In JavaScript, locate the closest element in an array based on its value, avoiding cases where the value exceeds the nearest element

Question

In JavaScript, locate the closest element in an array based on its value, avoiding cases where the value exceeds the nearest element

I am trying to find the nearest element in an array based on a given value, but I specifically want it to be the nearest one that is greater than or equal to the value. For example, if I have an array like [3000, 5000, 8000], when I search for a number below or equal to 3000, it should return 3000. However, if I search for 3001, it should return 5000 instead.

This is the code I currently have:

let array = [3000, 5000, 8000];

function closestNumArray(array, num) {
    var x = array.reduce(function(min, max) {
      return (Math.abs(max - num) < Math.abs(min - num) ? max : min);
    });

    return x;
}

closesNumArray(array, 3001) // returns 3000

I would like it to return 5000, or the nearest next element based on the value, but it currently returns 3000.

Thank you!

javascript arrays

Answer 1

Answer №1

To find the closest number in a sorted array, a simple while loop can be used. If the loop reaches the end of the array (due to the value being too large), it will return the last value in the array.

let array = [3000, 5000, 8000];

const findClosestNumber = function(array, val) {
  let i = 0;
  while (i < array.length && array[i] < val) i++;
  return array[i] || array[i-1];
}

console.log(findClosestNumber(array, 2999));
console.log(findClosestNumber(array, 3000));
console.log(findClosestNumber(array, 3001));
console.log(findClosestNumber(array, 5000));
console.log(findClosestNumber(array, 5001));
console.log(findClosestNumber(array, 8000));
console.log(findClosestNumber(array, 8001));

Answer 2

To find the closest number in a sorted array, a simple while loop can be used. If the loop reaches the end of the array (due to the value being too large), it will return the last value in the array.

let array = [3000, 5000, 8000];

const findClosestNumber = function(array, val) {
  let i = 0;
  while (i < array.length && array[i] < val) i++;
  return array[i] || array[i-1];
}

console.log(findClosestNumber(array, 2999));
console.log(findClosestNumber(array, 3000));
console.log(findClosestNumber(array, 3001));
console.log(findClosestNumber(array, 5000));
console.log(findClosestNumber(array, 5001));
console.log(findClosestNumber(array, 8000));
console.log(findClosestNumber(array, 8001));

Answer 3

Answer №2

Discovering the lower_bound of an element is what you need to do. This involves finding the position in a sorted array where your target element should be inserted.

let numbers = [3000, 5000, 8000];


numbers.sort();

console.log(numbers)

let goal = 3001;
let result = -1;
for(let i=0;i<numbers.length;i++){
if(goal<=numbers[i]){
result = i;
        break;
}
}

if(result == -1){
console.log("Element not found");
}
else{
console.log(numbers[result]); // -1 for when all elements are smaller than target
}

Answer 4

Discovering the lower_bound of an element is what you need to do. This involves finding the position in a sorted array where your target element should be inserted.

let numbers = [3000, 5000, 8000];


numbers.sort();

console.log(numbers)

let goal = 3001;
let result = -1;
for(let i=0;i<numbers.length;i++){
if(goal<=numbers[i]){
result = i;
        break;
}
}

if(result == -1){
console.log("Element not found");
}
else{
console.log(numbers[result]); // -1 for when all elements are smaller than target
}

Answer 5

Answer №3

Finding the lowest element in O(log(n)) time complexity using binary search.

let numbers = [3000, 5000, 8000];

numbers.sort();

let key = 10000;

function findLowest(numbers, key){
    let low = 0, high = numbers.length;

    while(low < high){
        let mid = Math.floor((low+high)/2);

        if(key<=numbers[mid]){
            high = mid;
        }
        else{
            low = mid+1;
        }
    }
    return low;
}

let result = findLowest(numbers, key);
if(result == numbers.length){
    console.log('Element not found')
}
else{
    console.log(numbers[result]);
}

Answer 6

Finding the lowest element in O(log(n)) time complexity using binary search.

let numbers = [3000, 5000, 8000];

numbers.sort();

let key = 10000;

function findLowest(numbers, key){
    let low = 0, high = numbers.length;

    while(low < high){
        let mid = Math.floor((low+high)/2);

        if(key<=numbers[mid]){
            high = mid;
        }
        else{
            low = mid+1;
        }
    }
    return low;
}

let result = findLowest(numbers, key);
if(result == numbers.length){
    console.log('Element not found')
}
else{
    console.log(numbers[result]);
}

Answer 7

Answer №4

It is not necessary for the array to be sorted.

(function(array, num) { 
   return array.reduce(function(nearest, current) { 
                          return (Math.abs(current - num) < Math.abs(nearest - num))  
                                     ? current : nearest; 
                       }); 
})([3000, 5000, 8000], 4001)

The result is 5000 because 4001 is closer to 5000 than 3000.

The choice of naming the arguments in the reduce function as "min" and "max" might have led to confusion.

Answer 8

It is not necessary for the array to be sorted.

(function(array, num) { 
   return array.reduce(function(nearest, current) { 
                          return (Math.abs(current - num) < Math.abs(nearest - num))  
                                     ? current : nearest; 
                       }); 
})([3000, 5000, 8000], 4001)

The result is 5000 because 4001 is closer to 5000 than 3000.

The choice of naming the arguments in the reduce function as "min" and "max" might have led to confusion.

In JavaScript, locate the closest element in an array based on its value, avoiding cases where the value exceeds the nearest element

Answer №1

Answer №2

Answer №3

Answer №4

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