I have been desperately stuck on this question on Kattis here. The basic idea is that you are given a general formula for a valid two variable recursive function and then given that formula you are to evaluate the function of this function at various inputs. I have trawled through this question for over an hour and I was wondering if anyone could point me in the right direction (or give a counter-example) of why my answer is wrong. The code I used which got me a Wrong Answer verdict is shown below.
#include<bits/stdc++.h>
using namespace std;
long long dp[105][105];
int a[25];
int b[25];
long long rec(int x_in, int y_in, int coeffs, long long c) {
if(dp[x_in][y_in]!=-1){
return dp[x_in][y_in];
}
else {
long long ans = c;
for(int i=0; i<coeffs; ++i) {
int x_cell = max(0, x_in-a[i]);
int y_cell = max(0, y_in-b[i]);
ans+=rec(x_cell, y_cell, coeffs, c);
}
dp[x_in][y_in] = ans;
return dp[x_in][y_in];
}
}
int main() {
int n;
scanf("%d", &n);
string ex;
getline(cin, ex);
for(int caseno=0; caseno<n; ++caseno) {
memset(dp, -1, sizeof(dp));
if(caseno>0) printf("\n");
string coeffs_list;
getline(cin, coeffs_list);
int pairs = 0;
long long c, d;
char * dup = strdup(coeffs_list.c_str());
char * token = strtok(dup, " ") ;
while(token != NULL){
a[pairs] = atoi(token);
token = strtok(NULL, " ") ;
b[pairs] = atoi(token);
token = strtok(NULL, " ") ;
pairs++;
}
free(dup);
c= (long long) a[pairs-1];
d = (long long) b[pairs-1];
pairs--;
dp[0][0] = d;
for(int i=1; i<105; ++i) {dp[0][i] = d; dp[i][0]=d;}
string inputs;
getline(cin, inputs);
int x_i, y_i;
char * dup2 = strdup(inputs.c_str());
char * token2 = strtok(dup2, " ") ;
while(token2 != NULL){
x_i = atoi(token2);
token2 = strtok(NULL, " ") ;
y_i = atoi(token2);
token2 = strtok(NULL, " ") ;
printf("%lld\n", rec(x_i, y_i, pairs, c));
}
free(dup2);
}
}
As you can see the basic idea of what i did was to construct a dp table and evaluate the function accordingly. Thank you in advance for anyone that could help me over here.
The function can grow very fast, thus overflow a 64bit integer (you can check this by using assert(rec(...) >= 0) (not guaranteed but likely to fail for an overflow)). Use a custom BigInteger implementation or switch to Java/Python.
e.g. for n=20, a_i=1, b_i=0: f(x,y)=20*f(x-1,y)+c = O(20^x*(c+d))
This is also indicated by most solutions under stats using Java: https://open.kattis.com/problems/generalizedrecursivefunctions/statistics https://baylor.kattis.com/problems/generalizedrecursivefunctions/statistics
I am trying to write a simple nonogram solver, in a kind of bruteforce way, but I am stuck on a relatively easy task. Let's say I have a row with clues [2,3] that has a length of 10
so the solutions are:
$$-$$$----
$$--$$$---
$$---$$$--
$$----$$$-
$$-----$$$
-$$----$$$
--$$---$$$
---$$--$$$
----$$-$$$
-$$---$$$-
--$$-$$$--
I want to find all the possible solutions for a row
I know that I have to consider each block separately, and each block will have an availible space of n-(sum of remaining blocks length + number of remaining blocks) but I do not know how to progress from here
Well, this question already have a good answer, so think of this one more as an advertisement of python's prowess.
def place(blocks,total):
if not blocks: return ["-"*total]
if blocks[0]>total: return []
starts = total-blocks[0] #starts = 2 means possible starting indexes are [0,1,2]
if len(blocks)==1: #this is special case
return [("-"*i+"$"*blocks[0]+"-"*(starts-i)) for i in range(starts+1)]
ans = []
for i in range(total-blocks[0]): #append current solutions
for sol in place(blocks[1:],starts-i-1): #with all possible other solutiona
ans.append("-"*i+"$"*blocks[0]+"-"+sol)
return ans
To test it:
for i in place([2,3,2],12):
print(i)
Which produces output like:
$$-$$$-$$---
$$-$$$--$$--
$$-$$$---$$-
$$-$$$----$$
$$--$$$-$$--
$$--$$$--$$-
$$--$$$---$$
$$---$$$-$$-
$$---$$$--$$
$$----$$$-$$
-$$-$$$-$$--
-$$-$$$--$$-
-$$-$$$---$$
-$$--$$$-$$-
-$$--$$$--$$
-$$---$$$-$$
--$$-$$$-$$-
--$$-$$$--$$
--$$--$$$-$$
---$$-$$$-$$
This is what i got:
#include <iostream>
#include <vector>
#include <string>
using namespace std;
typedef std::vector<bool> tRow;
void printRow(tRow row){
for (bool i : row){
std::cout << ((i) ? '$' : '-');
}
std::cout << std::endl;
}
int requiredCells(const std::vector<int> nums){
int sum = 0;
for (int i : nums){
sum += (i + 1); // The number + the at-least-one-cell gap at is right
}
return (sum == 0) ? 0 : sum - 1; // The right-most number don't need any gap
}
bool appendRow(tRow init, const std::vector<int> pendingNums, unsigned int rowSize, std::vector<tRow> &comb){
if (pendingNums.size() <= 0){
comb.push_back(init);
return false;
}
int cellsRequired = requiredCells(pendingNums);
if (cellsRequired > rowSize){
return false; // There are no combinations
}
tRow prefix;
int gapSize = 0;
std::vector<int> pNumsAux = pendingNums;
pNumsAux.erase(pNumsAux.begin());
unsigned int space = rowSize;
while ((gapSize + cellsRequired) <= rowSize){
space = rowSize;
space -= gapSize;
prefix.clear();
prefix = init;
for (int i = 0; i < gapSize; ++i){
prefix.push_back(false);
}
for (int i = 0; i < pendingNums[0]; ++i){
prefix.push_back(true);
space--;
}
if (space > 0){
prefix.push_back(false);
space--;
}
appendRow(prefix, pNumsAux, space, comb);
++gapSize;
}
return true;
}
std::vector<tRow> getCombinations(const std::vector<int> row, unsigned int rowSize) {
std::vector<tRow> comb;
tRow init;
appendRow(init, row, rowSize, comb);
return comb;
}
int main(){
std::vector<int> row = { 2, 3 };
auto ret = getCombinations(row, 10);
for (tRow r : ret){
while (r.size() < 10)
r.push_back(false);
printRow(r);
}
return 0;
}
And my output is:
$$-$$$----
$$--$$$---
$$---$$$--
$$----$$$--
$$-----$$$
-$$-$$$----
-$$--$$$--
-$$---$$$-
-$$----$$$-
--$$-$$$--
--$$--$$$-
--$$---$$$
---$$-$$$-
---$$--$$$
----$$-$$$
For sure, this must be absolutely improvable.
Note: i did't test it more than already written case
Hope it works for you
I noticed while doing some F# experiments that if write my own reduce function for Array that it performs much better than the built in reduce. For example:
type Array with
static member inline fastReduce f (values : 'T[]) =
let mutable result = Unchecked.defaultof<'T>
for i in 0 .. values.Length-1 do
result <- f result values.[i]
result
This seems to behave identically to the built in Array.reduce but is ~2x faster for simple f
Is the built in one more flexible in some way?
By looking at the generated IL code it's easier to understand what's happening.
Using the built-in Array.reduce:
let reducer (vs : int []) : int = Array.reduce (+) vs
Gives the following equivalent C# (reverse engineered from the IL code using ILSpy)
public static int reducer(int[] vs)
{
return ArrayModule.Reduce<int>(new Program.BuiltIn.reducer#31(), vs);
}
Array.reduce looks like this:
public static T Reduce<T>(FSharpFunc<T, FSharpFunc<T, T>> reduction, T[] array)
{
if (array == null)
{
throw new ArgumentNullException("array");
}
int num = array.Length;
if (num == 0)
{
throw new ArgumentException(LanguagePrimitives.ErrorStrings.InputArrayEmptyString, "array");
}
OptimizedClosures.FSharpFunc<T, T, T> fSharpFunc = OptimizedClosures.FSharpFunc<T, T, T>.Adapt(reduction);
T t = array[0];
int num2 = 1;
int num3 = num - 1;
if (num3 >= num2)
{
do
{
t = fSharpFunc.Invoke(t, array[num2]);
num2++;
}
while (num2 != num3 + 1);
}
return t;
}
Notice that it invoking the reducer function f is a virtual call which typically the JIT:er struggles to inline.
Compare to your fastReduce function:
let reducer (vs : int []) : int = Array.fastReduce (+) vs
The reverse-engineered C# code:
public static int reducer(int[] vs)
{
int num = 0;
for (int i = 0; i < vs.Length; i++)
{
num += vs[i];
}
return num;
}
A lot more efficient as the virtual call is now gone. It seems that in this case F# inlines both the code for fastReduce as well as (+).
There's some kind of cut-off in F# as more complex reducer functions won't be inlined. I am unsure on the exact details.
Hope this helps
A side-note; Unchecked.defaultOf returns null values for class types in .NET such as string. I prefer LanguagePrimitives.GenericZero.
PS. A common trick for the real performance hungry is to loop towards 0. In F# that doesn't work for for-expressions because of a slight performance bug in how for-expressions are generated. In those case you can try to implement the loop using tail-recursion.
I'm trying to put my first steps into Scala, and to practice I took a look at the google code jam storecredit excersize. I tried it in java first, which went well enough, and now I'm trying to port it to Scala. Now with the java collections framework, I could try to do a straight syntax conversion, but I'd end up writing java in scala, and that kind of defeats the purpose. In my Java implementation, I have a PriorityQueue that I empty into a Deque, and pop the ends off untill we have bingo. This all uses mutable collections, which give me the feeling is very 'un-scala'. What I think would be a more functional approach is to construct a datastructure that can be traversed both from highest to lowest, and from lowest to highest. Am I on the right path? Are there any suitable datastructures supplied in the Scala libraries, or should I roll my own here?
EDIT: full code of the much simpler version in Java. It should run in O(max(credit,inputchars)) and has become:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
public class StoreCredit {
private static BufferedReader in;
public static void main(String[] args) {
in = new BufferedReader(new InputStreamReader(System.in));
try {
int numCases = Integer.parseInt(in.readLine());
for (int i = 0; i < numCases; i++) {
solveCase(i);
}
} catch (IOException e) {
e.printStackTrace();
}
}
private static void solveCase(int casenum) throws NumberFormatException,
IOException {
int credit = Integer.parseInt(in.readLine());
int numItems = Integer.parseInt(in.readLine());
int itemnumber = 0;
int[] item_numbers_by_price = new int[credit];
Arrays.fill(item_numbers_by_price, -1); // makes this O(max(credit,
// items)) instead of O(items)
int[] read_prices = readItems();
while (itemnumber < numItems) {
int next_price = read_prices[itemnumber];
if (next_price <= credit) {
if (item_numbers_by_price[credit - next_price] >= 0) {
// Bingo! DinoDNA!
printResult(new int[] {
item_numbers_by_price[credit - next_price],
itemnumber }, casenum);
break;
}
item_numbers_by_price[next_price] = itemnumber;
}
itemnumber++;
}
}
private static int[] readItems() throws IOException {
String line = in.readLine();
String[] items = line.split(" "); // uh-oh, now it's O(max(credit,
// inputchars))
int[] result = new int[items.length];
for (int i = 0; i < items.length; i++) {
result[i] = Integer.parseInt(items[i]);
}
return result;
}
private static void printResult(int[] result, int casenum) {
int one;
int two;
if (result[0] > result[1]) {
one = result[1];
two = result[0];
} else {
one = result[0];
two = result[1];
}
one++;
two++;
System.out.println(String.format("Case #%d: %d %d", casenum + 1, one,
two));
}
}
I'm wondering what you are trying to accomplish using sophisticated data structures such as PriorityQueue and Deque for a problem such as this. It can be solved with a pair of nested loops:
for {
i <- 2 to I
j <- 1 until i
if i != j && P(i-1) + P(j - 1) == C
} println("Case #%d: %d %d" format (n, j, i))
Worse than linear, better than quadratic. Since the items are not sorted, and sorting them would require O(nlogn), you can't do much better than this -- as far as I can see.
Actually, having said all that, I now have figured a way to do it in linear time. The trick is that, for every number p you find, you know what its complement is: C - p. I expect there are a few ways to explore that -- I have so far thought of two.
One way is to build a map with O(n) characteristics, such as a bitmap or a hash map. For each element, make it point to its index. One then only has to find an element for which its complement also has an entry in the map. Trivially, this could be as easily as this:
val PM = P.zipWithIndex.toMap
val (p, i) = PM find { case (p, i) => PM isDefinedAt C - p }
val j = PM(C - p)
However, that won't work if the number is equal to its complement. In other words, if there are two p such that p + p == C. There are quite a few such cases in the examples. One could then test for that condition, and then just use indexOf and lastIndexOf -- except that it is possible that there is only one p such that p + p == C, in which case that wouldn't be the answer either.
So I ended with something more complex, that tests the existence of the complement at the same time the map is being built. Here's the full solution:
import scala.io.Source
object StoreCredit3 extends App {
val source = if (args.size > 0) Source fromFile args(0) else Source.stdin
val input = source getLines ()
val N = input.next.toInt
1 to N foreach { n =>
val C = input.next.toInt
val I = input.next.toInt
val Ps = input.next split ' ' map (_.toInt)
val (_, Some((p1, p2))) = Ps.zipWithIndex.foldLeft((Map[Int, Int](), None: Option[(Int, Int)])) {
case ((map, None), (p, i)) =>
if (map isDefinedAt C - p) map -> Some(map(C - p) -> (i + 1))
else (map updated (p, i + 1), None)
case (answer, _) => answer
}
println("Case #%d: %d %d" format (n, p1, p2))
}
}
How do you sort an array of strings naturally in different programming languages? Post your implementation and what language it is in in the answer.
Here's how you can get explorer-like behaviour in Python:
#!/usr/bin/env python
"""
>>> items = u'a1 a003 b2 a2 a10 1 10 20 2 c100'.split()
>>> items.sort(explorer_cmp)
>>> for s in items:
... print s,
1 2 10 20 a1 a2 a003 a10 b2 c100
>>> items.sort(key=natural_key, reverse=True)
>>> for s in items:
... print s,
c100 b2 a10 a003 a2 a1 20 10 2 1
"""
import re
def natural_key(astr):
"""See http://www.codinghorror.com/blog/archives/001018.html"""
return [int(s) if s.isdigit() else s for s in re.split(r'(\d+)', astr)]
def natural_cmp(a, b):
return cmp(natural_key(a), natural_key(b))
try: # use explorer's comparison function if available
import ctypes
explorer_cmp = ctypes.windll.shlwapi.StrCmpLogicalW
except (ImportError, AttributeError):
# not on Windows or old python version
explorer_cmp = natural_cmp
if __name__ == '__main__':
import doctest; doctest.testmod()
To support Unicode strings, .isdecimal() should be used instead of .isdigit().
.isdigit() may also fail (return value that is not accepted by int()) for a bytestring on Python 2 in some locales e.g., '\xb2' ('²') in cp1252 locale on Windows.
JavaScript
Array.prototype.alphanumSort = function(caseInsensitive) {
for (var z = 0, t; t = this[z]; z++) {
this[z] = [], x = 0, y = -1, n = 0, i, j;
while (i = (j = t.charAt(x++)).charCodeAt(0)) {
var m = (i == 46 || (i >=48 && i <= 57));
if (m !== n) {
this[z][++y] = "";
n = m;
}
this[z][y] += j;
}
}
this.sort(function(a, b) {
for (var x = 0, aa, bb; (aa = a[x]) && (bb = b[x]); x++) {
if (caseInsensitive) {
aa = aa.toLowerCase();
bb = bb.toLowerCase();
}
if (aa !== bb) {
var c = Number(aa), d = Number(bb);
if (c == aa && d == bb) {
return c - d;
} else return (aa > bb) ? 1 : -1;
}
}
return a.length - b.length;
});
for (var z = 0; z < this.length; z++)
this[z] = this[z].join("");
}
Source
For MySQL, I personally use code from a Drupal module, which is available at hhttp://drupalcode.org/project/natsort.git/blob/refs/heads/5.x-1.x:/natsort.install.mysql
Basically, you execute the posted SQL script to create functions, and then use ORDER BY natsort_canon(field_name, 'natural')
Here's a readme about the function:
http://drupalcode.org/project/natsort.git/blob/refs/heads/5.x-1.x:/README.txt
Here's a cleanup of the code in the article the question linked to:
def sorted_nicely(strings):
"Sort strings the way humans are said to expect."
return sorted(strings, key=natural_sort_key)
def natural_sort_key(key):
import re
return [int(t) if t.isdigit() else t for t in re.split(r'(\d+)', key)]
But actually I haven't had occasion to sort anything this way.
If the OP is asking about idomatic sorting expressions, then not all languages have a natural expression built in. For c I'd go to <stdlib.h> and use qsort. Something on the lines of :
/* non-functional mess deleted */
to sort the arguments into lexical order. Unfortunately this idiom is rather hard to parse for those not used the ways of c.
Suitably chastened by the downvote, I actually read the linked article. Mea culpa.
In anycase the original code did not work, except in the single case I tested. Damn. Plain vanilla c does not have this function, nor is it in any of the usual libraries.
The code below sorts the command line arguments in the natural way as linked. Caveat emptor as it is only lightly tested.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
int naturalstrcmp(const char **s1, const char **s2);
int main(int argc, char **argv){
/* Sort the command line arguments in place */
qsort(&argv[1],argc-1,sizeof(char*),
(int(*)(const void *, const void *))naturalstrcmp);
while(--argc){
printf("%s\n",(++argv)[0]);
};
}
int naturalstrcmp(const char **s1p, const char **s2p){
if ((NULL == s1p) || (NULL == *s1p)) {
if ((NULL == s2p) || (NULL == *s2p)) return 0;
return 1;
};
if ((NULL == s2p) || (NULL == *s2p)) return -1;
const char *s1=*s1p;
const char *s2=*s2p;
do {
if (isdigit(s1[0]) && isdigit(s2[0])){
/* Compare numbers as numbers */
int c1 = strspn(s1,"0123456789"); /* Could be more efficient here... */
int c2 = strspn(s2,"0123456789");
if (c1 > c2) {
return 1;
} else if (c1 < c2) {
return -1;
};
/* the digit strings have equal length, so compare digit by digit */
while (c1--) {
if (s1[0] > s2[0]){
return 1;
} else if (s1[0] < s2[0]){
return -1;
};
s1++;
s2++;
};
} else if (s1[0] > s2[0]){
return 1;
} else if (s1[0] < s2[0]){
return -1;
};
s1++;
s2++;
} while ( (s1!='\0') || (s2!='\0') );
return 0;
}
This approach is pretty brute force, but it is simple and can probably be duplicated in any imperative language.
I just use StrCmpLogicalW. It does exactly what Jeff is wanting, since it's the same API that explorer uses. Admittedly, it's not portable.
In C++:
bool NaturalLess(const wstring &lhs, const wstring &rhs)
{
return StrCmpLogicalW(lhs.c_str(), rhs.c_str()) < 0;
}
vector<wstring> strings;
// ... load the strings
sort(strings.begin(), strings.end(), &NaturalLess);
Just a link to some nice work in Common Lisp by Eric Normand:
http://www.lispcast.com/wordpress/2007/12/human-order-sorting/
In C, this solution correctly handles numbers with leading zeroes:
#include <stdlib.h>
#include <ctype.h>
/* like strcmp but compare sequences of digits numerically */
int strcmpbynum(const char *s1, const char *s2) {
for (;;) {
if (*s2 == '\0')
return *s1 != '\0';
else if (*s1 == '\0')
return 1;
else if (!(isdigit(*s1) && isdigit(*s2))) {
if (*s1 != *s2)
return (int)*s1 - (int)*s2;
else
(++s1, ++s2);
} else {
char *lim1, *lim2;
unsigned long n1 = strtoul(s1, &lim1, 10);
unsigned long n2 = strtoul(s2, &lim2, 10);
if (n1 > n2)
return 1;
else if (n1 < n2)
return -1;
s1 = lim1;
s2 = lim2;
}
}
}
If you want to use it with qsort, use this auxiliary function:
static int compare(const void *p1, const void *p2) {
const char * const *ps1 = p1;
const char * const *ps2 = p2;
return strcmpbynum(*ps1, *ps2);
}
And you can do something on the order of
char *lines = ...;
qsort(lines, next, sizeof(lines[0]), compare);
In C++ I use this example code to do natural sorting. The code requires the boost library.
Note that for most such questions, you can just consult the Rosetta Code Wiki. I adapted my answer from the entry for sorting integers.
In a system's programming language doing something like this is generally going to be uglier than with a specialzed string-handling language. Fortunately for Ada, the most recent version has a library routine for just this kind of task.
For Ada 2005 I believe you could do something along the following lines (warning, not compiled!):
type String_Array is array(Natural range <>) of Ada.Strings.Unbounded.Unbounded_String;
function "<" (L, R : Ada.Strings.Unbounded.Unbounded_String) return boolean is
begin
--// Natural ordering predicate here. Sorry to cheat in this part, but
--// I don't exactly grok the requirement for "natural" ordering. Fill in
--// your proper code here.
end "<";
procedure Sort is new Ada.Containers.Generic_Array_Sort
(Index_Type => Natural;
Element_Type => Ada.Strings.Unbounded.Unbounded_String,
Array_Type => String_Array
);
Example use:
using Ada.Strings.Unbounded;
Example : String_Array := (To_Unbounded_String ("Joe"),
To_Unbounded_String ("Jim"),
To_Unbounded_String ("Jane"),
To_Unbounded_String ("Fred"),
To_Unbounded_String ("Bertha"),
To_Unbounded_String ("Joesphus"),
To_Unbounded_String ("Jonesey"));
begin
Sort (Example);
...
end;
Python, using itertools:
def natural_key(s):
return tuple(
int(''.join(chars)) if isdigit else ''.join(chars)
for isdigit, chars in itertools.groupby(s, str.isdigit)
)
Result:
>>> natural_key('abc-123foo456.xyz')
('abc-', 123, 'foo', 456, '.xyz')
Sorting:
>>> sorted(['1.1.1', '1.10.4', '1.5.0', '42.1.0', '9', 'banana'], key=natural_key)
['1.1.1', '1.5.0', '1.10.4', '9', '42.1.0', 'banana']
My implementation on Clojure 1.1:
(ns alphanumeric-sort
(:import [java.util.regex Pattern]))
(defn comp-alpha-numerical
"Compare two strings alphanumerically."
[a b]
(let [regex (Pattern/compile "[\\d]+|[a-zA-Z]+")
sa (re-seq regex a)
sb (re-seq regex b)]
(loop [seqa sa seqb sb]
(let [counta (count seqa)
countb (count seqb)]
(if-not (not-any? zero? [counta countb]) (- counta countb)
(let [c (first seqa)
d (first seqb)
c1 (read-string c)
d1 (read-string d)]
(if (every? integer? [c1 d1])
(def result (compare c1 d1)) (def result (compare c d)))
(if-not (= 0 result) result (recur (rest seqa) (rest seqb)))))))))
(sort comp-alpha-numerical ["a1" "a003" "b2" "a10" "a2" "1" "10" "20" "2" "c100"])
Result:
("1" "2" "10" "20" "a1" "a2" "a003" "a10" "b2" "c100")
For Tcl, the -dict (dictionary) option to lsort:
% lsort -dict {a b 1 c 2 d 13}
1 2 13 a b c d
php has a easy function "natsort" to do that,and I implements it by myself:
<?php
$temp_files = array('+====','-==',"temp15-txt","temp10.txt",
"temp1.txt","tempe22.txt","temp2.txt");
$my_arr = $temp_files;
natsort($temp_files);
echo "Natural order: ";
print_r($temp_files);
echo "My Natural order: ";
usort($my_arr,'my_nat_func');
print_r($my_arr);
function is_alpha($a){
return $a>='0'&&$a<='9' ;
}
function my_nat_func($a,$b){
if(preg_match('/[0-9]/',$a)){
if(preg_match('/[0-9]/',$b)){
$i=0;
while(!is_alpha($a[$i])) ++$i;
$m = intval(substr($a,$i));
$i=0;
while(!is_alpha($b[$i])) ++$i;
$n = intval(substr($b,$i));
return $m>$n?1:($m==$n?0:-1);
}
return 1;
}else{
if(preg_match('/[0-9]/',$b)){
return -1;
}
return $a>$b?1:($a==$b?0:-1);
}
}
Java solution:-
This can be achieved by implementing new Comparator<String> and pass it to Collections.sort(list, comparator) method.
#Override
public int compare(String s1, String s2) {
int len1 = s1.length();
int len2 = s2.length();
int lim = Math.min(len1, len2);
char v1[] = s1.toCharArray();
char v2[] = s2.toCharArray();
int k = 0;
while (k < lim) {
char c1 = v1[k];
char c2 = v2[k];
if (c1 != c2) {
if(this.isInteger(c1) && this.isInteger(c2)) {
int i1 = grabContinousInteger(v1, k);
int i2 = grabContinousInteger(v2, k);
return i1 - i2;
}
return c1 - c2;
}
k++;
}
return len1 - len2;
}
private boolean isInteger(char c) {
return c >= 48 && c <= 57; // ascii value 0-9
}
private int grabContinousInteger(char[] arr, int k) {
int i = k;
while(i < arr.length && this.isInteger(arr[i])) {
i++;
}
return Integer.parseInt(new String(arr, k, i - k));
}