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### Five Ways to Solve the DNA Pairing Challenge in JavaScript

December 30, 2019

Computer Science is awesome and amazing, trust me! There is always more than one way to come to a solution to a given problem in most cases.

In this tutorial, we will learn how to solve the Free Code Camp DNA Pairing Challenge in five different ways.

## Algorithm Challenge

The DNA strand is missing the pairing element. Take each character, get its pair, and return the results as a 2d array. Base pairs are a pair of AT and CG. Match the missing element to the provided character. Return the provided character as the first element in each array. For example, for the input GCG, return [[“G”, “C”], [“C”,“G”],[“G”, “C”]] The character and its pair are paired up in an array, and all the arrays are grouped into one encapsulating array.

function pairElement(str) {
return str;
}

pairElement("GCG");

## Provided Test Cases

• pairElement(“ATCGA”) should return [[“A”,“T”],[“T”,“A”],[“C”,“G”],[“G”,“C”],[“A”,“T”]]
• pairElement(“TTGAG”) should return [[“T”,“A”],[“T”,“A”],[“G”,“C”],[“A”,“T”],[“G”,“C”]]
• pairElement(“CTCTA”) should return [[“C”,“G”],[“T”,“A”],[“C”,“G”],[“T”,“A”],[“A”,“T”]]

## Understanding the Problem

As you may have read in the challenge description above, the goal of this exercise is to return the missing trand into a 2d array. In biology class, we learned about DNA base pairs (need a refresher? Wikipedia is your friend). They are A - T and C - G, and they go both ways. So every time we have:

• A string we return an array [‘A’, ‘T’]
• T string we return an array [‘T’, ‘A’]
• C string we return an array [‘C’, ‘G’]
• G string we return an array [‘G’, ‘C’]

## 1. Using For Loop, and If Statement

For this solution, we will loop over the parameter passed to the function and use if statement to return the correct pair.

function pairElement(str) {
// Step 1. Declare the variable of type array that will encapsulate other paired arrays
const arrDNA = [];

// Step 2. Create the FOR loop with initializer less then str.length
for (let i = 0; i < str.length; i += 1) {
// Step 3. Use if statement to evaluate baise pair and push it to arrDNA
if (str[i] === 'A') arrDNA.push([str[i], 'T']);
if (str[i] === 'T') arrDNA.push([str[i], 'A']);
if (str[i] === 'C') arrDNA.push([str[i], 'G']);
if (str[i] === 'G') arrDNA.push([str[i], 'C']);
}

/* Here "GCG"'s length equals 3
For each iteration: i = 0 and arrDNA = [[str[i], 'corresponding pair']]
First iteration:  i = 0        arrDNA = [['G', 'C']]
Second iteration: i = 1        arrDNA = [['G', 'C'], ['C', 'G']]
Third iteration:  i = 2        arrDNA = [['G', 'C'], ['C', 'G'], ['G', 'C']]

End of the FOR Loop*/

// Step 4. Return the 2D array
return arrDNA;
}

pairElement("GCG");

function pairElement(str) {

const arrDNA = [];

for (let i = 0; i < str.length; i += 1) {

if (str[i] === 'A') arrDNA.push([str[i], 'T']);
if (str[i] === 'T') arrDNA.push([str[i], 'A']);
if (str[i] === 'C') arrDNA.push([str[i], 'G']);
if (str[i] === 'G') arrDNA.push([str[i], 'C']);

}

return arrDNA;
}

pairElement("GCG");

## 2. Using For Loop, CharAt(), and If Statement

In this solution, we will make use of the traditional for loop and if statements once more in combination with the String object’s charAt() method. This method (String.prototype.charAt()) returns the character at the specified index in a string.

function pairElement(str) {
// Step 1. Create an empty array that will encapsulate other paired arrays
const arrDNA = [];

// Step 2. Iterate through the str with a FOR loop
for (let i = 0; i < str.length; i += 1) {
// Step 3. Use if statement to evaluate base pair and push it to arrDNA

// If the current str character is X create an array of current str with its corresponding pair and push the array to arrDNA

if (str.chartAt(i) === 'A') // if A
arrDNA.push([str[i], 'T']); // ...push [A - T]
else if (chartAt(i) === 'T') // if T
arrDNA.push([str[i], 'A']); //...push [T - A]
else if (chartAt(i) === 'C') // if C
arrDNA.push([str[i], 'G']); // ...push [C - G]
else if (chartAt(i) === 'G') // if G
arrDNA.push([str[i], 'C']); // ...push [G - C]

}

// Step 4. Return the 2D array
return arrDNA;
}

pairElement("GCG");

function pairElement(str) {
const arrDNA = [];

for (let i = 0; i < str.length; i += 1) {

if (str.chartAt(i) === 'A')
arrDNA.push([str[i], 'T']);
else if (chartAt(i) === 'T')
arrDNA.push([str[i], 'A']);
else if (chartAt(i) === 'C')
arrDNA.push([str[i], 'G']);
else if (chartAt(i) === 'G')
arrDNA.push([str[i], 'C']);

}

return arrDNA;
}

pairElement("GCG");

## 3. Using For…of

The for...of creates a loop iterating over iterable objects (built-in String, Array, Array-like objects).

function pairElement(str) {
// Step 1. Create an empty array that will encapsulate other paired arrays
const arrDNA = [];

// Step 2. Create an object of base pair
const basePair = {
'A': 'T',
'T': 'A',
'C': 'G',
'G': 'C'
}

// Step 3. Iterate through the str with a for of loop
for (const letter of str) {
// Step 4. Create an array of letter with its corresponding pair and  push to arrDNA
arrDNA.push([letter, basePair[letter]]);
}

// Step 5. Return the 2D array
return arrDNA;
}

pairElement("GCG");

function pairElement(str) {
const arrDNA = [];

const basePair = {
'A': 'T',
'T': 'A',
'C': 'G',
'G': 'C'
}

for (const letter of str) {
arrDNA.push([letter, basePair[letter]]);
}

return arrDNA;
}

pairElement("GCG");

## 4. Using Split and Map

Let’s try to resolve using String.prototype.split() and Array.prototype.map(). The first method (split()) is used to convert a string into an array. The map() method creates a new array with the results of calling a function for every array element.

function pairElement(str) {
// Step 1. Create an object of base pair
const basePair = {
'A': 'T',
'T': 'A',
'C': 'G',
'G': 'C'
}
// Step 2. convert the str into an array with split and store the result into arrStr variable
const arrStr = str.split('');

/* Step 3. Map through the arrStr and return an array of current value and it baise
Keep the result of mapping under arrDNA variable
*/
const arrDNA = arrStr.map(letter => [letter, basePair[letter]])

// Step 4. Return the 2D array
return arrDNA;
}

pairElement("GCG");

function pairElement(str) {
const basePair = {
'A': 'T',
'T': 'A',
'C': 'G',
'G': 'C'
}
const arrStr = str.split('');

const arrDNA = arrStr.map(letter => [letter, basePair[letter]])

return arrDNA;
}

pairElement("GCG");

or even better use split() map() in one line

function pairElement(str) {
const basePair = {
'A': 'T',
'T': 'A',
'C': 'G',
'G': 'C'
}

return str.split('').map(letter => [letter, basePair[letter]]);
}

pairElement("GCG");

## 5. Using Split, ForEach, and Switch

In this solution will take help of split(), forEach(), and switch. we have already discussed split() in another solution above. Let talk a bit about the remaining two:

• array.forEach(): this method executes a provided function once for each array element
• switch: is similar to if, it gives a more descriptive way to compare a value with multiple variants.
function pairElement(str) {
// Step 1. Create an empty array that will encapsulate other paired arrays
const arrDNA = [];

// Step 2. convert the str into an array with split and store the result into arrStr variable
const arrStr = str.split('');

// Step 3. Loop through arrStr using forEach
arrStr.forEach(x => {
/* Step 4. Use switch statement to test x and push the corresponding array to arrDNA */
switch (x) {
case "G": // in case x = G
arrDNA.push(["G","C"]); // ...push ["G","C"] to arrDNA
break // break tells the script to run from the case where the criterion is met
case "C":
arrDNA.push(["C","G"]);
break;
case "T":
arrDNA.push(["T","A"]);
break;
case "A":
arrDNA.push(["A","T"]);
break;
}
});

// Step 5. Return the 2D array
return arrDNA;
}
pairElement("GCG");

function pairElement(str) {
const arrDNA = [];

const arrStr = str.split('');

arrStr.forEach(x => {

switch (x) {
case "G":
arrDNA.push(["G","C"]);
break
case "C":
arrDNA.push(["C","G"]);
break;
case "T":
arrDNA.push(["T","A"]);
break;
case "A":
arrDNA.push(["A","T"]);
break;
}
});

return arrDNA;
}
pairElement("GCG");

## Wrap up

That is it for this tutorial. We have used five different ways to solve the DNA Pairing challenge that is available on FreeCodeCamp.

Which of these solutions is fine for you? Do you have any other solutions? Share with us. Among all these solutions if I have to chose just one, I’d go for the 3rd one that makes use of for...of.