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Add DNA_Matching problem to (L-I).
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function dnaMatch(dna1, dna2) {
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// Check if the lengths of the two DNA sequences are different
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if (dna1.length !== dna2.length) {
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return false; // Different lengths, not a match
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}
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// Iterate through each character of the DNA sequences
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for (let i = 0; i < dna1.length; i++) {
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// Compare characters at the current position
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if (dna1[i] !== dna2[i]) {
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return false; // Characters don't match, not a match
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}
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}
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// If we reach this point, all characters match, so it's a match
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return true;
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}
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// Test cases
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console.log(dnaMatch("ACGT", "ACGT")); // true
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console.log(dnaMatch("ACGT", "ACTG")); // false
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console.log(dnaMatch("AACT", "AACC")); // false

L-I/0011 DNA_Matching/README.md

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# 0011 DNA_Matching ( L-I )
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## Problem
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You are given two DNA sequences as strings, representing the genetic code of two individuals. Each DNA sequence consists of a series of characters, where each character can be one of four options: 'A' (adenine), 'C' (cytosine), 'G' (guanine), or 'T' (thymine).
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Write a JavaScript function called dnaMatch that takes these two DNA sequences as input and returns true if they are a match and false otherwise. Two DNA sequences are considered a match if they have the same characters in the same order.
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**Example**
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```
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If the input DNA sequences are "ACGT" and "ACGT", the function should return true because they are an exact match.
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If the input DNA sequences are "ACGT" and "ACTG", the function should return false because the order of characters is different.
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If the input DNA sequences are "AACT" and "AACC", the function should return true because some characters do not match.
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```
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**Here's an example of how the function should work:**
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```javascript
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console.log(dnaMatch("ACGT", "ACGT")); // true
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console.log(dnaMatch("ACGT", "ACTG")); // false
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console.log(dnaMatch("AACT", "AACC")); // false
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```
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## Solutions
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```javascript
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function dnaMatch(dna1, dna2) {
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// Check if the lengths of the two DNA sequences are different
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if (dna1.length !== dna2.length) {
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return false; // Different lengths, not a match
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}
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// Iterate through each character of the DNA sequences
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for (let i = 0; i < dna1.length; i++) {
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// Compare characters at the current position
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if (dna1[i] !== dna2[i]) {
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return false; // Characters don't match, not a match
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}
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}
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// If we reach this point, all characters match, so it's a match
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return true;
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}
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// Test cases
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console.log(dnaMatch("ACGT", "ACGT")); // true
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console.log(dnaMatch("ACGT", "ACTG")); // false
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console.log(dnaMatch("AACT", "AACC")); // false
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```
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## How it works
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1. The `dnaMatch` function takes two input arguments: `dna1` and `dna2`, which are the two DNA sequences we want to compare.
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2. First, it checks if the lengths of `dna1` and `dna2` are different using the `if (dna1.length !== dna2.length)` conditional statement. If they have different lengths, the function immediately returns `false` because sequences of different lengths cannot be a match.
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3. If the lengths are the same, the function proceeds to compare the characters of the two DNA sequences.
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4. It enters a `for` loop that iterates through the characters of the sequences. The loop is controlled by the variable `i`, which starts at 0 and goes up to `dna1.length - 1`. This loop allows us to compare characters at the same position in both sequences.
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5. Inside the loop, the code compares the characters at the current position `i` using the conditional statement `if (dna1[i] !== dna2[i])`. If the characters at position `i` are not the same in both sequences, the function immediately returns `false` because it found a mismatch.
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6. If the loop completes without finding any mismatches, meaning it has compared all characters in both sequences, the function returns `true`. This indicates that the two DNA sequences are identical and, therefore, a match.
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## References
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- [ChatGPT](https://chat.openai.com/)
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## Problem Added By
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- [Tipchan](https://github.com/tsongtheng)
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## Contributing
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Pull requests are welcome. For major changes, please open an issue first to discuss what you would like to change.
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Please make sure to update tests as appropriate.

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