Best Way to Get Objects with Highest Property Value

Question

I have the following multidimensional array of student objects:

var students = [
{name: \"Jack\", age: \"NYN\", attempts: 3, wrong: 2},
{name: \"Phil\", age: \"N         


        

Answer 1

The most straight-forward algorithm to achieve your goal would be:

1. Find the maximum value for wrong within students array.
2. Use filter to leave only the relevant students (with the property wrong equals the maximum).

var students = [{name: "Jack", age: "NYN", attempts: 3, wrong: 2},{name: "Phil", age: "NNNY", attempts: 4, wrong: 3},{name: "Tom", age: "", attempts: 0, wrong: 0},{name: "Lucy", age: "YYNY", attempts: 4, wrong: 1},{name: "Ben", age: "NYNN", attempts: 4, wrong: 3},{name: "Hardest", age: "NNN", attempts: 3, wrong: 3}];

// Find the maximum 'wrong' value
let maxValue = 0;
for(let student of students) {
  if(student.wrong > maxValue) {
    maxValue = student.wrong;
  }
}

// filter out the students with 'wrong' value different than the maximum
let onlyMax = students.filter(item => item.wrong == maxValue);
console.log(onlyMax);

Note all the algorithm does is iterating the array twice, resulting in run-time of O(2n) = O(n).

---

The more general solution allows one to find the items with maximum value of property in an objects array:

var students = [{name: "Jack", age: "NYN", attempts: 3, wrong: 2},{name: "Phil", age: "NNNY", attempts: 4, wrong: 3},{name: "Tom", age: "", attempts: 0, wrong: 0},{name: "Lucy", age: "YYNY", attempts: 4, wrong: 1},{name: "Ben", age: "NYNN", attempts: 4, wrong: 3},{name: "Hardest", age: "NNN", attempts: 3, wrong: 3}];

function filterByMax(arr, property) {
  // Find the maximum 'wrong' value
  let maxValue = 0;
  for(let item of arr) {
    if(item[property] > maxValue) {
      maxValue = item[property];
    }
  }
  
  // filter out the students with 'wrong' value different than the maximum
  return arr.filter(item => item[property] == maxValue);
}

console.log(filterByMax(students, 'wrong'));

Answer 2

Your code will also fail if the "wrong" property values are in ascending order like

var students = [
    {name: "Jack", age: "NYN", attempts: 3, wrong: 2},
    {name: "Phil", age: "NNNY", attempts: 4, wrong: 3},
    {name: "Tom", age: "", attempts: 0, wrong: 0},
    {name: "Lucy", age: "YYNY", attempts: 4, wrong: 1},
    {name: "Ben", age: "NYNN", attempts: 4, wrong: 3},
    {name: "Hardest", age: "NNN", attempts: 3, wrong: 3}
    {name: "Mad", age: "NYN", attempts: 3, wrong: 5},
]

The result will include Jack, Phil, Ben, Hardest & Mad

---

You have to discard the previous results once you find new person with greater "wrong" value, see the snippet below...

var s2 = "Jack:NYN,Phil:NNNY,Tom:,Lucy:YYNY,Ben:NYNN,Hardest:NNN";
var s2Arr = s2.split(','); // convert string to an array
var s2MdArr = s2Arr.map(function(e) {return e.split(':'); }); // convert to MD array
var totalWrongAnswers = 0;

for(i=0; i < s2Arr.length; i++) {
    var attempts = s2MdArr[i][1].length;
	var noWrong = (s2MdArr[i][1].match(/N/g) || []).length;
	s2MdArr[i].push(attempts); // add to array[i][2]
	s2MdArr[i].push(noWrong); // add to array[i][3]
	totalWrongAnswers += noWrong; // update total wrong
}

var s2ArrObj = s2MdArr.map(function(e) { return {name: e[0], age: e[1], attempts: e[2], wrong: e[3]} }); // create objects in MD Array

    var firstPerson = s2ArrObj[0]; // initialise so can make a comparison
    var person = firstPerson;
    var peopleMostNeedingHelp = [];
// update person to the person with the highest no. of wrong answers
function something() {
    for (i = 0; i < s2ArrObj.length; i++) {
        // for each person
        if (s2ArrObj[i].wrong > person.wrong) {
            // discard previous results and create new list with single new person only
            person = s2ArrObj[i];
            // update person variable so can compare next person
            peopleMostNeedingHelp = [person];
        }
        else if (s2ArrObj[i].wrong == person.wrong) {
            // add the person to list
            person = s2ArrObj[i];
            // update person variable so can compare next person
            peopleMostNeedingHelp.push(person);
        }
    }
}

something();
console.log(peopleMostNeedingHelp);

Answer 3

You could reduce the array by checking the wrong property.

var students = [{ name: "Jack", age: "NYN", attempts: 3, wrong: 2 }, { name: "Phil", age: "NNNY", attempts: 4, wrong: 3 }, { name: "Tom", age: "", attempts: 0, wrong: 0 }, { name: "Lucy", age: "YYNY", attempts: 4, wrong: 1 }, { name: "Ben", age: "NYNN", attempts: 4, wrong: 3 }, { name: "Hardest", age: "NNN", attempts: 3, wrong: 3 }],
    topWrong = students.reduce((r, o) => {
        if (!r || o.wrong > r[0].wrong) {
            return [o];
        }
        if (o.wrong === r[0].wrong) {
            r.push(o);
        }
        return r;
    }, undefined);
    
console.log(topWrong);

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Answer 4

You have to reset the peopleMostNeedingHelp when a new student with a higher number of errors is discovered:

let mostNeedingHelp = [students[0]];

for(const student of students.slice(1)) {
  if(student.errors === mostNeedingHelp[0].errors) {
    mostNeedingHelp.push(student);
  } else if(student.errors >= mostNeedingHelp[0].errors) {
    mostNeedingHelp = [student]; // <<<<
  }
}

This can be shortified with reduce:

const mostNeedingHelp = students.slice(1).reduce((arr, student) => 
 arr[0].errors === student.errors ? arr.concat(student) : arr[0].errors < student.errors ? [student] : arr, [students[0]]);