Question

How to find the second largest number in array in JavaScript

Answer 1

function secondLargest(arr) {
    if (arr.length < 2) return null; // Handle edge case for arrays with less than 2 elements
    
    let sortedArr = [...arr].sort((a, b) => b - a);
    
    return sortedArr[1];
}

let numbers = [3, 14, 14, 1, 1, 1, 90, 2, 6, 86, 7];

console.log(secondLargest(numbers)); 



/*
run:

86

*/

 

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Answer 2

function secondLargest(arr) {
    if (arr.length < 2) return null; // Handle edge case for arrays with less than 2 elements
    
    let first = -Infinity, second = -Infinity;
    for (let num of arr) {
        if (num > first) {
            second = first;
            first = num;
        } else if (num > second && num < first) {
            second = num;
        }
    }
    
    return second;
}

let numbers = [3, 14, 14, 1, 1, 1, 90, 2, 6, 86, 7];

console.log(secondLargest(numbers)); 



/*
run:

86

*/

 

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Answer 3

function secondLargest(arr) {
    if (arr.length < 2) return null; // Handle edge case for arrays with less than 2 elements
    
    let uniqueArr = [...new Set(arr)];
    if (uniqueArr.length < 2) return null; // Handle edge case for arrays with less than 2 unique elements
    uniqueArr.sort((a, b) => b - a);
    
    return uniqueArr[1];
}

let numbers = [3, 14, 14, 1, 1, 1, 90, 2, 6, 86, 7];

console.log(secondLargest(numbers)); 



/*
run:

86

*/

 

70+ SQL courses for beginners and professionals