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  1. Prepare
  2. Tutorials
  3. 10 Days of Javascript
  4. Day 3: Arrays
  5. Discussions

Day 3: Arrays

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1474 Discussions

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  • pavankalyan1899
     + 0 comments

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

    }

  • anshulmankar84
     + 0 comments

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

  • habiba0helmy
     + 0 comments

    function getSecondLargest(nums) { nums.sort((a,b)=>{ return b-a;
    }) for(var i=0; i nums[i+1]) return nums[i+1]; } }

  • amanhaidry
     + 0 comments

    Simple solution in single loop

    function getSecondLargest(nums) {
    // Complete the function
    let largest = nums[0];
    let secondLargest= nums[0];
    let n= nums.length ;
    
    for(let i=0; i<n; i++){
        if(nums[i]>largest){
            secondLargest=largest;
            largest= nums[i];
        }else if(nums[i]>secondLargest && nums[i]!= largest){
            secondLargest= nums[i];
        }
    }
    return secondLargest;
}

  • slavynskyi_denys
     + 0 comments

    So, O(n) is the best we could expect here. 

    function getSecondLargest(nums) {
    // Complete the function
    let max = -Infinity, sndMax = -Infinity;
    for (let num of nums) {
        if (num > max) {
            sndMax = max;
            max = num;
        } else if (num > sndMax && num !== max) {
            sndMax = num;
        }
    }
    return sndMax;
}