Write a function countMatches that searches the substring in the given string and returns how many times the substring appears in the string.
I've been stuck on this awhile now (6+ hours) and would really appreciate any help I can get. I would really like to understand this better.
int countMatches(string str, string comp)
{
int small = comp.length();
int large = str.length();
int count = 0;
// If string is empty
if (small == 0 || large == 0) {
return -1;
}
// Increment i over string length
for (int i = 0; i < small; i++) {
// Output substring stored in string
for (int j = 0; j < large; j++) {
if (comp.substr(i, small) == str.substr(j, large)) {
count++;
}
}
}
cout << count << endl;
return count;
}
When I call this function from main, with countMatches("Hello", "Hello"); I get the output of 5. Which is completely wrong as it should return 1. I just want to know what I'm doing wrong here so I don't repeat the mistake and actually understand what I am doing.
I figured it out. I did not need a nested for loop because I was only comparing the secondary string to that of the string. It also removed the need to take the substring of the first string. SOOO... For those interested, it should have looked like this:
int countMatches(string str, string comp)
{
int small = comp.length();
int large = str.length();
int count = 0;
// If string is empty
if (small == 0 || large == 0) {
return -1;
}
// Increment i over string length
for (int i = 0; i < large; i++) {
// Output substring stored in string
if (comp == str.substr(i, small)) {
count++;
}
}
cout << count << endl;
return count;
}
The usual approach is to search in place:
std::string::size_type pos = 0;
int count = 0;
for (;;) {
pos = large.find(small, pos);
if (pos == std::string::npos)
break;
++count;
++pos;
}
That can be tweaked if you're not concerned about overlapping matches (i.e., looking for all occurrences of "ll" in the string "llll", the answer could be 3, which the above algorithm will give, or it could be 2, if you don't allow the next match to overlap the first. To do that, just change ++pos to pos += small.size() to resume the search after the entire preceding match.
The problem with your function is that you are checking that:
Hello is substring of Hello
ello is substring of ello
llo is substring of llo
...
of course this matches 5 times in this case.
What you really need is:
For each position i of str
check if the substring of str starting at i and of length = comp.size() is exactly comp.
The following code should do exactly that:
size_t countMatches(const string& str, const string& comp)
{
size_t count = 0;
for (int j = 0; j < str.size()-comp.size()+1; j++)
if (comp == str.substr(j, comp.size()))
count++;
return count;
}
Related
i already studied c++ in school and during the last days i have been doing the beginner c++ course of codecademy. On codecademy there is an exercise in which i have to identify palindrome words and return true or false. I haven't been able to resolve it so i saw the solution and it was:
#include <iostream>
// Define is_palindrome() here:
bool is_palindrome(std::string text) {
std::string reversed_text = "";
for (int i = text.size() - 1; i >= 0; i--) {
reversed_text += text[i];
}
if (reversed_text == text) {
return true;
}
return false;
}
int main() {
std::cout << is_palindrome("madam") << "\n";
std::cout << is_palindrome("ada") << "\n";
std::cout << is_palindrome("lovelace") << "\n";
}
My only doubt is with this line:
for (int i = text.size() - 1; i >= 0; i--) {
reversed_text += text[i];
i know it has to do with index values but i can't understand why it has a -1.
Could somebody explain this to me?
i thanks in advance whoever read this post. i'm sorry for my english or my poor using of stacksoverflow, i'm italian and that's my first time using this site.
for (int i = text.size() - 1; i >= 0; i--) {
reversed_text += text[i];
text is basically the string that you receive as input via function. size() is function that returns the size of the string i.e text.size() so in our test cases it will return
5 for madam
3 for ada
8 for lovelace
If you think about the strings as an array with exact above size then the index range will become
0-4 for madam
0-2 for ada
0-7 for lovelace
So that's why the text.size()-1 is using as the starting index of loop. text.size() will return the actual size of string and then minus 1 to get the index of last character in string.
so behind the scene, your loop iteration will look something like below
for (int i = 4; i >= 0; i--) { //for madam
}
//aca
for (int i = 2; i >= 0; i--) {
}
//lovelace
for (int i = 7; i >= 0; i--) {
}
I hope it clear out your confusion.
Thanks,
i know it has to do with index values but i can't understand why it has a -1.
If a string is n characters long, the characters in it are indexed from 0 to n−1.
Since the loop works with characters from the end of the string to the beginning, it starts with index text.size() - 1.
However, the solution you have shown is nominally inefficient. There is no reason to make a reversed copy of the string. It suffices merely to test whether each character in the first half of the string equals the character in the reflected position:
bool is_palindrome(std::string text)
{
size_t e = text.size();
for (int i = 0; i < e/2; ++i)
if (text[i] != text[e-1-i])
return false;
return true;
}
If using a for loop to reverse the string is confusing, you could also use the reverse function
std::string reversed_text = text;
reverse(reversed_text.begin(),reversed_text.end());
which just helps flip the entire string reversed_text and can achieve the same result in a simpler way.
Doing some work with timing different algorithms, however my brute force implementation which I have found numerous times on different sites is sometimes returning more results than, say, Notepad++ search or VSCode search. Not sure what I am doing wrong.
The program opens a txt file with a DNA strand string of length 10000000 and searches and counts the number of occurrences of the string passed in via command line.
Algorithm:
int main(int argc, char *argv[]) {
// read in dna strand
ifstream file("dna.txt");
string dna((istreambuf_iterator<char>(file)), istreambuf_iterator<char>());
dna.c_str();
int dnaLength = dna.length();
cout << "DNA Strand Length: " << dnaLength << endl;
string pat = argv[1];
cout << "Pattern: " << pat << endl;
// algorithm
int M = pat.length();
int N = dnaLength;
int localCount = 0;
for (int i = 0; i <= N - M; i++) {
int j;
for (j = 0; j < M; j++) {
if (dna.at(i + j) != pat.at(j)) {
break;
}
}
if (j == M) {
localCount++;
}
}
The difference might be because your algorithm also counts overlapping results, while a quick check with Notepad++ shows that it does not.
Example:
Let dna be "FooFooFooFoo"
And your pattern "FooFoo"
What result do you expect? Notepad++ shows 2 (one starts at position 1, the second at position 7 (after the first).
Your algorithm will find 3 (position 1, 4 and 7)
In your algorithm, the index i increase by 1 every loop. This may cause double counting for some searching pattern. For eaxample, search for ABAB in the text ... ABABABABABAB .... The answer may be 5 times in your methods, and it would be 3 times if each character is not allowed to be double counted. Which answer you want?
To avoid double counting, you may rewrite the index i to a while loop:
i = 0;
while (i < M)
{
for (j = 0; j < M; j++) {
if (dna.at(i + j) != pat.at(j)) {
break;
}
}
if (j == M) {
localCount++;
i += M;
}
else ++i;
}
Or, you can employ the function std::string::find(const string&, int p=0). The first argument is the pattern to look for, and the second the position to start search:
int pos = 0, count=0;
pos = dna.find(pat); // initial serach start from pos=0;
while( pos != std::string::npos) { // while not end of string
++count;
pos = dna.find(pat, pos + M); // start search from pos+M
}
These two methods provide a self-confirmation for confidence.
First of all, I know there are many duplicates of this question but I have tried and tried and non have been able to solve my issue.
I have the following string
string s = "asdfqasdfp";
I need to loop through the string and find which sub string appears more than once. so in this case its
asdf
I have made the following code but I do not know why it doesn't work. I start from the full string and go down one at a time. I should get occurence value of 2.
int t = s.length();
for (int i = 0; i < s.length(); i++) {
string str = s.substr(0, t);
int occurence = 0;
size_t start = 0;
while ((start = s.find(str, start)) != string::npos) {
++occurence;
start += str.length();
}
if (occurence > 1) {
cout << occurence;
}
else {
--t;
}
}
EDIT: I only want the largest substring that the string contains, in this case
"asdf"
Here's a fixed version of your code, including Daniel's suggestions (thanks! Daniel's demo)
for (size_t t = s.length(); t >= 1; --t) {
for (size_t i = 0; (i + t) <= s.length(); i++) {
std::string str = s.substr(i, t);
size_t occurence = 0;
size_t start = 0;
while ((start = s.find(str, start)) != std::string::npos) {
++occurence;
start += str.length();
}
if (occurence > 1) {
std::cout << str << " " << occurence << std::endl;
return 0;
}
}
}
You need to
loop over t, and decrement it at the end of the i loop rather than inside
limit i to s.length() - t, so that there's always a t-length string to take as str
fix your substr to start at i not 0
You can also stop at the first time you find a duplicate, since this will be a largest duplicate (e.g. if there are two pairs of duplicates of length 4 it will find one of them, but it doesn't sound like you need both). You should also use size_t throughout as your integer type since that's what's used by the string functions here.
Looks ok. Do you need an << endl; to see output in your terminal?
Given a string S.We need to tell if we can make it to palindrome by removing exactly one letter from it or not.
I have a O(N^2) approach by modifying Edit Distance method.Is their any better way ?
My Approach :
int ModifiedEditDistance(const string& a, const string& b, int k) {
int i, j, n = a.size();
int dp[MAX][MAX];
memset(dp, 0x3f, sizeof dp);
for (i = 0 ; i < n; i++)
dp[i][0] = dp[0][i] = i;
for (i = 1; i <= n; i++) {
int from = max(1, i-k), to = min(i+k, n);
for (j = from; j <= to; j++) {
if (a[i-1] == b[j-1]) // same character
dp[i][j] = dp[i-1][j-1];
// note that we don't allow letter substitutions
dp[i][j] = min(dp[i][j], 1 + dp[i][j-1]); // delete character j
dp[i][j] = min(dp[i][j], 1 + dp[i-1][j]); // insert character i
}
}
return dp[n][n];
}
How to improve space complexity as max size of string can go upto 10^5.
Please help.
Example : Let String be abc then answer is "NO" and if string is "abbcbba then answer is "YES"
The key observation is that if the first and last characters are the same then you needn't remove either of them; which is to say that xSTRINGx can be turned into a palindrome by removing a single letter if and only if STRING can (as long as STRING is at least one character long).
You want to define a method (excuse the Java syntax--I'm not a C++ coder):
boolean canMakePalindrome(String s, int startIndex, int endIndex, int toRemove);
which determines whether the part of the string from startIndex to endIndex-1 can be made into a palindrome by removing toRemove characters.
When you consider canMakePalindrome(s, i, j, r), then you can define it in terms of smaller problems like this:
If j-i is 1 then return true; if it's 0 then return true if and only if r is 0. The point here is that a 1-character string is a palindrome regardless of whether you remove a character; a 0-length string is a palindrome, but can't be made into one by removing a character (because there aren't any to remove).
If s[i] and s[j-1] are the same, then it's the same answer as canMakePalindrome(s, i+1, j-1, r).
If they're different, then either s[i] or s[j-1] needs removing. If toRemove is zero, then return false, because you haven't got any characters left to remove. If toRemove is 1, then return true if either canMakePalindrome(s, i+1, j, 0) or canMakePalindrome(s, i, j-1, 0). This is because you're now testing whether it's already a palindrome if you remove one of those two characters.
Now this can be coded up pretty easily, I think.
If you wanted to allow for removal of more than one character, you'd use the same idea, but using dynamic programming. With only one character to remove, dynamic programming will reduce the constant factor, but won't reduce the asymptotic time complexity (linear in the length of the string).
Psudocode (Something like this I havn't tested it at all).
It is based on detecting the conditions that you CAN remove a character, ie
There is exactly 1 wrong character
It is a palendrome (0 mismatch)
O(n) in time, O(1) in space.
bool foo(const std::string& s)
{
int i = 0;
int j = s.size()-1;
int mismatch_count = 0;
while (i < j)
{
if (s[i]==s[j])
{
i++; j--;
}
else
{
mismatch_count++;
if (mismatch_count > 1) break;
//override first preference if cannot find match for next character
if (s[i+1] == s[j] && ((i+2 >= j-1)||s[i+2]==s[j-1]))
{
i++;
}
else if (s[j-1]==s[i])
{
j--;
}
else
{
mismatch_count++; break;
}
}
}
//can only be a palendrome if you remove a character if there is exactly one mismatch
//or if a palendrome
return (mismatch_count == 1) || (mismatch_count == 0);
}
Here's a (slightly incomplete) solution which takes O(n) time and O(1) space.
// returns index to remove to make a palindrome; string::npos if not possible
size_t willYouBeMyPal(const string& str)
{
size_t toRemove = string::npos;
size_t len = str.length();
for (size_t c1 = 0, c2 = len - 1; c1 < c2; ++c1, --c2) {
if (str[c1] != str[c2]) {
if (toRemove != string::npos) {
return string::npos;
}
bool canRemove1 = str[c1 + 1] == str[c2];
bool canRemove2 = str[c1] == str[c2 - 1];
if (canRemove1 && canRemove2) {
abort(); // TODO: handle the case where both conditions are true
} else if (canRemove1) {
toRemove = c1++;
} else if (canRemove2) {
toRemove = c2--;
} else {
return string::npos;
}
}
}
// if str is a palindrome already, remove the middle char and it still is
if (toRemove == string::npos) {
toRemove = len / 2;
}
return toRemove;
}
Left as an exercise is what to do if you get this:
abxyxcxyba
The correct solution is:
ab_yxcxyba
But you might be led down a bad path:
abxyxcx_ba
So when you find the "next" character on both sides is a possible solution, you need to evaluate both possibilities.
I wrote a sample with O(n) complexity that works for the tests I threw at it. Not many though :D
The idea behind it is to ignore the first and last letters if they are the same, deleting one of them if they are not, and reasoning what happens when the string is small enough. The same result could be archived with a loop instead of the recursion, which would save some space (making it O(1)), but it's harder to understand and more error prone IMO.
bool palindrome_by_1(const string& word, int start, int end, bool removed = false) // Start includes, end excludes
{
if (end - start == 2){
if (!removed)
return true;
return word[start] == word[end - 1];
}
if (end - start == 1)
return true;
if (word[start] == word[end - 1])
return palindrome_by_1(word, start + 1, end - 1, removed);
// After this point we need to remove a letter
if (removed)
return false;
// When two letters don't match, try to eliminate one of them
return palindrome_by_1(word, start + 1, end, true) || palindrome_by_1(word, start, end - 1, true);
}
Checking if a single string is palindrome is O(n). You can implement a similar algorithm than moves two pointers, one from the start and another from the end. Move each pointer as long as the chars are the same, and on the first mismatch try to match which char you can skip, and keep moving both pointers as long as the rest chars are the same. Keep track of the first mismatch. This is O(n).
I hope my algorithm will pass without providing code.
If a word a1a2....an can be made a palindrome by removing ak, we can search for k as following:
If a1 != an, then the only possible k would be 1 or n. Just check if a1a2....an-1 or a2a3....an is a palindrome.
If a1 == an, next step is solving the same problem for a2....an-1. So we have a recursion here.
public static boolean pal(String s,int start,int end){
if(end-start==1||end==start)
return true;
if(s.charAt(start)==s.charAt(end))
return pal(s.substring(start+1, end),0,end-2);
else{
StringBuilder sb=new StringBuilder(s);
sb.deleteCharAt(start);
String x=new String(sb);
if(x.equals(sb.reverse().toString()))
return true;
StringBuilder sb2=new StringBuilder(s);
sb2.deleteCharAt(end);
String x2=new String(sb2);
if(x2.equals(sb2.reverse().toString()))
return true;
}
return false;
}
I tried the following,f and b are the indices at which characters do not match
int canwemakepal(char *str)//str input string
{
long int f,b,len,i,j;
int retval=0;
len=strlen(str);
f=0;b=len-1;
while(str[f]==str[b] && f<b)//continue matching till we dont get a mismatch
{
f++;b--;
}
if(f>=b)//if the index variable cross over each other, str is palindrome,answer is yes
{
retval=1;//true
}
else if(str[f+1]==str[b])//we get a mismatch,so check if removing character at str[f] will give us a palindrome
{
i=f+2;j=b-1;
while(str[i]==str[j] && i<j)
{
i++;j--;
}
if(i>=j)
retval=1;
else
retval=0;
}
else if(str[f]==str[b-1])//else check the same for str[b]
{
i=f+1;j=b-2;
while(str[i]==str[j] && i<j)
{
i++;j--;
}
if(i>=j)
retval=1;
else
retval=0;
}
else
retval=0;
return retval;
}
I created this solution,i tried with various input giving correct result,still not accepted as correct solution,Check it n let me know if m doing anything wrong!! Thanks in advance.
public static void main(String[] args)
{
Scanner s = new Scanner(System.in);
int t = s.nextInt();
String result[] = new String[t];
short i = 0;
while(i < t)
{
String str1 = s.next();
int length = str1.length();
String str2 = reverseString(str1);
if(str1.equals(str2))
{
result[i] = "Yes";
}
else
{
if(length == 2)
{
result[i] = "Yes";
}
else
{
int x = 0,y = length-1;
int counter = 0;
while(x<y)
{
if(str1.charAt(x) == str1.charAt(y))
{
x++;
y--;
}
else
{
counter ++;
if(str1.charAt(x) == str1.charAt(y-1))
{
y--;
}
else if(str1.charAt(x+1) == str1.charAt(y))
{
x++;
}
else
{
counter ++;
break;
}
}
}
if(counter >= 2)
{
result[i] = "No";
}
else
result[i]="Yes";
}
}
i++;
} // Loop over
for(int j=0; j<i;j++)
{
System.out.println(result[j]);
}
}
public static String reverseString(String original)
{
int length = original.length();
String reverse = "";
for ( int i = length - 1 ; i >= 0 ; i-- )
reverse = reverse + original.charAt(i);
return reverse;
}
I was solving a question, with which I am having some problems:
Complete the function getEqualSumSubstring, which takes a single argument. The single argument is a string s, which contains only non-zero digits.
This function should print the length of longest contiguous substring of s, such that the length of the substring is 2*N digits and the sum of the leftmost N digits is equal to the sum of the rightmost N digits. If there is no such string, your function should print 0.
int getEqualSumSubstring(string s) {
int i=0,j=i,foundLength=0;
for(i=0;i<s.length();i++)
{
for(j=i;j<s.length();j++)
{
int temp = j-i;
if(temp%2==0)
{
int leftSum=0,rightSum=0;
string tempString=s.substr(i,temp);
for(int k=0;k<temp/2;k++)
{
leftSum=leftSum+tempString[k]-'0';
rightSum=rightSum+tempString[k+(temp/2)]-'0';
}
if((leftSum==rightSum)&&(leftSum!=0))
if(s.length()>foundLength)
foundLength=s.length();
}
}
}
return(foundLength);
}
The problem is that this code is working for some samples and not for the others. Since this is an exam type question I don't have the test cases either.
This code works
int getEqualSumSubstring(string s) {
int i=0,j=i,foundLength=0;
for(i=0;i<s.length();i++)
{
for(j=i;j<s.length();j++)
{
int temp = j-i+1;
if(temp%2==0)
{
int leftSum=0,rightSum=0;
string tempString=s.substr(i,temp);
// printf("%d ",tempString.length());
for(int k=0;k<temp/2;k++)
{
leftSum=leftSum+tempString[k]-48;
rightSum=rightSum+tempString[k+(temp/2)]-48;
}
if((leftSum==rightSum)&&(leftSum!=0))
if(tempString.length()>foundLength)
foundLength=tempString.length();
}
}
}
return(foundLength);
}
The temp variable must be j-i+1. Otherwise the case where the whole string is the answer will not be covered. Also, we need to make the change suggested by Scott.
Here's my solution that I can confirm works. The ones above didn't really work for me - they gave me compile errors somehow. I got the same question on InterviewStreet, came up with a bad, incomplete solution that worked for 9/15 of the test cases, so I had to spend some more time coding afterwards.
The idea is that instead of caring about getting the left and right sums (which is what I initially did as well), I will get all the possible substrings out of each half (left and right half) of the given input, sort and append them to two separate lists, and then see if there are any matches.
Why?
Say the strings "423" and "234" have the same sum; if I sorted them, they would both be "234" and thus match. Since these numbers have to be consecutive and equal length, I no longer need to worry about having to add them up as numbers and check.
So, for example, if I'm given 12345678, then on the left side, the for-loop will give me:
[1,12,123,1234,2,23,234,3,34]
And on the right:
[5,56,567,5678,...]
And so forth.
However, I'm only taking substrings of a length of at least 2 into account.
I append each of these substrings, sorted by converting into a character array then converting back into a string, into ArrayLists.
So now that all this is done, the next step is to see if there are identical strings of the same numbers in these two ArrayLists. I simply check each of temp_b's strings against temp_a's first string, then against temp_a's second string, and so forth.
If I get a match (say, "234" and "234"), I'll set the length of those matching substrings as my tempCount (tempCount = 3). I also have another variable called 'count' to keep track of the greatest length of these matching substrings (if this was the first occurrence of a match, then count = 0 is overwritten by tempCount = 3, so count = 3).
As for the odd/even string length with the variable int end, the reason for this is because in the line of code s.length()/2+j, is the length of the input happened to be 11, then:
s.length() = 11
s.length()/2 = 11/5 = 5.5 = 5
So in the for-loop, s.length()/2 + j, where j maxes out at s.length()/2, would become:
5 + 5 = 10
Which falls short of the s.length() that I need to reach for to get the string's last index.
This is because the substring function requires an end index of one greater than what you'd put for something like charAt(i).
Just to demonstrate, an input of "47582139875" will generate the following output:
[47, 457, 4578, 24578, 57, 578, 2578, 58, 258, 28] <-- substrings from left half
[139, 1389, 13789, 135789, 389, 3789, 35789, 789, 5789, 578] <-- substrings from right half
578 <-- the longest one that matched
6 <-- the length of '578' x 2
public static int getEqualSumSubtring(String s){
// run through all possible length combinations of the number string on left and right half
// append sorted versions of these into new ArrayList
ArrayList<String> temp_a = new ArrayList<String>();
ArrayList<String> temp_b = new ArrayList<String>();
int end; // s.length()/2 is an integer that rounds down if length is odd, account for this later
for( int i=0; i<=s.length()/2; i++ ){
for( int j=i; j<=s.length()/2; j++ ){
// only account for substrings with a length of 2 or greater
if( j-i > 1 ){
char[] tempArr1 = s.substring(i,j).toCharArray();
Arrays.sort(tempArr1);
String sorted1 = new String(tempArr1);
temp_a.add(sorted1);
//System.out.println(sorted1);
if( s.length() % 2 == 0 )
end = s.length()/2+j;
else // odd length so we need the extra +1 at the end
end = s.length()/2+j+1;
char[] tempArr2 = s.substring(i+s.length()/2, end).toCharArray();
Arrays.sort(tempArr2);
String sorted2 = new String(tempArr2);
temp_b.add(sorted2);
//System.out.println(sorted2);
}
}
}
// For reference
System.out.println(temp_a);
System.out.println(temp_b);
// If the substrings match, it means they have the same sum
// Keep track of longest substring
int tempCount = 0 ;
int count = 0;
String longestSubstring = "";
for( int i=0; i<temp_a.size(); i++){
for( int j=0; j<temp_b.size(); j++ ){
if( temp_a.get(i).equals(temp_b.get(j)) ){
tempCount = temp_a.get(i).length();
if( tempCount > count ){
count = tempCount;
longestSubstring = temp_a.get(i);
}
}
}
}
System.out.println(longestSubstring);
return count*2;
}
Heres my solution to this question including tests. I've added an extra function just because I feel it makes the solution way easier to read than the solutions above.
#include <string>
#include <iostream>
using namespace std;
int getMaxLenSumSubstring( string s )
{
int N = 0; // The optimal so far...
int leftSum = 0, rightSum=0, strLen=s.size();
int left, right;
for(int i=0;i<strLen/2+1;i++) {
left=(s[i]-int('0')); right=(s[strLen-i-1]-int('0'));
leftSum+=left; rightSum+=right;
if(leftSum==rightSum) N=i+1;
}
return N*2;
}
int getEqualSumSubstring( string s ) {
int maxLen = 0, substrLen, j=1;
for( int i=0;i<s.length();i++ ) {
for( int j=1; j<s.length()-i; j++ ) {
//cout<<"Substring = "<<s.substr(i,j);
substrLen = getMaxLenSumSubstring(s.substr(i,j));
//cout<<", Len ="<<substrLen;
if(substrLen>maxLen) maxLen=substrLen;
}
}
return maxLen;
}
Here are a few tests I ran. Based upon the examples above they seem right.
int main() {
cout<<endl<<"Test 1 :"<<getEqualSumSubstring(string("123231"))<<endl;
cout<<endl<<"Test 2 :"<<getEqualSumSubstring(string("986561517416921217551395112859219257312"))<<endl;
cout<<endl<<"Test 3:"<<getEqualSumSubstring(string("47582139875"))<<endl;
}
Shouldn't the following code use tempString.length() instead of s.length()
if((leftSum==rightSum)&&(leftSum!=0))
if(s.length()>foundLength)
foundLength=s.length();
Below is my code for the question... Thanks !!
public class IntCompl {
public String getEqualSumSubstring_com(String s)
{
int j;
int num=0;
int sum = 0;
int m=s.length();
//calculate String array Length
for (int i=m;i>1;i--)
{
sum = sum + m;
m=m-1;
}
String [] d = new String[sum];
int k=0;
String ans = "NULL";
//Extract strings
for (int i=0;i<s.length()-1;i++)
{
for (j=s.length();j>=i+1;k++,j--)
{
num = k;
d[k] = s.substring(i,j);
}
k=num+1;
}
//Sort strings in such a way that the longest strings precede...
for (int i=0; i<d.length-1; i++)
{
for (int h=1;h<d.length;h++)
{
if (d[i].length() > d[h].length())
{
String temp;
temp=d[i];
d[i]=d[h];
d[h]=temp;
}
}
}
// Look for the Strings with array size 2*N (length in even number) and such that the
//the sum of left N numbers is = to the sum of right N numbers.
//As the strings are already in decending order, longest string is searched first and break the for loop once the string is found.
for (int x=0;x<d.length;x++)
{
int sum1=0,sum2=0;
if (d[x].length()%2==0 && d[x].length()<49)
{
int n;
n = d[x].length()/2;
for (int y=0;y<n;y++)
{
sum1 = sum1 + d[x].charAt(y)-'0';
}
for (int y=n;y<d[x].length();y++)
{
sum2 = sum2 + d[x].charAt(y)-'0';
}
if (sum1==sum2)
{
ans = d[x];
break;
}
}
}
return ans;
}
}
Here is the complete Java Program for this question.
Complexity is O(n^3)
This can however be solved in O(n^2).For O(n^2) complexity solution refer to this link
import java.util.Scanner;
import static java.lang.System.out;
public class SubStringProblem{
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
out.println("Enter the Digit String:");
String s = sc.nextLine();
int n = (new SubStringProblem()).getEqualSumSubString(s);
out.println("The longest Sum SubString is "+n);
}
public int getEqualSumSubString(String s){
int N;
if(s.length()%2==0)
{
//String is even
N = s.length();
}
else{
//String is odd
N=s.length()-1;
}
boolean flag =false;
int sum1,sum2;
do{
for(int k=0;k<=s.length()-N;k++){
sum1=0;
sum2=0;
for(int i =k,j=k+N-1;i<j;i++,j--)
{
sum1=sum1 + Integer.parseInt(s.substring(i,i+1));
sum2+=Integer.parseInt(s.substring(j,j+1));
}
if(sum1==sum2){
return N;
}
}
N-=2;
flag =true;
}while(N>1);
return -1;
}
}
What is your rationale for the number 48 on these two lines?
for(int k=0;k<temp/2;k++)
{
leftSum=leftSum+tempString[k]-48;
rightSum=rightSum+tempString[k+(temp/2)]-48;
}
I am just overly curious and would like to hear the reasoning behind it, because I have a similar solution, but without the 48 and it still works. However, I added the 48 an still got the correct answer.
Simple solution. O(n*n). s - input string.
var longest = 0;
for (var i = 0; i < s.length-1; i++) {
var leftSum = rightSum = 0;
for (var j = i, k = i+1, l = 2; j >=0 && k < s.length; j--, k++, l+=2) {
leftSum += parseInt(s[j]);
rightSum += parseInt(s[k]);
if (leftSum == rightSum && l > longest) {
longest = l;
}
}
}
console.log(longest);