Discovering which elements in a numeric array are larger than the preceding elements

Question

Discovering which elements in a numeric array are larger than the preceding elements

array: const numbers = [158,164,742,99,155,250,240,87]

To display in the console only those numbers that are greater than the previous ones: console.log(164 742 155 250)

const numbers = [158, 164, 742, 99, 155, 250, 240, 87]
function findBiggerNumbers(numbers) {
  for (let i = 1; i < -1; i++) {
    if (numbers[i] > numbers[i - 1]) console.log(numbers[i])
  }
}

What is the solution to this problem? Appreciate any help.

javascript arrays

Answer 1

Answer №1

Employ .flatMap() and compare the current values with future values:

x < a[i+1] ? a[i+1] : []

I have noticed that there seems to be limited utility for using .flatMap() as a filter.

Therefore, why not opt for .map() instead of .flatMap()? Since .flatMap() essentially serves as an equivalent but with an additional step? This supplementary step enables me to devise a more straightforward logic where I can utilize simpler algorithms without having to concern myself with common side effects associated with handling arrays, such as obtaining empties. Here's a comparison between example #1 using .flatMap() and another version utilizing .map().

const data = [158, 164, 742, 99, 155, 250, 240, 87];

let resA = data.flatMap((x, i, a) => x < a[i + 1] ? a[i + 1] : []);

console.log(resA);

let resB = data.map((x, i, a) => x < a[i + 1] ? a[i + 1] : '');

console.log(resB);

Both .map() and .flatMap() necessitate returning data after each iteration, yet due to .flatMap() containing a built-in .flat() method, it allows for returning an empty array that equates to nothing, which is superior to returning an empty string ('') or undefined.

What about using .filter(), which undeniably appears to be the most straightforward solution? Let's compare Vectorjohn's callback here with my callback:

i > 0 && x > a[i - 1] // Vectorjohn, employing .filter()

x < a[i + 1] ? a[i + 1] : [] // My approach, utilizing .flatMap()

When utilizing .filter(), there is typically no need to worry about overshooting zero and triggering a return of

undefined</code, since <code>.filter()

solely returns truthy data and never falsey data. Hence, his callback might simply be the reverse of my callback by eliminating i > 0 &&:

x > a[i - 1] // .filter() will return x

x < a[i + 1] ? a[i + 1] : [] // .flatMap() will yield a[i + 1]

My approach doesn't automatically remove undefined, making it slightly more verbose than Vj's (once corrected). However, .flatMap() offers far greater versatility compared to .filter(). While .filter() consistently returns the current value (x) on each iteration, .flatMap() has the ability to yield anything (a[i + 1]), multiple items (["any", x]), or nothing ([]) during each iteration.

Illustrative Example 1

const data = [158, 164, 742, 99, 155, 250, 240, 87];

let res = data.flatMap((x, i, a) => x < a[i + 1] ? a[i + 1] : []);

console.log(res);

Answer 2

Employ .flatMap() and compare the current values with future values:

x < a[i+1] ? a[i+1] : []

I have noticed that there seems to be limited utility for using .flatMap() as a filter.

Therefore, why not opt for .map() instead of .flatMap()? Since .flatMap() essentially serves as an equivalent but with an additional step? This supplementary step enables me to devise a more straightforward logic where I can utilize simpler algorithms without having to concern myself with common side effects associated with handling arrays, such as obtaining empties. Here's a comparison between example #1 using .flatMap() and another version utilizing .map().

const data = [158, 164, 742, 99, 155, 250, 240, 87];

let resA = data.flatMap((x, i, a) => x < a[i + 1] ? a[i + 1] : []);

console.log(resA);

let resB = data.map((x, i, a) => x < a[i + 1] ? a[i + 1] : '');

console.log(resB);

Both .map() and .flatMap() necessitate returning data after each iteration, yet due to .flatMap() containing a built-in .flat() method, it allows for returning an empty array that equates to nothing, which is superior to returning an empty string ('') or undefined.

What about using .filter(), which undeniably appears to be the most straightforward solution? Let's compare Vectorjohn's callback here with my callback:

i > 0 && x > a[i - 1] // Vectorjohn, employing .filter()

x < a[i + 1] ? a[i + 1] : [] // My approach, utilizing .flatMap()

When utilizing .filter(), there is typically no need to worry about overshooting zero and triggering a return of

undefined</code, since <code>.filter()

solely returns truthy data and never falsey data. Hence, his callback might simply be the reverse of my callback by eliminating i > 0 &&:

x > a[i - 1] // .filter() will return x

x < a[i + 1] ? a[i + 1] : [] // .flatMap() will yield a[i + 1]

My approach doesn't automatically remove undefined, making it slightly more verbose than Vj's (once corrected). However, .flatMap() offers far greater versatility compared to .filter(). While .filter() consistently returns the current value (x) on each iteration, .flatMap() has the ability to yield anything (a[i + 1]), multiple items (["any", x]), or nothing ([]) during each iteration.

Illustrative Example 1

const data = [158, 164, 742, 99, 155, 250, 240, 87];

let res = data.flatMap((x, i, a) => x < a[i + 1] ? a[i + 1] : []);

console.log(res);

Answer 3

Answer №2

When using the array filter method, a function is executed for each element in the array. Only elements that return true from the function are included in the final result. The current element's index is passed as an argument to the function, allowing you to access the previous element within your filter logic.

data.filter((value, i) => i > 0 && value > data[i - 1])

filter can also take the original array as a third argument, which can be useful for making the filter callback more reusable:

const biggerThanPrevious = (value, i, array) => i > 0 && value > array[i - 1]
// ... elsewhere in the code
const data = [158,164,742,99,155,250,240,87]
console.log(data.filter(biggerThanPrevious))

Answer 4

When using the array filter method, a function is executed for each element in the array. Only elements that return true from the function are included in the final result. The current element's index is passed as an argument to the function, allowing you to access the previous element within your filter logic.

data.filter((value, i) => i > 0 && value > data[i - 1])

filter can also take the original array as a third argument, which can be useful for making the filter callback more reusable:

const biggerThanPrevious = (value, i, array) => i > 0 && value > array[i - 1]
// ... elsewhere in the code
const data = [158,164,742,99,155,250,240,87]
console.log(data.filter(biggerThanPrevious))

Answer 5

Answer №3

You were so close to getting it right with your forloop.

IMPORTANT: Avoid using negative values in your loops unless specifically needed to iterate backwards.

const numbers = [158, 164, 742, 99, 155, 250, 240, 87];
var result = [];
for(let i = 0; i < numbers.length; i++){
  if(numbers[i] > numbers[i - 1]){
    result.push(numbers[i]);
  }
}

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

Answer 6

You were so close to getting it right with your forloop.

IMPORTANT: Avoid using negative values in your loops unless specifically needed to iterate backwards.

const numbers = [158, 164, 742, 99, 155, 250, 240, 87];
var result = [];
for(let i = 0; i < numbers.length; i++){
  if(numbers[i] > numbers[i - 1]){
    result.push(numbers[i]);
  }
}

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

Answer 7

Answer №4

const dataSet = [158, 164, 742, 99, 155, 250, 240, 87]

let outputNumbers = ''

dataSet.forEach(num => {
  const indx = dataSet.indexOf(num)

  if (num > dataSet[indx - 1]) {
    outputNumbers += `${num} `
    // you may also simply log the number using console.log(num), but I prefer to append to a string
  } 
})

console.log(outputNumbers)

Answer 8

const dataSet = [158, 164, 742, 99, 155, 250, 240, 87]

let outputNumbers = ''

dataSet.forEach(num => {
  const indx = dataSet.indexOf(num)

  if (num > dataSet[indx - 1]) {
    outputNumbers += `${num} `
    // you may also simply log the number using console.log(num), but I prefer to append to a string
  } 
})

console.log(outputNumbers)

Answer 9

Answer №5

This situation seems like a perfect fit for utilizing Array.prototype.reduce().

Set the second parameter of the reduce function (the accumulator/carry value) as an object with two keys that represent the resulting array and the value from the previous iteration.

let data = [158, 164, 742, 99, 155, 250, 240, 87];

let result = data.reduce(function(accumulator, currValue) {
  if (currValue > accumulator.lastValue) {
    accumulator.greaterThanPrev.push(currValue);
  }
  accumulator.lastValue = currValue;
  return accumulator;

}, {
  greaterThanPrev: [],
  lastValue: 4294967295 // BACK-COMPAT. MAX INT.
});

console.log(result.greaterThanPrev)

In each iteration, compare the current value with the 'lastValue' in the object passed through the reduce function. If it's larger, add it to the 'greaterThanPrev' array and update the 'lastValue'. Return this updated object for the next iteration. Once the reduce function completes, the final object will contain all the desired values in 'greaterThanPrev'.

There might be cleaner solutions out there, but this one came to mind quickly.

Answer 10

This situation seems like a perfect fit for utilizing Array.prototype.reduce().

Set the second parameter of the reduce function (the accumulator/carry value) as an object with two keys that represent the resulting array and the value from the previous iteration.

let data = [158, 164, 742, 99, 155, 250, 240, 87];

let result = data.reduce(function(accumulator, currValue) {
  if (currValue > accumulator.lastValue) {
    accumulator.greaterThanPrev.push(currValue);
  }
  accumulator.lastValue = currValue;
  return accumulator;

}, {
  greaterThanPrev: [],
  lastValue: 4294967295 // BACK-COMPAT. MAX INT.
});

console.log(result.greaterThanPrev)

In each iteration, compare the current value with the 'lastValue' in the object passed through the reduce function. If it's larger, add it to the 'greaterThanPrev' array and update the 'lastValue'. Return this updated object for the next iteration. Once the reduce function completes, the final object will contain all the desired values in 'greaterThanPrev'.

There might be cleaner solutions out there, but this one came to mind quickly.

Discovering which elements in a numeric array are larger than the preceding elements

Answer №1

Illustrative Example 1

Answer №2

Answer №3

Answer №4

Answer №5

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