function getSecondLargest(nums) {

// Complete the function

let max=nums[0];
let secondmax = 0;

for (let i=0;i<nums.length-1;i++)
{
    if(nums[i]>nums[i+1] && nums[i]>max){
        max = nums[i];
    }
    else if(nums[i+1]>max ){
        max=nums[i+1];
    }
}

//10,
for(let i=0;i<nums.length-1;i++){
    if(nums[i]>nums[i+1] && 
    Number(nums[i])!=Number(max) && Number(nums[i+1])!=
    Number(max) && nums[i]>secondmax)
    {
        secondmax=nums[i];

    }

    else if(Number(nums[i+1])!=
    Number(max)&& nums[i+1]>secondmax){secondmax=nums[i+1];}


}
return secondmax;

}

Says constraint of n can be 1, but it MUST be at least 2 for this exercise.

**Simple and Short Solution **

function getSecondLargest(nums) {
nums.sort((a,b) => a-b )
const c = [...new Set(nums)]
return c.at(-2)
}

function getSecondLargest(nums) { // Complete the function let fun = nums.sort(function(a,b){return b-a}) let rem = [] for (let i = 0;i

}

function getSecondLargest(nums) { let maxVal = Math.max(...nums); nums = nums.filter(item => item !== maxVal) let maxVal2 = Math.max(...nums); return maxVal2; }