findStrongNumbersRange()

Step 1: Understanding the Problem Statement

We are required to write a function that will find the strong numbers within a given range. A strong number is defined as a number that equals the sum of the factorial of its digits. For example, 145 is strong because 1! + 4! + 5! = 145.

Let's write this function in JavaScript:

function findStrong(n) {
	...
}

Step 2: Calculate Factorial

First, we would need a helper function to calculate the factorial of a number. This can be done using a simple for loop:

function factorial(n) {
	let f = 1;
	for(let i = 1; i <= n; i++) {
		f = f * i;
	}
	return f;
}

Step 3: Check Whether a Number is Strong

We then need a function to tell us if a given number is strong or not. This will involve converting the number to a string, so that we can access each digit, calculating the factorial of each digit, and checking if the sum equals the original number:

function isStrong(n) {
	let sum = 0;
	let numStr = n.toString();
	for(let digit of numStr) {
		sum += factorial(parseInt(digit));
	}
	return sum === n;
}

Step 4: Find all Strong Numbers in Range

Now we can write our main function. This function will iterate from 1 to n and return all strong numbers within this range:

function findStrong(n) {
	let strongNumbers = [];
	for(let i = 1; i <= n; i++) {
		if(isStrong(i)) {
			strongNumbers.push(i);
		}
	}
	return strongNumbers;
}

Step 5: Conclusion

By wrapping up all these steps, we get our final implementation of the function to find all strong numbers within a given range.

function factorial(n) {
	let f = 1;
	for(let i = 1; i <= n; i++) {
		f = f * i;
	}
	return f;
}

function isStrong(n) {
	let sum = 0;
	let numStr = n.toString();
	for(let digit of numStr) {
		sum += factorial(parseInt(digit));
	}
	return sum === n;
}

function findStrong(n) {
	let strongNumbers = [];
	for(let i = 1; i <= n; i++) {
		if(isStrong(i)) {
			strongNumbers.push(i);
		}
	}
	return strongNumbers;
}

This function works correctly, but note that it is not very efficient. The time complexity is O(n^2) due to nested loops.