Coding problem · Easy · Basics

Second Largest Number

Easy Algorithms & data structures

Return the second-largest distinct value in nums. If all values are equal (or there's only one), return -1. Values are non-negative.

Example 1

Input: nums = [4,9,2,9,7]
Output: 7
Explanation: 9 is the largest; the next distinct value is 7.

Example 2

Input: nums = [5,5]
Output: -1

Hints

Hint 1

Track two variables, the largest and second largest, in one pass.

Hint 2

When you find a new largest, the old largest becomes the second largest. Skip values equal to the largest.

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Solution and explanation

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Time O(n) · Space O(1)

Keep first and second. For each value: if it's bigger than first, shift first into second; else if it's between them (and not equal to first), update second.

Reference solution

function secondLargest(nums) {
  let first = -1, second = -1;
  for (const n of nums) {
    if (n > first) { second = first; first = n; }
    else if (n < first && n > second) second = n;
  }
  return second;
}

function secondLargest(nums: number[]): number {
  let first = -1, second = -1;
  for (const n of nums) {
    if (n > first) { second = first; first = n; }
    else if (n < first && n > second) second = n;
  }
  return second;
}

def second_largest(nums: list[int]) -> int:
    first = second = -1
    for n in nums:
        if n > first:
            first, second = n, first
        elif second < n < first:
            second = n
    return second

<?php

function secondLargest(array $nums): int
{
    $first = $second = -1;
    foreach ($nums as $n) {
        if ($n > $first) {
            $second = $first;
            $first = $n;
        } elseif ($n < $first && $n > $second) {
            $second = $n;
        }
    }
    return $second;
}

class Solution {
    public int secondLargest(int[] nums) {
        int first = -1, second = -1;
        for (int n : nums) {
            if (n > first) { second = first; first = n; }
            else if (n < first && n > second) second = n;
        }
        return second;
    }
}

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