Return Max number from each sub-array of array into new array

Question

I am doing a freecodecamp algorithm scripting problem, and I've gone as far as my understanding will allow. The challenge at hand is to take an array full of 4 sub-arrays of numbers, search through it with javascript, and return a new array with the max value of each sub-array. I've seen their solution and understand it decently, but I've been working on my own solution using nested for loops and a ternary operator. I'll display their solution first, then my faulty solution, where it is saying that the function with arguments is undefined.

Below is the code:

Their Solution:

function largestOfFour(arr) {
  var results = [];
  for (var n = 0; n < arr.length; n++) {
    var largestNumber = arr[n][0];
    for (var sb = 1; sb < arr[n].length; sb++) {
      if (arr[n][sb] > largestNumber) {
        largestNumber = arr[n][sb];
      }
    }

    results[n] = largestNumber;
  }

  return results;
}

The solution that I am working on (currently doesn't work):

function largestOfFour(arr) {
  var maxNum = 0;
  var results = [];
  for (let i = 0; i < arr.length; i++) {
    for (let j = 0; j < arr[i].length; j++) {
      arr[i][j] > maxNum ? results.push(arr[i][j]) : delete arr[i][j]; 
    }
  }
}

For example,

console.log(largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]));

should display an array of [5, 27, 39, 1001].

Your function doesn't return anything, so the array your create dies when the function exits. — WillardSolutions
– WillardSolutions, Commented Apr 25, 2019 at 13:25
It's not clear what you're asking here. Your best bet here is to use your debugger, step through the code, and watch what happens. — T.J. Crowder
– T.J. Crowder, Commented Apr 25, 2019 at 13:27
@Oversought - Here's the documentation for Chrome. It's similar in other browsers. — T.J. Crowder
– T.J. Crowder, Commented Apr 25, 2019 at 13:59

Ashutosh Hans · Accepted Answer · 2019-04-25 14:48:18Z

1

function largestOfFour(arr) {
  //var maxNum = 0; updated
  var results = [];
  for (let i = 0; i < arr.length; i++) {
    var maxNum = -Infinity; //updated
    for (let j = 0; j < arr[i].length; j++) {
      if(arr[i][j] > maxNum){
        maxNum = arr[i][j];
      }
    }
    results.push(maxNum);
  }
 return results;
}
largestOfFour([[2,4,6],[45,56,78]]); //[6, 78]

hey, you can modify it this way. You might have missed updating the maxNum and after innerLoop ends, we can push the maximum in the results array.

edited Apr 25, 2019 at 14:48

answered Apr 25, 2019 at 13:37

Ashutosh Hans

93 bronze badges

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5 Comments

Oversought Over a year ago

Yea, I did miss that. Thank you. :)

Oversought Over a year ago

Although your solution isn't working completely. I am not sure why, hopefully someone can explain.

Oversought Over a year ago

Apparently the var maxNum = 0 should be put at the beginning of the first for loop. I don't know why, but that seems to be the consensus above.

Ashutosh Hans Over a year ago

@Oversought, yep I got it, thanks for pointing out. It is because if the max of the previous sub array is greater than all the elements of the next sub array it will push the previous max only. Hope you get me. I have updated the answer

Oversought Over a year ago

dope thx, appreciate it, I posted my solution below that was made with everyone's help. Hopefully it is of use to people.

Niraj Kaushal · Accepted Answer · 2019-04-29 08:46:31Z

0

Note: You can solve this problem with two lines of code in ES6

You can see this working example to see where you are doing wrong,

function largestOfFour(arr) {
 //var maxNum = 0;
  var results = [];
  for (let i = 0; i < arr.length; i++) {
    var maxNum = null;
    for (let j = 0; j < arr[i].length; j++) { 
      //You can use ternary operator here 
      if (maxNum) {
        if (arr[i][j] > maxNum) {
          maxNum = arr[i][j];
        }
      }
      else {
        maxNum = arr[i][j];
      }
      
    }
    results.push(maxNum);
  }
  
  return results;
}

console.log(largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1], [-1, -45, -65, -3]]));

edited Apr 29, 2019 at 8:46

answered Apr 25, 2019 at 13:34

Niraj Kaushal

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8 Comments

Oversought Over a year ago

Thank you for showing me how I can make my line of logic for coding out this algorithm work. I appreciate it, rather than just saying I am wrong completely and to do it a different way. Could you explain how moving where maxNum is defined changes it to the correct solution?

Oversought Over a year ago

How would I get this to work if they used a negative number? I have some ideas but don't know hot to elucidate them easily. Obviously setting var maxNum = 0 is this issue with negative numbers here.

Niraj Kaushal Over a year ago

@Oversought, You have to reset maxNum(maxNum=0) for each subarray(if subarray contains positive numbers) to get maxNum of that subarray.

Oversought Over a year ago

I'm so confused with this website. Looks like someone downvoted your solution. That's not right. It works, doesn't it? Just doesn't work if all numbers in subarray are negative, but I'm working on that atm.

Niraj Kaushal Over a year ago

@Oversought I have updated my answer and that will work with negative numbers too....Every youtube videos(having more than 1000 views) have dislike counts...I hope you got my point :)

|

Oversought · Accepted Answer · 2019-04-25 14:32:07Z

0

HERE IS MY SOLUTION WITH HELP OF PEOPLE IN THREAD:

function largestOfFour(arr) {
  var results = [];
  for (let i = 0; i < arr.length; i++) {
    var maxNum = undefined;
    for (let j = 0; j < arr[i].length; j++) {
      if (maxNum) {
        arr[i][j] > maxNum ? maxNum = arr[i][j] : delete arr[i][j];
      } else {
        maxNum = arr[i][j];
      }
    }
    results.push(maxNum);
  }
  return results;
}

answered Apr 25, 2019 at 14:32

Oversought

778 bronze badges

1 Comment

Oversought Over a year ago

It may not be the most elegant, but it follows my initial line of logic and understanding, so that's good.

Oversought · Accepted Answer · 2019-04-28 17:47:50Z

I went back through the Basic Algorithm Scripting section to get a more thorough re-examination of the problems, as to learn the material better. The second time I re-wrote this, I ended up using a very similar if not exact-same approach to a solution given in this thread. Some things might be a little different, so I thought I would post my second solution. Again, this can be solved in two lines of code using ES6, but that's not as fun. :)

SOLUTION:

function largestOfFour(arr) {
  var finalArr = [];

  for (let i = 0; i < arr.length; i++) {
    var maxCounter = -Infinity;
    for (let j = 0; j < arr[i].length; j++) {
      if (arr[i][j] > maxCounter) {
        maxCounter = arr[i][j];
      }
    }
    finalArr.push(maxCounter);
  }

  return finalArr;
}

You can see here that -Infinity is used as a 'number' instead of undefined, as in my first posted solution. I think that cleared up some unnecessary space of code as well, and I like how it works logically more too.

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