}

// Accepted solution for LeetCode #414: Third Maximum Number
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #414: Third Maximum Number
// 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);
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.