I am writing a function to find a number in an array like I have 999 the greatest number but that is not present in array so I want the only one number which is nearby 999 like 899 or 951 or 50 even if 5 is bigger than other number which is present in the array if 999 is not there how I can do that please help.
let a = [5, 5, 9, 5, 6, 888, 98, 2, 6];
a.map(function (a) {
return a >= 900;
});I just need one number in output
If You Want The Closest Left Side Number :
let a = [5, 5, 9, 5, 6, 888, 98, 2, 6];
let target = 885; // Our Target
function closestLeftSide(array,target){
let filtered = a.filter(x=>x<=target); // discard all bigger numbers
let closest = filtered[0]; // initialaze first closest
for(number of filtered){
closest = Math.abs(target-number)<=Math.abs(target-closest)?number:closest;
}
return closest
}
console.log(closestLeftSide(a,target))if you want to find the most closest number in array to a given number :
let a = [5, 5, 9, 5, 6, 888, 98, 2, 6];
let target = 4;
let closest = a[0];
for(number of a){
closest = (Math.abs(target - number) <= Math.abs(target - closest))? number:closest;
}
console.log(closest)You can .reduce() your list updating a "running result" as you iterate over all elements in the array, and only update it if it's closer to your desired value than the previous running result:
function findClosest(list, targetValue) {
const better = (running, element) =>
Math.abs(targetValue - element) < Math.abs(targetValue - running);
return list.reduce(
(running, element) => better(running, element) ? element : running,
list[0]
);
}
const someList = [5, 5, 1200, 9, 5, 6, 888, 98, 2, 6];
const someValue = 900;
console.log(
`Closest value to ${someValue}: ${findClosest(someList, someValue)}`
);If you want to maximum number that is less than an upper bound, then:
const target = 700;
let a = [5, 5, 9, 5, 6, 888, 98, 2, 6];
console.log(findMaxWithUpperBound(a, target));
function findMaxWithUpperBound(list: number[], upperBound: number) {
let max = list[0];
list.forEach((n) => {
if (n > max && n < upperBound) {
max = n;
}
});
return max;
}