What to use instead of push?

Question

In the code below, I use "push" to fill and empty array. I need help to write the above code so that it comes out the same way, without using "push". I have been challenged to this by the book, "Head First JavaScript Programming". I have tried, but I am stumped.

let scores = [60, 50, 60, 58, 54, 54, 58, 50, 52, 54, 48, 69, 34, 55, 51, 52, 44, 51, 69,
  64, 66, 55, 52, 61, 46, 31, 57, 52, 44, 18, 41, 53, 55, 61, 51, 44
]
var highScore = 0
var output


for (var i = 0; i < scores.length; i++) {
  output = `Bubbles solution # ${i} score: ${scores[i]}<br>`
  document.write(output)
  if (scores[i] > highScore) {
    highScore = scores[i]
  }
}
let bestSolutions = []
for (var i = 0; i < scores.length; i++) {
  if (scores[i] == highScore) {
    bestSolutions.push([i])
  }
}


document.write(`Bubbles Tests: ${scores.length}<br>`)
document.write(`Highest Bubble Score: ${highScore}<br>`)
document.write(`Solutions with highest score: #${bestSolutions[0]} and #${bestSolutions[1]}`)

Answer 1

If I understand this correctly you're looking to find all max values in an array of numbers. If that is the case, you can use the reduce and filter methods as an alternative to using push.

function findMaxes(arr) {
    const max = Math.max.apply(null, arr);
    return arr.filter(n => n == max);
}

For example:

findMaxes([3, 5, 1, 1, 4, 5]) == [5, 5]
findMaxes([-1, -1, -1]) == [-1, -1, -1]

If you want to find the positions of all maxes:

function findAllMaxPositions(arr) {
    const max = Math.max.apply(null, arr);
    return arr.map((e, index) => [e, index])
              .filter(pair => pair[0] == max)
              .map(e => e[1]);
}

Answer 2

Ok, one approach is the following : If you want your bestscores array to contain the indices of the elements in scores that are equal to highscore, you can define bestcores as the map of scores through the function that takes a tuple (value,index) and return index if value == highscore otherwise null.

Examples of maps

const a = [2, 3, 5, 0, 6];

const b = a.map(value => value * 2); // [4, 6, 10, 0, 12];
const c = a.map((value, index) => index); //[0, 1, 2, 3, 4];
const d = a.map((value, index) => value + index); //[2, 4, 7, 3, 10];

const e = a.map((value,index) => value >= 3 ? index : null);
// [null, 1, 2, null, 4]. We can deal with those nulls with a filter
const f = e.filter(value => value != null) // [1,2,4]

Altogether in your case

const highscores = scores.map((value,index) => value == highscore ? index : null)
  .filter(value => value != null);

This gives you a high level approach but is a bit computationally inefficient as it actually does a double iteration over the scores array. A manual iteration over your array and a push is actually more efficient.

Answer 3

You can try this approach:

• Use Math.max.apply to find the highest score.
• Use a loop to find all matching values and store their indexes.

let scores = [ 60, 50, 60, 58, 54, 54, 58, 50, 52, 54, 48, 69, 34, 55, 51, 52, 44, 51, 69, 64, 66, 55, 52, 61, 46, 31, 57, 52, 44, 18, 41, 53, 55, 61, 51, 44 ]
var highScore = Math.max.apply(null, scores)
var bestSolutions = scores.reduce((acc, score, index) =>
  score === highScore ? [ ...acc, index] : acc, []
);

document.write(`Bubbles Tests: ${scores.length}<br>`)
document.write(`Highest Bubble Score: ${highScore}<br>`)
document.write(`Solutions with highest score: #${bestSolutions[0]} and #${bestSolutions[1]}`)

---

Algo based:

• Create 2 variables, highScore and bestSolutions.

• Loop over Array.

• For every iteration, make 2 checks:

• If current score is greater than highScore

  • If yes, initialize highScore with current score and initialize bestSolutions with current index.

• If highScore is equal to current score

  • If yes, push index to current bestSolutions

let scores = [ 60, 50, 60, 58, 54, 54, 58, 50, 52, 54, 48, 69, 34, 55, 51, 52, 44, 51, 69, 64, 66, 55, 52, 61, 46, 31, 57, 52, 44, 18, 41, 53, 55, 61, 51, 44 ];

let highScore = 0;
let bestSolutions = [];

for(let i = 0; i< scores.length; i++) {
  if (scores[i] > highScore) {
    highScore = scores[i];
    bestSolutions = [ i ];
  } else if (scores[i] === highScore) {
    bestSolutions.push(i)
  }
}

document.write(`Bubbles Tests: ${scores.length}<br>`)
document.write(`Highest Bubble Score: ${highScore}<br>`)
document.write(`Solutions with highest score: #${bestSolutions[0]} and #${bestSolutions[1]}`)

Reference:

• Find the min/max element of an Array in JavaScript

Answer 4

You can use ( ) rather than [ ] in bestSolutions.push(i)

var bestSolutions = [];
for (let i = 0; i < scores.length; i++) {
    if (scores[i] == highScore) {
        bestSolutions.push(i)
    }

}

Answer 5

A very obvious alternative would be

bestSolutions[bestSolutions.length] = [i];

Explanation: At any time, in bestSolutions you have exactly bestSolutions.length number of elements. The first has the index of 0, the second has the index of 1, ..., the last is having the index of bestSolutions.length - 1. Hence, the index of the n+1th element will be bestSolutions.length. After assigning the new value to this index, bestSolutions.length will equal its former value + 1.

The edge-case in the above is when bestSolutions is empty, in which case it does not have first or last elements, but the suggested code will work for this case as well, because in this case bestSolutions.length will be 0, exactly the element you want to initialize.

let scores = [60, 50, 60, 58, 54, 54, 58, 50, 52, 54, 48, 69, 34, 55, 51, 52, 44, 51, 69,
  64, 66, 55, 52, 61, 46, 31, 57, 52, 44, 18, 41, 53, 55, 61, 51, 44
]
var highScore = 0
var output


for (var i = 0; i < scores.length; i++) {
  output = `Bubbles solution # ${i} score: ${scores[i]}<br>`
  document.write(output)
  if (scores[i] > highScore) {
    highScore = scores[i]
  }
}
let bestSolutions = []
for (var i = 0; i < scores.length; i++) {
  if (scores[i] == highScore) {
    bestSolutions[bestSolutions.length] = [i];
  }
}


document.write(`Bubbles Tests: ${scores.length}<br>`)
document.write(`Highest Bubble Score: ${highScore}<br>`)
document.write(`Solutions with highest score: #${bestSolutions[0]} and #${bestSolutions[1]}`)