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Leetcode longest common prefix [closed]

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I am having a hard time following this solution. I understand we set a prefix, and loop through the remainder of the array and keep chopping the prefix until prefix fully exists in each string but why are we doing strs[i].indexOf(output) != 0 in the while loop?
public String longestCommonPrefix(String[] strs) {
if(strs.length == 0) {
return "";
}
String output = strs[0];
for(int i = 1; i < strs.length; i++) {
while(strs[i].indexOf(output) != 0) {
output = output.substring(0, output.length() - 1);
}
}
return output;
}

Much more understandable (and efficient) would be to use "not startWith."
while (!strs[i].startsWith(output)) {
output = output.substring(0, output.length() - 1);
}
indexOf would also search at other positions for a substring equal to output.
This would be more readable as "as long as strs[i] does not start with the prefix, shorten the prefix by 1. An empty prefix (output) would exit the loop. (You might then also break out of the for loop.)

!= 0 means that the prefix string did not start at the beginning of the string. If the index > 0 it was further into the string. If it was -1 it didn't exist in the string at all. If it was == 0 it started at the beginning.
Notice that the while loop keeps backing up using substring until a prefix matches the beginning. Then it exits the while loop. Then it continues to see if the next string contains the first and backs up until they share an equal prefix. This continues until either a longest common prefix is returned or an empty string.

Can't understand the logic behind this anagram problem's solution [closed]

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Given two strings, a and b, that may or may not be of the same length, determine the minimum number of character deletions required to make a and b anagrams. Any characters can be deleted from either of the strings.
This is my first time preparing for competitive programming and understanding the logic behind the two for loops is quite hard for me.
String str1 = s.next();
String str2 = s.next();
char []c1 = str1.toCharArray();
char []c2 = str2.toCharArray();
int []cnt1 = new int[26];
int []cnt2 = new int[26];
int len1 = str1.length();
for (int i = 0; i < len1; i++) {
cnt1[c1[i] - 97]++;
}
int len2 = str2.length();
for (int i = 0; i < len2; i++) {
cnt2[c2[i] - 97]++;
}
int cnt = 0;
for (int i = 0; i < 26; i++) {
cnt += Math.abs(cnt2[i] - cnt1[i]);
}
System.out.println(cnt);

This snippet goes over each string and counts the number of occurrences each letter has in it (and store the counters in an array for better performance).
It then goes over the two arrays of counters, and for each letter subtracts the counters for both strings (in absolute value). The difference is the number of that character that should be removed. These differences are summed, and the result is the answer.

Okay, this is what the program is doing with two for loops.
Imagine 'cnt1' as English alphabet 'A' to 'Z' written up on a paper from left to right, and so is 'cnt2'. First for loop is tick marking a letter on the paper should it be found in 'string1', and so does second for 'string2'.
Now, you have two papers with 'A' to 'Z' written up on them from left to right, and after two 'for loops' have executed, each of the paper has tick marks on those letters that were present in respective string inputs.
Now, if a letter is ticked on both the papers, leave it alone, and should you find any of the letter ticked in one paper (i.e. in the array), and is not ticked in the other array, then count it as a letter to be deleted.
By the time you have scanned both the papers like this from left to right, you'd have number of letters that need to be removed in totality from both the papers.
Lets see how it is implemented in code. Default initial values of primitive array is all zeros, and the act of 'tick marking' a letter on paper is achieved by changing that particular index to '1'.
So, by the time the first two for loops end, each of 'cnt1' and 'cnt2' arrays would have '1' in it randomly. If both the arrays have '1' or '0' in it for a given index, you need not count it, should they be different i.e. difference of that particular index for both the arrays is '1' (that's why you see Math.abs being used), then that is a letter to be deleted from first string, or the second.
edit: For competitive exams, you should be able to visualize the solution first, and then find an optimum one. Computers only add speed to the solution found. They don't think, we make them think :)
Hope you could visualize the solution first and are still getting used to programing. All the best!

The output of this loop should be less than 1, but I get very large numbers [closed]

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This program is supposed to count the number of characters "c" and "g" in genes in the DNA string and then take that number and divide it by the length of each gene. The number of cs and gs is always < gene.length(), therefore the output should be something like 0.65555, 0.35657 etc, but I get large numbers like 141, etc. Not sure what is wrong with this loop.
public void testfile(){
String dnaTest = "aaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaacccttaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctaaccctcacccttctaact";
int counter = 0;
for(String gene : s1.data()) {
for (char cORg : gene.toCharArray()) {
if ((cORg == 'c') || (cORg == 'g')) {
counter ++;
}
System.out.print(gene +" ");
}
float cgRatioGenes = counter/gene.length();
System.out.println("cgRatio: " + cgRatioGenes);
}
}
}
If you spot the error, let me know. Thanks!
EDIT
Even without the parentesis at the end of the DNA string and with the closing bracket, the loop was not producing the results I expected. Therefore, it is not off topic.

Two problems:
First, you never reinitialize counter when you start the loop again. Move that declaration inside the loop so that each repetition starts with a counter of zero.
If you make that change, all your results will be zero though, because you're diving two integers, which will truncate the results. Cast one to float, so that it keeps the decimal part. See this question for more information on the problem
for(String gene : s1.data()) {
int counter = 0; //Moved inside the for loop, so that it always starts at 0
for (char cORg : gene.toCharArray()) {
if ((cORg == 'c') || (cORg == 'g')) {
counter ++;
}
System.out.print(gene +" ");
}
//Floating point division, instead of integer division
float cgRatioGenes = ((float)counter)/gene.length();
System.out.println("cgRatio: " + cgRatioGenes);
}

One potential problem is here
float cgRatioGenes = counter/gene.length();
As gene.length() is an integer value the ratio is not computed correctly. Instead, you should cast one of them to float like this
float cgRatioGenes = ((float)counter)/gene.length();
In addition, the counter variable should probably be initialized to zero for each gene (unless you want to count the c/g values over all genes).
This probably does not explain the behavior you are observing, but it is not possible to figure it out unless a complete working example is given.

It's a little unclear what the exact intent of this code is, but my guess is that you're using one int counter for every gene in s1.data(). I assume you want to count the number of valid characters per gene, not in the entire pool.
If you do want to count for the entire pool, the problem is that you're peforming gene.length outside of the for, which should honestly throw a compiler error unless you have a gene defined somewhere else as well.
Additionally, you're dividing two ints for your answer, which will yield an int. Case one of your variables to float to get a decimal answer.

Finding all 3 character length substrings in a string [closed]

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I am trying to find all three letter substrings from a string in Java.
For example from the string "example string" I should get "exa", "xam", "amp", "mpl", "ple", "str", "tri", "rin", "ing".
I tried using the Java Regular expression "([a-zA-Z]){3}" but I only got "exa", "mpl", "str", "ing".
Can someone tell me a regex or method to correct this.

Implementing Juvanis' idea somewhat, iterate to get your substrings, then use a regular expression to make sure the substring is all letters:
String s = "example string";
for (int i = 0; i <= s.length() - 3; i++) {
String substr = s.substring(i, i + 3);
if (substr.matches("[a-zA-Z]+")) { System.out.println(substr); }
}

When a character is consumed in one regex, it cannot be used in other regexes. In your example, a is consumed in exa so amp will not be listed as output. You should try traditional iterative approach. It is easier to implement.

try this
Matcher m = Pattern.compile("([a-zA-Z]){3}").matcher("example string");
for (int i = 0; m.find(i); i = m.start() + 1) {
System.out.print(m.group() + " ");
}
output
exa xam amp mpl ple str tri rin ing

This can be done using regex as follows:
Find the position of all matches for the string using the regex \w(?=\w\w). This will give you the start index of the first character of each required sub-string.
In this case, you would get: 0, 1, 2, 3, 4, 8, 9, 10 and 11.
Get what you need by taking the sub-strings starting from each position going upto that plus 2.
In this case, that would mean, my_string.substring(0,3), my_string.substring(1,4) and so on, as the begin index parameter is inclusive while the end index parameter is exclusive.

Java function to output every case possible in a string [closed]

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I am trying to make a function that outputs every case possible of a string. The function must output each variation of a string and keep a counter for that variation. For example in no particular order:
C.d
C>d
C>D
C.D
c.d
c.D
c>D
c>d
So far I have this:
public int allCase(String data)
{
int count=0; // counter for the amount of case changes completed
int size= data.length();// length of incoming string
char c[]= data.toLowerCase().toCharArray();
double maxpos=Math.pow(2,size);
System.out.println("Maximum possibilities= "+maxpos);
for(int i=0;i<size;i++)
{
if (c[i]> 33 && c[i]<64) // if the character is special characters !##$%^&*()_<>?,./
{ // prints characters in front and behind of special character
System.out.println( data.substring(0,i)+((char)(c[i]+16))+data.substring(i+1));
}
else{
// output the string variation
}
count++;
}
return count;
}

You can handle the alphabetic characters as a group, adding or subtracting 32, but the rest of the mappings aren't regular enough to beat a table lookup.
Keep two parallel strings:
shifted = "ABCDEFGHIJKLMNOPQRSTUVWXYZ~!##$%^&*()_+|<>?:\"{}";
unshifted = "abcdefghijklmnopqrstuvwxyz`1234567890-=\,./;'[]";
Then, find each character in one or the other string with .indexOf(). Find the opposite shift at the same index in the other string. If a character isn't in either string, then it's a space or another character that isn't part of a shifted/unshifted pair. The total number of strings to generate is then 2^(number of chars found in one of those strings).
Speaking of powers of 2, using Math.pow() is a lousy way to compute small powers of 2. Use 1L<<n instead of Math.pow(n) for integers 0 <= n <= 62, or even 1<<n for 0<=n<=30 if you're able to live with only being able to print out a billion or so strings.
You can make an iterative version, looping from an index value of 0 to (2^n)-1, where n the length of the input string, and then loop for k=0 to (n-1) testing bit k of the outer loop index value to see whether to print upper or lowercase version of the character. I didn't see it before, but Hot Licks has a comment about using this approach. To perform that test, observe that index&(1<<k) is nonzero if and only if bit k is set in (index).
A recursive version is much simpler, but I suspect you'll learn about that later.