# Counting the number of 1s upto n in JavaScript

We are required to write a JavaScript function that takes in a positive integer, say num.

The task of our function is to count the total number of 1s that appears in all the positive integers upto n (including n, if it contains any 1).

Then the function should finally return this count.

For example −

If the input number is −

const num = 31;

Then the output should be −

const output = 14;

because 1 appears in,

1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 31

## Example

Following is the code −

const num = 31;
const countOnes = (num = 1) => {
if(num <= 0 ){
return 0
};
let sum = 0
num += '';
let helper = p => {
let leftNum = 0
let rightNum = 0
let di = num[p]
if(p>0){
leftNum = parseInt(num.slice(0,p))
}
if(p+1 < num.length){
rightNum = parseInt(num.slice(p+1))
}
if(di > 1){
sum += (leftNum+1)*(10**(num.length-1-p))
} else if(di == 0){
sum += (leftNum)*(10**(num.length-1-p))
} else{
sum += (leftNum)*(10**(num.length-1-p)) + rightNum + 1
}
}
for(let i =0; i < num.length; i++){
helper(i)
};
return sum;
};
console.log(countOnes(num));

## Output

Following is the console output −

14

Updated on: 27-Jan-2021

174 Views 