## How to calculate a mismatch score between n number of strings more efficiently?

Suppose I have a vector that contains `n`

strings, where the strings can be length 5...n. Each string must be compared with each string character by character. If there is a mismatch, the score is increased by one. If there is a match, the score does not increase. Then I will store the resulting scores in a matrix.

I have implemented this in the following way:

for (auto i = 0u; i < vector.size(); ++i) { // vector.size() x vector.size() matrix std::string first = vector[i]; //horrible naming convention for (auto j = 0u; j < vector.size(); ++j) { std::string next = vector[j]; int score = 0; for (auto k = 0u; k < sizeOfStrings; ++k) { if(first[k] == second[k]) { score += 0; } else { score += 1; } } //store score into matrix } }

I am not happy with this solution because it is `O(n^3)`

. So I have been trying to think of other ways to make this more efficient. I have thought about writing another function that would replace the innards of our `j`

for loop, however, that would still be `O(n^3)`

since the function would still need a `k`

loop.

I have also thought about a queue, since I only care about `string[0]`

compared to `string[1]`

to `string[n]`

. `String[1]`

compared to `string[2]`

to `string[n]`

. `String[2]`

compared to `string[3]`

to `string[n]`

, etc. So my solutions have unnecessary computations since each string is comparing to every other string. The problem with this, is I am not really sure how to build my matrix out of this.

I have finally, looked into the std template library, however `std::mismatch`

doesn't seem to be what I am looking for, or `std::find`

. What other ideas do you guys have?

I don't think you can easily get away from O(n^3) comparisons, but you can easily implement the change you talk about. Since the comparisons only need to be done one way (i.e. comparing string[1] to string[2] is the same as comparing string[2] to string[1]), as you point out, you don't need to iterate through the entire array each time and can change the start value of your inner loop to be the current index of your outer loop:

for (auto i = 0u; i < vector.size(); ++i) { // vector.size() x vector.size() matrix std::string first = vector[i]; //horrible naming convention for (auto j = i; j < vector.size(); ++j) {

To store it in a matrix, setup your `i`

x `j`

matrix, initialize it to all zeroes and simply store each score in `M[i][j]`

for (auto k = 0u; k < sizeOfStrings; ++k) { if (first[k] != second[k]) { M[i][j]++; } }

**MPEG-7 Audio and Beyond: Audio Content Indexing and Retrieval,** That is, inserts, deletes and mismatches do not change the score of the match, having a Baeza-Yates describes in (Baeza-Yates, 1992) an efficient algorithm for string The direct measure is robust against single note failures and can be M is equal to the sum of the lengths N +M. This is significantly more efficient than a It is much more efficient. But My sequences are too short: PDict would not let me set the max.mismatch parameter to 5. As the sequences from A and B are exactly of the same length, I do not need searches or alignments.

If you have n strings each of length m, then no matter what (even with your queue idea), you have to do at least (n-1)+(n-2)+...+(1)=n(n-1)/2 string comparisons, so you'll have to do (n(n-1)/2)*m char comparisons. So no matter what, your algorithm is going to be O(mn^2).

**Combinatorial Algorithms: 20th International Workshop, IWOCA 2009, ,** We are given a text string of length n and a pattern string of length m over and we want to either compute the number of mismatches for each possible more efficient but cannot be used for extremely long patterns (of length close to the length of the text). All results for Hamming distance apply to k-mismatches as well. Hamming Distance between two strings You are given two strings of equal length, you have to find the Hamming Distance between these string. Where the Hamming distance between two strings of equal length is the number of positions at which the corresponding character are different.

**In brief how to calculate mismatch? Part 2,** In the last label it appears 2 more results: Average power lossis this a full string will statistically be RMS / sqrt(n), where n is the number of modules in series. The new tool aims to evaluate the mismatch in voltage between strings. between the sum of the MPP powers of each string, and the effective This calculator determines the short-circuit current density of a solar cell under two separate spectra. It can be used to quantify 'spectral mismatch' between a solar cell illuminated by sunlight and by an IV tester. The calculator can also be used to evaluate the spectrum generated by a combination of LEDs, lasers and xenon-arc lamps. Disclaimer

The other answers that say this is at least O(mn^2) or O(n^3) are incorrect. This can be done in O(mn) time where m is string size and n is number of strings.

For simplicity we'll start with the assumption that all characters are ascii.

You have a data structure:

int counts[m][255]

where counts[x][y] is the number of strings that have ascii character y at index x in the string.

Now, if you did not restrict to ascii, then you would need to use a std::map

map counts[m]

But it works the same way, at index m in counts you have a map in which each entry in the map y,z tells you how many strings z use character y at index m. You would also want to choose a map with constant time lookups and constant time insertions to match the complexity.

Going back to ascii and the array

int counts[m][255] // start by initializing this array to all zeros

First initialize the data structure:

m is size of strings, vec is a std::vector with the strings

for (int i = 0; i < vec.size(); i++) { std::string str = vec[i]; for(int j = 0; j < m; j++) { counts[j][str[j]]++; } }

Now that you have this structure, you can calculate the scores easily:

for (int i = 0; i < vec.size(); i++) { std::string str = vec[i]; int score = 0; for(int j = 0; j < m; j++) { score += counts[j][str[j]] - 1; //subtracting 1 gives how many other strings have that same char at that index } std::cout << "string \"" << str << "\" has score " << score; }

As you can see by this code, this is O(m * n)

**Algorithms -- ESA 2004: 12th Annual European Symposium, Bergen, ,** A text location is considered a match if the distance between it and the pattern, under Cole and Hariharan [5] presented an O(nk4/m + n + m) algorithm for this problem. If only mismatches are counted for the distance metric, we get the Hamming A great amount of work was done on finding efficient algorithms for string Hi all, I have a question on how bam-readcount calculates the "average mismatch quality sum":. Assume there are 3 aligned reads (with length 100) at a locus, read1 has 3 mismatches, the base qualities are 2,2,2 respectively. read2 has 2 mismatches, the base qualities are 10,10 and read3 is a perfect alignment with no mismatch.

**Hamming Distance between two strings,** Hamming Distance between two strings · Split the given string into Primes : Digit DP string · Check whether an array of strings can correspond to a particular number X in which we traverse the strings and count the mismatch at corresponding position. function to calculate Hamming distance Time complexity : O(n). I have two strings. I need to compare them and get an exact percentage of match, ie. "four score and seven years ago" TO "for scor and sevn yeres ago" Well, I first started by comparing every word to every word, tracking every hit, and percentage = count \ numOfWords. Nope, didn't take into account misspelled words.

**Levenshtein distance,** In information theory, linguistics and computer science, the Levenshtein distance is a string metric for measuring the difference between two sequences. Informally, the Levenshtein distance between two words is the minimum number of single-character edits The Levenshtein distance between two strings is no greater than the sum of Hamming distance and similarity between two strings. Hamming distance is one of the most common ways to measure the similarity between two strings of the same length. Hamming distance is a position-by-position comparison that counts the number of positions in which the corresponding characters in the string are different.

**[PDF] On String Matching with Mismatches,** In. Step 2, we compute, using marking, the number of matches contributed by the characters 3 and 4 to each alignment between T and P. We get Hello, I am fairly new to Power BI and am even newer to the use of DAX functions. For years I have worked in Excel and know how I would do this there, but am finding the same formula I would use there does not work in Power BI. Basically, I need the DAX equivalent of the COUNTIF function (such as