Expand the following number into its expanded form

I have been working on a code exercise that involves creating a function expandedForm to handle a number parameter. The examples provided below should help clarify the task at hand.

expandedForm(12); // Should return '10 + 2'
expandedForm(42); // Should return '40 + 2'
expandedForm(70304); // Should return '70000 + 300 + 4'

This is my initial solution:

function expandedForm(num) {
  // Your code here
  let numStr = num.toString().split('');
  
  for(let i = 0 ; i < numStr; i++ ){
      
      for(let y = numStr.length; y > 1; y--){
         numStr[i] += '0'; 
         // console.log(y);  use this to debug y, and no y value print out from console
      }
  }
  
  return numStr.join('+')
}

console.log(expandedForm(23));

When I test expandedForm(23), the result is '2+3' and the value of y isn't printed to the console. Can anyone identify what's wrong with my approach? Thank you.


Solution

Thank you to everyone who provided feedback. It was pointed out that my y variable initialization in the for loop was incorrect, as well as the condition i < numStr (silly mistake).

After reviewing my code and taking inspiration from some suggestions, here is my final solution:

 function expandedForm(num) {
      // Your code here
      let numStr = num.toString().split('');
      
      for(let i = 0 ; i < numStr.length; i++ ){
          
          for(let y = numStr.length - i; y > 1; y--){
             numStr[i] += '0'; 
             // console.log(y);  use this to debug y, and no y value print out from console
          }
      }
      
     
      numStr = numStr.filter(value => !value.startsWith(0));
      return numStr.join(' + ')
    }

    console.log(expandedForm(23));

Answer №1

Initially, it's important to note that numStr is an array, hence the comparison should not be done as i < numStr (0 < ["2", "3"]).

Another thing to consider is that the range value of y remains unchanged in every iteration of the for loop when using

for(let y = numStr.length; y > 1; y--)
.

A more appropriate solution with minimal modifications to your existing code could be:

function expandedForm(num) {
  // Your code here
  let numStr = num.toString().split('');
  
  for(let i = 0 ; i < numStr.length; i++ ){
      for(let y = numStr.length - i; y > 1; y--){
         numStr[i] += '0'; 
      }
  }
  
  return numStr.join('+')
}

console.log(expandedForm(23));

Apologies for my poor English skills :(

Answer №2

function convertToExpandedForm(number) {
    return number.toString()
                .split("")
                .reverse()
                .map((digit, index) => digit * Math.pow(10, index))
                .filter(num => num > 0)
                .reverse()
                .join(" + ");
}

console.log(convertToExpandedForm(70304));

This piece of code demonstrates a series of methods that can tackle the given issue effectively.

Answer №3

function convertNumberToExpandedForm(input) {
  let str = String(input).split('');
  
  for(let index = 0; index < str.length; index++ ){
      for(let j = str.length - index; j > 1; j--){
        if(str[index] == '0'){
          index++;
        }else{
         str[index] += '0'; 
        }
      }
  }
  
  return str.filter(item => item != '0').join(' + ');
}

Answer №4

In order to properly expand the number in its expanded form, it is crucial to adjust the second loop variable 'y' by initializing it as (numStr.length - i). This adjustment ensures that each iteration reduces the zeroes accurately, unlike in a scenario where the expansion remains constant.

Additionally, to achieve the desired result and eliminate redundant zeros in expressions like "70000 + 0+ 300 + 0 + 4", a filtering process should be applied using the following code snippet:

function expandedForm(num) {
  // Your code here
  let numStr = num.toString().split('');

  for(let i = 0; i < numStr.length; i++ ){
      if(numStr[i] != 0){
          for(let y = (numStr.length - i); y > 1; y--){
              numStr[i] += '0'; 
          }
      }
  }

  numStr = numStr.filter(value => value !== '0');
  return numStr.join('+')
}

console.log(expandedForm(23));

By implementing these adjustments and applying the specified filter function, the correct expanded form of the number can be printed successfully.

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