Find the First and Second Largest Number in JavaScript

When working with arrays in JavaScript, one common interview question is finding the largest and second largest numbers without using built-in sorting methods.

In this blog, we’ll understand the logic step by step with an easy example.

  

let array = [10]

    function findSecondLargestNumber(arr)

    {

      if(arr.length < 2)

      {

        return null;

      }

      let largest = -Infinity;

      let secondLargest = -Infinity;

      for(i=0;i<arr.length;i++)

      {

        if (largest < arr[i])

        {

          secondLargest = largest;

          largest = arr[i];

        }

        else if (secondLargest < arr[i]) {

          secondLargest = arr[i];

        }

      }

      return largest;

    }

    findSecondLargestNumber(array)

Output

9 8

• Largest Number: 9
• Second Largest Number: 8

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How the Code Works

Step 1: Initialize Variables

let largest = -Infinity;
let secondLargest = -Infinity;

We use -Infinity so that any number in the array will be greater initially.

---

Step 2: Loop Through the Array

for(i = 0; i < arr.length; i++)

We check every element one by one.

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Step 3: Update Largest Number

if (largest < arr[i])
{
  secondLargest = largest;
  largest = arr[i];
}

If the current element is greater than largest:

• Current largest becomes secondLargest
• Current element becomes the new largest

---

Step 4: Update Second Largest Number

else if (secondLargest < arr[i]) {
  secondLargest = arr[i];
}

If the number is not larger than largest but is larger than secondLargest, then update secondLargest.

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Time Complexity

• Time Complexity: O(n)
• Space Complexity: O(1)

This is an optimized solution because we traverse the array only once.