Finding the largest number that is less than a specified variable within an array

Question

Finding the largest number that is less than a specified variable within an array

I have a JavaScript array and a variable, set up like this;

var values = [0, 1200, 3260, 9430, 13220],
    targetValue = 4500;

How can I efficiently find the largest value in the array that is less than or equal to the given variable?

In the provided example, the desired result would be 3260.

One approach could be like the following;

values.sort((a, b) => b - a);
let result = values.find(val => val <= targetValue);

However, it's important to consider cases where the selected value is the last one in the array. Additionally, the solution should be concise for such a simple task.

Is there a more efficient way to achieve this without unnecessary verbosity?

(and yes, using a jQuery loop here was not ideal, but it serves as a simplified example)

javascript arrays

Answer 1

Answer №1

If you want to achieve the same outcome in a more elegant way, consider using a combination of array filtering and apply(). This method provides a cleaner and easier-to-read solution. By utilizing filter(), you can extract elements from array 'a' that do not meet the condition specified by the predicate function. Then, by employing apply(), Math.max is called with each extracted element as an argument.

let a = [1, 2, 3, 4, 5];
let b = 4;
let result = Math.max.apply(Math, a.filter(function(x){return x <= b}));

The resulting value will be 4.

Answer 2

If you want to achieve the same outcome in a more elegant way, consider using a combination of array filtering and apply(). This method provides a cleaner and easier-to-read solution. By utilizing filter(), you can extract elements from array 'a' that do not meet the condition specified by the predicate function. Then, by employing apply(), Math.max is called with each extracted element as an argument.

let a = [1, 2, 3, 4, 5];
let b = 4;
let result = Math.max.apply(Math, a.filter(function(x){return x <= b}));

The resulting value will be 4.

Answer 3

Answer №2

let maximumValue = Number.MIN_VALUE;
for (let index = 0; index < array.length; index++) { 
  if (array[index] <= limit && array[index] > maximumValue) { 
    maximumValue = array[index]; 
  } 
}

In my opinion, the method mentioned above is straightforward, easy to understand, and not overly wordy. Another approach would be to utilize the reduce function, like this:

let maxVal = array.reduce(function (previousValue, currentValue) { return currentValue <= limit ? Math.max(previousValue, currentValue) : previousValue }, Number.MIN_VALUE);

Answer 4

let maximumValue = Number.MIN_VALUE;
for (let index = 0; index < array.length; index++) { 
  if (array[index] <= limit && array[index] > maximumValue) { 
    maximumValue = array[index]; 
  } 
}

In my opinion, the method mentioned above is straightforward, easy to understand, and not overly wordy. Another approach would be to utilize the reduce function, like this:

let maxVal = array.reduce(function (previousValue, currentValue) { return currentValue <= limit ? Math.max(previousValue, currentValue) : previousValue }, Number.MIN_VALUE);

Answer 5

Answer №3

Using the grep() function in jQuery

 var numbers = [10, 200, 560, 980, 1220],
 target = 300; 
 var result= Math.max.apply( Math,$.grep(numbers,function(num){return num<=target}));
 document.write(result)

SEE IT IN ACTION

Answer 6

Using the grep() function in jQuery

 var numbers = [10, 200, 560, 980, 1220],
 target = 300; 
 var result= Math.max.apply( Math,$.grep(numbers,function(num){return num<=target}));
 document.write(result)

SEE IT IN ACTION

Answer 7

Answer №4

Is there always a sorted array? If that's the case, you may be able to optimize your code in this way (make sure to verify the indexes, as I have not done so):

let start = 0;
let finish = arr.length;
while ((finish - start) > 1) {
    let currentIdx = Math.floor((start + finish) / 2);;
    if (arr[currentIdx] < target) {
        start = currentIdx;
    } else if (arr[currentIdx] > target){
        finish = currentIdx;
    } else {
        start = finish = currentIdx;
    }
}
let maximum = arr[start];

Answer 8

Is there always a sorted array? If that's the case, you may be able to optimize your code in this way (make sure to verify the indexes, as I have not done so):

let start = 0;
let finish = arr.length;
while ((finish - start) > 1) {
    let currentIdx = Math.floor((start + finish) / 2);;
    if (arr[currentIdx] < target) {
        start = currentIdx;
    } else if (arr[currentIdx] > target){
        finish = currentIdx;
    } else {
        start = finish = currentIdx;
    }
}
let maximum = arr[start];

Answer 9

Answer №5

let nearestValue = null;
$.each(arrayData, function() {
    if (nearestValue == null || Math.abs(this - target) < Math.abs(nearestValue - target)) {
        nearestValue = this;
    }
});

A jQuery alternative in case it is needed.

Answer 10

let nearestValue = null;
$.each(arrayData, function() {
    if (nearestValue == null || Math.abs(this - target) < Math.abs(nearestValue - target)) {
        nearestValue = this;
    }
});

A jQuery alternative in case it is needed.

Answer 11

Answer №6

To achieve the desired outcome, you can use the following function:

function findMaximum(arr, maxValue){
  var result;
  for(var index = 0, length=arr.length; index<length; index++){
    if(arr[index] <= maxValue) result=arr[index];
    else return result;
  }
  return false;
}

Answer 12

To achieve the desired outcome, you can use the following function:

function findMaximum(arr, maxValue){
  var result;
  for(var index = 0, length=arr.length; index<length; index++){
    if(arr[index] <= maxValue) result=arr[index];
    else return result;
  }
  return false;
}

Finding the largest number that is less than a specified variable within an array

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

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