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# Code Wars - Sum(n)Strings

## Breaking down code war challenges in a simple and understandable way.

# The Problem Statement

Complete the square sum function so that it squares each number passed into it and then sums the results together.

For example, for [1, 2, 2] it should return 9 because 1^2 + 2^2 + 2^2 = 9.

# Analysis -Breaking down the question

Since it’s an array and we need to square each element - the map function will be perfect.

Next, we can use reduce function to sum the squared array.

Let’s see the solution in action.

# The Syntax for solving the problem

The Map Function

```
map((element)=> {//your code here}) //map function returns
//the new array without modifying the original array
```

The Reduce Function

```
reduce((previousValue, currentValue))=> {//Your code here } reduce function reduces
//the entire array to a single value.
```

# The Solution

Let’s write a basic function that takes an array as input as below

```
function squareSum(numbers){
}
```

Now, use the map function to square each element of the array.

```
function squareSum(numbers){
const squared = numbers.map((number)=> Math.pow(number, 2)) //[1,4,6]
//At this point, it takes each
//element and squares using the Math.pow
return squared;
}
squareSum([1,2,3])
```

We got the squared array and now let’s use the map function and add the array elements

```
function squareSum(numbers){
const squared = numbers.map((number)=> Math.pow(number, 2))
const sumSquared = squared.reduce((previous, current)=> previous + current, 0)
return sumSquared; //11
}
squareSum([1,2,3])
```

Here the 0 in the reduce function is the initial value. In the beginning, the previous value is 0(initial value), the current value will be 1 and adds 0 to 1 = 1

Next, the previous value becomes 1, the current value becomes 2 and adds 1 and 2 = 3, and so on until all the array elements are reduced to a single value.

We can minimize the above code to one line as follows

```
function squareSum(numbers) {
return numbers.map((item) => Math.pow(item, 2)).reduce((prev, curr) => prev + curr, 0);
}
squareSum([1,2,3]) 11
```

# Final Words

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