Question

How to find the second largest number in array in TypeScript

Answer 1

function secondLargest(arr: number[]): any {
    if (arr.length < 2) return null; // Handle edge case for arrays with less than 2 elements
     
    let sortedArr: number[] = [...arr].sort((a, b) => b - a);
     
    return sortedArr[1];
}
 
let numbers: number[] = [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: number[]): any {
    if (arr.length < 2) return null; // Handle edge case for arrays with less than 2 elements
     
    let first: number = -Infinity, second : number= -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: number[] = [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: number[]): any {
    if (arr.length < 2) return null; // Handle edge case for arrays with less than 2 elements
     
    let uniqueArr: number[] = [...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: number[] = [3, 14, 14, 1, 1, 1, 90, 2, 6, 86, 7];
 
console.log(secondLargest(numbers)); 
 
 
 
/*
run:
 
86
 
*/
 

 

70+ SQL courses for beginners and professionals