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I'm currently trying to solve an algorithm problem and due to being new to programming I feel like I won't be able to figure it out without some knowledge first. My intention is not to receive the answer to the problem but recommendations to process huge arrays in Java.
The problem is: you receive a number (3 > n < 200.000) and then receive a sorted array of numbers that has n length and then you receive another number (n2) (all the numbers can be hold by an int).
You have to find two numbers in the array that their sum will make n2-n.
You will always have a right answer and will always be two numbers.
Per example:
3
1 2 3
3
The answer would be 1 and 2 because 6 - 3 = 3;
My code works but it exceeds time and memory consumption. So I'm looking for someone to tell better ways to find those numbers than can also be applicable to improve my algorithm practices.
What I've done so far is to sum all the numbers in the array and obtain n2 - n; then create two indexes, the first 0 and the second (n2 - n) - 1; Then loop through while the first index is less than the second index. During the loop, performing a binary search of the first index and of the second index. If the binary search finds both numbers then return them as the answer.
public static BufferedReader br;
public static int numHechizos;
public static String line;
public static String[] split;
public static int damage;
public static int[] hechizos;
public static void main(String[] args) throws IOException {
br = new BufferedReader(new InputStreamReader(System.in));
numHechizos = Integer.parseInt(br.readLine());
while (numHechizos != 0){
caso(numHechizos);
numHechizos = Integer.parseInt(br.readLine());
}
}
public static void caso(int numHechizos) throws IOException {
line = br.readLine();
split = line.split(" ");
damage = Integer.parseInt(br.readLine());
int sum = 0;
hechizos = new int[numHechizos];
for (int i = 0; i < numHechizos; i++) {
int num = Integer.parseInt(split[i]);
hechizos[i] = num;
sum += num;
}
int find = sum - damage;
int n1 = 1; int n2 = find - 1;
while (n1 < n2){
int bn1 = Arrays.binarySearch(hechizos, n1);
int bn2 = Arrays.binarySearch(hechizos, n2);
if (bn1 > -1 && bn2 > -1){
System.out.println(n1 + " " + n2);
break;
}
n1++;
n2--;
}
System.gc();
}
Leet Code Question: You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.
For example, given: s="barfoothefoobarman" & words=["foo", "bar"], return [0,9].
How to implement it with suffix tree or with better approach ?
Here is my running solution with all 169 test cases passed:
public class Solution {
public List<Integer> findSubstring(String S, String[] words)
ArrayList<Integer> result = new ArrayList<Integer>();
HashMap<String, Integer> toFind = new HashMap<String, Integer>();
HashMap<String, Integer> found = new HashMap<String, Integer>();
int wordsTotalLength = words.length, wordLength = words[0].length();
//create a map to find the words
for (int i = 0; i < wordsTotalLength; i ++){
if (!toFind.containsKey(words[i])){
toFind.put(words[i], 1);
}
else{
toFind.put(words[i], toFind.get(words[i]) + 1);
}
}
for (int i = 0; i <= S.length() - (wordsTotalLength * wordLength); i ++){
found.clear();
int j;
for (j = 0; j < wordsTotalLength; j ++){
int k = i + j * wordLength;
String sub = S.substring(k, k + wordLength);
//if we donot find any matching word in dictionary , just break;
if (!toFind.containsKey(sub)) break;
//or else if the word is in 'toFind' the dictionary
// then put it in the 'found' dictionary
if(!found.containsKey(sub)){
found.put(sub, 1);
}
else{
found.put(sub, found.get(sub) + 1);
}
if (found.get(sub) > toFind.get(sub)) break;
}
//if there is a continuous match
//put it in the result array
if (j == wordsTotalLength) result.add(i);
}
return result;
}
}
If the input is 'abba' then the possible palindromes are a, b, b, a, bb, abba.
I understand that determining if string is palindrome is easy. It would be like:
public static boolean isPalindrome(String str) {
int len = str.length();
for(int i=0; i<len/2; i++) {
if(str.charAt(i)!=str.charAt(len-i-1) {
return false;
}
return true;
}
But what is the efficient way of finding palindrome substrings?
This can be done in O(n), using Manacher's algorithm.
The main idea is a combination of dynamic programming and (as others have said already) computing maximum length of palindrome with center in a given letter.
What we really want to calculate is radius of the longest palindrome, not the length.
The radius is simply length/2 or (length - 1)/2 (for odd-length palindromes).
After computing palindrome radius pr at given position i we use already computed radiuses to find palindromes in range [i - pr ; i]. This lets us (because palindromes are, well, palindromes) skip further computation of radiuses for range [i ; i + pr].
While we search in range [i - pr ; i], there are four basic cases for each position i - k (where k is in 1,2,... pr):
no palindrome (radius = 0) at i - k
(this means radius = 0 at i + k, too)
inner palindrome, which means it fits in range
(this means radius at i + k is the same as at i - k)
outer palindrome, which means it doesn't fit in range
(this means radius at i + k is cut down to fit in range, i.e because i + k + radius > i + pr we reduce radius to pr - k)
sticky palindrome, which means i + k + radius = i + pr
(in that case we need to search for potentially bigger radius at i + k)
Full, detailed explanation would be rather long. What about some code samples? :)
I've found C++ implementation of this algorithm by Polish teacher, mgr Jerzy Wałaszek.
I've translated comments to english, added some other comments and simplified it a bit to be easier to catch the main part.
Take a look here.
Note: in case of problems understanding why this is O(n), try to look this way:
after finding radius (let's call it r) at some position, we need to iterate over r elements back, but as a result we can skip computation for r elements forward. Therefore, total number of iterated elements stays the same.
Perhaps you could iterate across potential middle character (odd length palindromes) and middle points between characters (even length palindromes) and extend each until you cannot get any further (next left and right characters don't match).
That would save a lot of computation when there are no many palidromes in the string. In such case the cost would be O(n) for sparse palidrome strings.
For palindrome dense inputs it would be O(n^2) as each position cannot be extended more than the length of the array / 2. Obviously this is even less towards the ends of the array.
public Set<String> palindromes(final String input) {
final Set<String> result = new HashSet<>();
for (int i = 0; i < input.length(); i++) {
// expanding even length palindromes:
expandPalindromes(result,input,i,i+1);
// expanding odd length palindromes:
expandPalindromes(result,input,i,i);
}
return result;
}
public void expandPalindromes(final Set<String> result, final String s, int i, int j) {
while (i >= 0 && j < s.length() && s.charAt(i) == s.charAt(j)) {
result.add(s.substring(i,j+1));
i--; j++;
}
}
So, each distinct letter is already a palindrome - so you already have N + 1 palindromes, where N is the number of distinct letters (plus empty string). You can do that in single run - O(N).
Now, for non-trivial palindromes, you can test each point of your string to be a center of potential palindrome - grow in both directions - something that Valentin Ruano suggested.
This solution will take O(N^2) since each test is O(N) and number of possible "centers" is also O(N) - the center is either a letter or space between two letters, again as in Valentin's solution.
Note, there is also O(N) solution to your problem, based on Manacher's algoritm (article describes "longest palindrome", but algorithm could be used to count all of them)
I just came up with my own logic which helps to solve this problem.
Happy coding.. :-)
System.out.println("Finding all palindromes in a given string : ");
subPal("abcacbbbca");
private static void subPal(String str) {
String s1 = "";
int N = str.length(), count = 0;
Set<String> palindromeArray = new HashSet<String>();
System.out.println("Given string : " + str);
System.out.println("******** Ignoring single character as substring palindrome");
for (int i = 2; i <= N; i++) {
for (int j = 0; j <= N; j++) {
int k = i + j - 1;
if (k >= N)
continue;
s1 = str.substring(j, i + j);
if (s1.equals(new StringBuilder(s1).reverse().toString())) {
palindromeArray.add(s1);
}
}
}
System.out.println(palindromeArray);
for (String s : palindromeArray)
System.out.println(s + " - is a palindrome string.");
System.out.println("The no.of substring that are palindrome : "
+ palindromeArray.size());
}
Output:-
Finding all palindromes in a given string :
Given string : abcacbbbca
******** Ignoring single character as substring palindrome ********
[cac, acbbbca, cbbbc, bb, bcacb, bbb]
cac - is a palindrome string.
acbbbca - is a palindrome string.
cbbbc - is a palindrome string.
bb - is a palindrome string.
bcacb - is a palindrome string.
bbb - is a palindrome string.
The no.of substring that are palindrome : 6
I suggest building up from a base case and expanding until you have all of the palindomes.
There are two types of palindromes: even numbered and odd-numbered. I haven't figured out how to handle both in the same way so I'll break it up.
1) Add all single letters
2) With this list you have all of the starting points for your palindromes. Run each both of these for each index in the string (or 1 -> length-1 because you need at least 2 length):
findAllEvenFrom(int index){
int i=0;
while(true) {
//check if index-i and index+i+1 is within string bounds
if(str.charAt(index-i) != str.charAt(index+i+1))
return; // Here we found out that this index isn't a center for palindromes of >=i size, so we can give up
outputList.add(str.substring(index-i, index+i+1));
i++;
}
}
//Odd looks about the same, but with a change in the bounds.
findAllOddFrom(int index){
int i=0;
while(true) {
//check if index-i and index+i+1 is within string bounds
if(str.charAt(index-i-1) != str.charAt(index+i+1))
return;
outputList.add(str.substring(index-i-1, index+i+1));
i++;
}
}
I'm not sure if this helps the Big-O for your runtime, but it should be much more efficient than trying each substring. Worst case would be a string of all the same letter which may be worse than the "find every substring" plan, but with most inputs it will cut out most substrings because you can stop looking at one once you realize it's not the center of a palindrome.
I tried the following code and its working well for the cases
Also it handles individual characters too
Few of the cases which passed:
abaaa --> [aba, aaa, b, a, aa]
geek --> [g, e, ee, k]
abbaca --> [b, c, a, abba, bb, aca]
abaaba -->[aba, b, abaaba, a, baab, aa]
abababa -->[aba, babab, b, a, ababa, abababa, bab]
forgeeksskeegfor --> [f, g, e, ee, s, r, eksske, geeksskeeg,
o, eeksskee, ss, k, kssk]
Code
static Set<String> set = new HashSet<String>();
static String DIV = "|";
public static void main(String[] args) {
String str = "abababa";
String ext = getExtendedString(str);
// will check for even length palindromes
for(int i=2; i<ext.length()-1; i+=2) {
addPalindromes(i, 1, ext);
}
// will check for odd length palindromes including individual characters
for(int i=1; i<=ext.length()-2; i+=2) {
addPalindromes(i, 0, ext);
}
System.out.println(set);
}
/*
* Generates extended string, with dividors applied
* eg: input = abca
* output = |a|b|c|a|
*/
static String getExtendedString(String str) {
StringBuilder builder = new StringBuilder();
builder.append(DIV);
for(int i=0; i< str.length(); i++) {
builder.append(str.charAt(i));
builder.append(DIV);
}
String ext = builder.toString();
return ext;
}
/*
* Recursive matcher
* If match is found for palindrome ie char[mid-offset] = char[mid+ offset]
* Calculate further with offset+=2
*
*
*/
static void addPalindromes(int mid, int offset, String ext) {
// boundary checks
if(mid - offset <0 || mid + offset > ext.length()-1) {
return;
}
if (ext.charAt(mid-offset) == ext.charAt(mid+offset)) {
set.add(ext.substring(mid-offset, mid+offset+1).replace(DIV, ""));
addPalindromes(mid, offset+2, ext);
}
}
Hope its fine
public class PolindromeMyLogic {
static int polindromeCount = 0;
private static HashMap<Character, List<Integer>> findCharAndOccurance(
char[] charArray) {
HashMap<Character, List<Integer>> map = new HashMap<Character, List<Integer>>();
for (int i = 0; i < charArray.length; i++) {
char c = charArray[i];
if (map.containsKey(c)) {
List list = map.get(c);
list.add(i);
} else {
List list = new ArrayList<Integer>();
list.add(i);
map.put(c, list);
}
}
return map;
}
private static void countPolindromeByPositions(char[] charArray,
HashMap<Character, List<Integer>> map) {
map.forEach((character, list) -> {
int n = list.size();
if (n > 1) {
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
if (list.get(i) + 1 == list.get(j)
|| list.get(i) + 2 == list.get(j)) {
polindromeCount++;
} else {
char[] temp = new char[(list.get(j) - list.get(i))
+ 1];
int jj = 0;
for (int ii = list.get(i); ii <= list
.get(j); ii++) {
temp[jj] = charArray[ii];
jj++;
}
if (isPolindrome(temp))
polindromeCount++;
}
}
}
}
});
}
private static boolean isPolindrome(char[] charArray) {
int n = charArray.length;
char[] temp = new char[n];
int j = 0;
for (int i = (n - 1); i >= 0; i--) {
temp[j] = charArray[i];
j++;
}
if (Arrays.equals(charArray, temp))
return true;
else
return false;
}
public static void main(String[] args) {
String str = "MADAM";
char[] charArray = str.toCharArray();
countPolindromeByPositions(charArray, findCharAndOccurance(charArray));
System.out.println(polindromeCount);
}
}
Try out this. Its my own solution.
// Maintain an Set of palindromes so that we get distinct elements at the end
// Add each char to set. Also treat that char as middle point and traverse through string to check equality of left and right char
static int palindrome(String str) {
Set<String> distinctPln = new HashSet<String>();
for (int i=0; i<str.length();i++) {
distinctPln.add(String.valueOf(str.charAt(i)));
for (int j=i-1, k=i+1; j>=0 && k<str.length(); j--, k++) {
// String of lenght 2 as palindrome
if ( (new Character(str.charAt(i))).equals(new Character(str.charAt(j)))) {
distinctPln.add(str.substring(j,i+1));
}
// String of lenght 2 as palindrome
if ( (new Character(str.charAt(i))).equals(new Character(str.charAt(k)))) {
distinctPln.add(str.substring(i,k+1));
}
if ( (new Character(str.charAt(j))).equals(new Character(str.charAt(k)))) {
distinctPln.add(str.substring(j,k+1));
} else {
continue;
}
}
}
Iterator<String> distinctPlnItr = distinctPln.iterator();
while ( distinctPlnItr.hasNext()) {
System.out.print(distinctPlnItr.next()+ ",");
}
return distinctPln.size();
}
Code is to find all distinct substrings which are palindrome.
Here is the code I tried. It is working fine.
import java.util.HashSet;
import java.util.Set;
public class SubstringPalindrome {
public static void main(String[] args) {
String s = "abba";
checkPalindrome(s);
}
public static int checkPalindrome(String s) {
int L = s.length();
int counter =0;
long startTime = System.currentTimeMillis();
Set<String> hs = new HashSet<String>();
// add elements to the hash set
System.out.println("Possible substrings: ");
for (int i = 0; i < L; ++i) {
for (int j = 0; j < (L - i); ++j) {
String subs = s.substring(j, i + j + 1);
counter++;
System.out.println(subs);
if(isPalindrome(subs))
hs.add(subs);
}
}
System.out.println("Total possible substrings are "+counter);
System.out.println("Total palindromic substrings are "+hs.size());
System.out.println("Possible palindromic substrings: "+hs.toString());
long endTime = System.currentTimeMillis();
System.out.println("It took " + (endTime - startTime) + " milliseconds");
return hs.size();
}
public static boolean isPalindrome(String s) {
if(s.length() == 0 || s.length() ==1)
return true;
if(s.charAt(0) == s.charAt(s.length()-1))
return isPalindrome(s.substring(1, s.length()-1));
return false;
}
}
OUTPUT:
Possible substrings:
a
b
b
a
ab
bb
ba
abb
bba
abba
Total possible substrings are 10
Total palindromic substrings are 4
Possible palindromic substrings: [bb, a, b, abba]
It took 1 milliseconds
I recently participated in a competition where I was asked this question. Given an array with lengths what is the area of the biggest rectangle that can be made using ALL the lengths. The lengths can be added but not broken in between.
Example:
[ 4,2,4,4,6,8 ] given this array the best we can do is make a rectangle of sides 8 and 6 like this.
giving an area of 8 * 6 = 48.
I am a beginner and even after a long hard think about how to do it I am unable to get anywhere. I am not looking for a solution but any clue to nudge me in the right direction would be appreciated.
TIA
Edit: Somebody pointed out(comment deleted now) that its difficult to explain the solution with just hints and not posting some code. Kindly post code if necessary.
The problem is NP-Hard, thus the backtracking solution [or other exponential solution as suggested by #vhallac] will be your best shot, since there is not known [and if P!=NP, there is no existing] polynomial solution for this kind of problem.
NP-Hardness proof:
First, we know that a rectangle consists of 4 edges, that are equal in pairs [e1=e2,e3=e4].
We will show that if there is a polynomial algorithm A to this problem, we can also solve the Partition Problem, by the following algorithm:
input: a group of numbers S=a1,a2,...,an
output: true if and only if the numbers can be partitioned
algorithm:
sum <- a1 + a2 + .. + an
lengths <- a1, a2 , ... , an , (sum*5), (sum*5)
activate A with lengths.
if A answered there is any rectangle [solution is not 0], answer True
else answer False
Correctness:
(1) if there is a partition to S, let it be S1,S2, there is also a rectangle with edges: (sum*5),(sum*5),S1,S2, and the algorithm will yield True.
(2) if the algorithm yields True, there is a rectangle available in lengths, since a1 + a2 + ... + an < sum*5, there are 2 edges with length sum*5, since the 2 other edges must be made using all remaining lengths [as the question specified], each other edge is actually of length (a1 + a2 + ... + an)/2, and thus there is a legal partition to the problem.
Conclusion: There is a reduction PARTITION<=(p) this problem, and thus, this problem is NP-Hard
EDIT:
the backtracking solution is pretty simple, get all possible rectangles, and check each of them to see which is the best.
backtracking solution: pseudo-code:
getAllRectangles(S,e1,e2,e3,e4,sol):
if S == {}:
if legalRectangle(e1,e2,e3,e4):
sol.add((e1,e2,e3,e4))
else: //S is not empty
elem <- S[0]
getAllRectangles(S-elem,e1+elem,e2,e3,e4,sol)
getAllRectangles(S-elem,e1,e2+elem,e3,e4,sol)
getAllRectangles(S-elem,e1,e2,e3+elem,e4,sol)
getAllRectangles(S-elem,e1,e2,e3,e4+elem,sol)
getRectangle(S):
RECS <- new Set
getAllRectangles(S,{},{},{},{},RECS)
getBest(RECS)
EDIT2:
As discussed in the comments, this answer shows not only this is hard to find the BEST rectangle, it is also hard to find ANY rectangle, making this problem hard for heuristic solutions as well.
Here is one solution to the problem in Python. It is not optimized at all. I even check 2, 4 after I check 4,2, for example. But for showing how you can find a solution, I think it is good enough.
def all_but(lst, pos):
return lst[0:pos]+lst[pos+1:]
def find_sets_with_len(segs, l):
for i in range(0, len(segs)):
val = segs[i]
if (val == l):
yield [val], all_but(segs, i)
if (val < l):
for soln, rest in find_sets_with_len(all_but(segs, i), l - val):
yield [val]+soln, rest
def find_rect(segs, l1, l2):
for side1, rest1 in find_sets_with_len(segs, l1):
for side2, rest2 in find_sets_with_len(rest1, l1):
for side3, rest3 in find_sets_with_len(rest2, l2):
return [side1, side2, side3, rest3]
def make_rect(segs):
tot_len = sum(segs)
if (tot_len %2) == 0:
opt_len=tot_len/4
for l in range(opt_len, 0, -1):
sides = find_rect(segs, l, tot_len/2-l)
if sides is not None:
print(sides)
return sides
print("Can't find any solution")
make_rect([4,2,4,4,6,8])
The idea is simple: first, calculate the optimal length (that is, the length to make a square), then search everything starting off with the optimal length, and go down to 1 for one side. For each length, enumerate all sets for one side of the claculated length, then enumerate all sets for the opposite side (of the same length), then if I can find one more set of the remaining length (that is total_len/2 minus the side length I am looking at), then I've got the best solution. This happens in find_rect() function.
Well, I get little bit bored so play around with Java to have some experience, can be poorly coded and without tuning, as I am trying to increase my coding skill, comments are welcome. My computer able to answer me for small arrays:)
Output looks like:
Largest rectangle range is ; 48
-------------------------------------------------
input values; [ 4,2,4,4,6,8,9 ]
-------------------------------------------------
Array details of the rectangle:
A1: [ 6 ]
B1: [ 8 ]
A2: [ 2,4 ]
B2: [ 4,4 ]
combination.class using Kenneth algorithm;
import java.math.BigInteger;
public class Combination {
/**
* Burak
*/
private int[] a;
private int n;
private int r;
private BigInteger numLeft;
private BigInteger total;
public Combination (int n, int r) {
if (r > n) {
throw new IllegalArgumentException ();
}
if (n < 1) {
throw new IllegalArgumentException ();
}
this.n = n;
this.r = r;
a = new int[r];
BigInteger nFact = getFactorial (n);
BigInteger rFact = getFactorial (r);
BigInteger nminusrFact = getFactorial (n - r);
total = nFact.divide (rFact.multiply (nminusrFact));
reset ();
}
//------
// Reset
//------
public void reset () {
for (int i = 0; i < a.length; i++) {
a[i] = i;
}
numLeft = new BigInteger (total.toString ());
}
//------------------------------------------------
// Return number of combinations not yet generated
//------------------------------------------------
public BigInteger getNumLeft () {
return numLeft;
}
//-----------------------------
// Are there more combinations?
//-----------------------------
public boolean hasMore () {
return numLeft.compareTo (BigInteger.ZERO) == 1;
}
//------------------------------------
// Return total number of combinations
//------------------------------------
public BigInteger getTotal () {
return total;
}
//------------------
// Compute factorial
//------------------
private static BigInteger getFactorial (int n) {
BigInteger fact = BigInteger.ONE;
for (int i = n; i > 1; i--) {
fact = fact.multiply (new BigInteger (Integer.toString (i)));
}
return fact;
}
//--------------------------------------------------------
// Generate next combination (algorithm from Rosen p. 286)
//--------------------------------------------------------
public int[] getNext () {
if (numLeft.equals (total)) {
numLeft = numLeft.subtract (BigInteger.ONE);
return a;
}
int i = r - 1;
while (a[i] == n - r + i) {
i--;
}
a[i] = a[i] + 1;
for (int j = i + 1; j < r; j++) {
a[j] = a[i] + j - i;
}
numLeft = numLeft.subtract (BigInteger.ONE);
return a;
}
}
And main Combinator.class;
import java.util.*;
public class Combinator {
/**
* #param args
*/
private static int[] ad;
private static int[] bd;
private static String a1;
private static String a2;
private static String b1;
private static String b2;
private static int bestTotal =0;
public static void main(String[] args) {
int[] array={4,2,4,4,6,8,9};
getBestCombination(array, 1);
if(bestTotal <= 0){
System.out.println("System couldnt create any rectangle.");
}else{
System.out.println("Largest rectangle range is ; " + bestTotal);
System.out.println("-------------------------------------------------");
System.out.println("input values; " + parseArrayToString(array));
System.out.println("-------------------------------------------------");
System.out.println("Array details of the rectangle:");
System.out.println("A1: " + a1);
System.out.println("B1: " + b1);
System.out.println("A2: " + a2);
System.out.println("B2: " + b2);
}
}
private static void getBestCombination(int[] array, int level){
int[] a;
int[] b;
int bestPerimeter = getTotal(array,true);
Vector<Vector<Integer>> results = null;
for(int o=array.length-1;o>=1;o--){
for(int u=bestPerimeter;u>=1;u--){
results = Combinator.compute (array, o, u);
if(results.size() > 0){
for(int i=0;i<results.size();i++){
a = new int[results.elementAt(i).size()];
for(int j = 0;j<results.elementAt(i).size();j++){
a[j] = results.elementAt(i).elementAt(j);
}
b = removeItems(array, results.elementAt(i));
if(level == 1){
getBestCombination(a,2);
getBestCombination(b,3);
}else if(level == 2){
ad = a;
bd = b;
}else{
getBestCombination(a,4);
getBestCombination(b,4);
if(getTotal(ad, false) == getTotal(a, false) && getTotal(bd, false) == getTotal(b, false)){
if(bestTotal<(getTotal(ad, false)*getTotal(bd, false))){
bestTotal = getTotal(ad, false)*getTotal(bd, false);
a1 = parseArrayToString(ad);
a2 = parseArrayToString(a);
b1 = parseArrayToString(bd);
b2 = parseArrayToString(b);
}
}else if(getTotal(ad, false) == getTotal(b, false) && getTotal(bd, false) == getTotal(a, false)){
if(bestTotal<(getTotal(ad, false)*getTotal(bd, false))){
bestTotal = getTotal(ad, false)*getTotal(bd, false);
a1 = parseArrayToString(ad);
a2 = parseArrayToString(b);
b1 = parseArrayToString(bd);
b2 = parseArrayToString(a);
}
}
}
}
}
}
}
}
private static String parseArrayToString(int[] items){
String s = "[ ";
for(int i=0;i<items.length;i++){
if(i!=items.length-1){
s = s + items[i] + ",";
}else{
s = s + items[i];
}
}
s = s + " ]";
return s;
}
#SuppressWarnings("rawtypes")
private static int[] removeItems(int[] array, Vector items){
ArrayList<Integer> res = new ArrayList<Integer>();
for(int i=0;i<array.length;i++){
res.add(array[i]);
}
for(int u = 0;u<items.size();u++){
res.remove(items.elementAt(u));
}
int[] results = new int[res.size()];
for(int o=0;o<res.size();o++){
results[o] = res.get(o);
}
return results;
}
private static int getTotal(int[] array,boolean bestPerimeter){
int sum = 0;
for (int i = 0; i < array.length; i++) {
sum += array[i];
}
if(bestPerimeter == true){
if(sum%2!=0){
sum = sum -1;
}
sum = sum/2;
}
//System.out.println(sum);
return sum;
}
#SuppressWarnings("rawtypes")
private static int getSum (Vector v) {
int sum = 0;
Integer n;
for (int i = 0; i < v.size (); i++) {
n = (Integer) v.elementAt(i);
sum += n.intValue ();
}
return sum;
}
#SuppressWarnings({ "rawtypes", "unchecked" })
public static Vector<Vector<Integer>> compute (int[] array, int atATime, int desiredTotal) {
int[] indices;
Combination gen = new Combination (array.length, atATime);
Vector results = new Vector ();
Vector combination;
int sum;
Integer intObj;
while (gen.hasMore ()) {
combination = new Vector ();
indices = gen.getNext ();
for (int i = 0; i < indices.length; i++) {
intObj = new Integer (array[indices[i]]);
combination.addElement (intObj);
}
sum = getSum (combination);
if (sum == desiredTotal) {
Collections.sort (combination);
if (!results.contains (combination)) {
results.addElement (combination);
}
}
}
return results;
}
}
Find minimum window width in string x that contains all characters of another string y. For example:
String x = "coobdafceeaxab"
String y = "abc"
The answer should be 5, because the shortest substring in x that contains all three letters of y is "bdafc".
I can think of a naive solution with complexity O(n^2 * log(m)), where n = len(x) and m = len(y). Can anyone suggest a better solution? Thanks.
Update: now think of it, if I change my set to tr1::unordered_map, then I can cut the complexity down to O(n^2), because insertion and deletion should both be O(1).
time: O(n) (One pass)
space: O(k)
This is how I would do it:
Create a hash table for all the characters from string Y. (I assume all characters are different in Y).
First pass:
Start from first character of string X.
update hash table, for exa: for key 'a' enter location (say 1).
Keep on doing it until you get all characters from Y (until all key in hash table has value).
If you get some character again, update its newer value and erase older one.
Once you have first pass, take smallest value from hash table and biggest value.
Thats the minimum window observed so far.
Now, go to next character in string X, update hash table and see if you get smaller window.
Edit:
Lets take an example here:
String x = "coobdafceeaxab"
String y = "abc"
First initialize a hash table from characters of Y.
h[a] = -1
h[b] = -1
h[c] = -1
Now, Start from first character of X.
First character is c, h[c] = 0
Second character (o) is not part of hash, skip it.
..
Fourth character (b), h[b] = 3
..
Sixth character(a), enter hash table h[a] = 5.
Now, all keys from hash table has some value.
Smallest value is 0 (of c) and highest value is 5 (of a), minimum window so far is 6 (0 to 5).
First pass is done.
Take next character. f is not part of hash table, skip it.
Next character (c), update hash table h[c] = 7.
Find new window, smallest value is 3 (of b) and highest value is 7 (of c).
New window is 3 to 7 => 5.
Keep on doing it till last character of string X.
I hope its clear now.
Edit
There are some concerns about finding max and min value from hash.
We can maintain sorted Link-list and map it with hash table.
Whenever any element from Link list changes, it should be re-mapped to hash table.
Both these operation are O(1)
Total space would be m+m
Edit
Here is small visualisation of above problem:
For "coobdafceeaxab" and "abc"
step-0:
Initial doubly linked-list = NULL
Initial hash-table = NULL
step-1:
Head<->[c,0]<->tail
h[c] = [0, 'pointer to c node in LL']
step-2:
Head<->[c,0]<->[b,3]<->tail
h[c] = [0, 'pointer to c node in LL'], h[b] = [3, 'pointer to b node in LL'],
Step-3:
Head<->[c,0]<->[b,3]<->[a,5]<->tail
h[c] = [0, 'pointer to c node in LL'], h[b] = [3, 'pointer to b node in LL'], h[a] = [5, 'pointer to a node in LL']
Minimum Window => difference from tail and head => (5-0)+1 => Length: 6
Step-4:
Update entry of C to index 7 here. (Remove 'c' node from linked-list and append at the tail)
Head<->[b,3]<->[a,5]<->[c,7]<->tail
h[c] = [7, 'new pointer to c node in LL'], h[b] = [3, 'pointer to b node in LL'], h[a] = [5, 'pointer to a node in LL'],
Minimum Window => difference from tail and head => (7-3)+1 => Length: 5
And so on..
Note that above Linked-list update and hash table update are both O(1).
Please correct me if I am wrong..
Summary:
TIme complexity: O(n) with one pass
Space Complexity: O(k) where k is length of string Y
I found this very nice O(N) time complexity version here http://leetcode.com/2010/11/finding-minimum-window-in-s-which.html, and shortened it slightly (removed continue in a first while , which allowed to simplify condition for the second while loop). Note, that this solution allows for duplicates in the second string, while many of the above answers do not.
private static String minWindow(String s, String t) {
int[] needToFind = new int[256];
int[] hasFound = new int[256];
for(int i = 0; i < t.length(); ++i) {
needToFind[t.charAt(i)]++;
}
int count = 0;
int minWindowSize = Integer.MAX_VALUE;
int start = 0, end = -1;
String window = "";
while (++end < s.length()) {
char c = s.charAt(end);
if(++hasFound[c] <= needToFind[c]) {
count++;
}
if(count < t.length()) continue;
while (hasFound[s.charAt(start)] > needToFind[s.charAt(start)]) {
hasFound[s.charAt(start++)]--;
}
if(end - start + 1 < minWindowSize) {
minWindowSize = end - start + 1;
window = s.substring(start, end + 1);
}
}
return window;
}
Here's my solution in C++:
int min_width(const string& x, const set<char>& y) {
vector<int> at;
for (int i = 0; i < x.length(); i++)
if (y.count(x[i]) > 0)
at.push_back(i);
int ret = x.size();
int start = 0;
map<char, int> count;
for (int end = 0; end < at.size(); end++) {
count[x[at[end]]]++;
while (count[x[at[start]]] > 1)
count[x[at[start++]]]--;
if (count.size() == y.size() && ret > at[end] - at[start] + 1)
ret = at[end] - at[start] + 1;
}
return ret;
}
Edit: Here's an implementation of Jack's idea. It's the same time complexity as mine, but without the inner loop that confuses you.
int min_width(const string& x, const set<char>& y) {
int ret = x.size();
map<char, int> index;
set<int> index_set;
for (int j = 0; j < x.size(); j++) {
if (y.count(x[j]) > 0) {
if (index.count(x[j]) > 0)
index_set.erase(index[x[j]]);
index_set.insert(j);
index[x[j]] = j;
if (index.size() == y.size()) {
int i = *index_set.begin();
if (ret > j-i+1)
ret = j-i+1;
}
}
}
return ret;
}
In Java it can be implemented nicely with LinkedHashMap:
static int minWidth(String x, HashSet<Character> y) {
int ret = x.length();
Map<Character, Integer> index = new LinkedHashMap<Character, Integer>();
for (int j = 0; j < x.length(); j++) {
char ch = x.charAt(j);
if (y.contains(ch)) {
index.remove(ch);
index.put(ch, j);
if (index.size() == y.size()) {
int i = index.values().iterator().next();
if (ret > j - i + 1)
ret = j - i + 1;
}
}
}
return ret;
}
All operations inside the loop take constant time (assuming hashed elements disperse properly).
There is an O(n solution to this problem). It very well described in this article.
http://www.leetcode.com/2010/11/finding-minimum-window-in-s-which.html
Hope it helps.
This is my solution in C++, just for reference.
Update: originally I used std::set, now I change it to tr1::unordered_map to cut complexity down to n^2, otherwise these two implementations look pretty similar, to prevent this post from getting too long, I only list the improved solution.
#include <iostream>
#include <tr1/unordered_map>
#include <string>
using namespace std;
using namespace std::tr1;
typedef tr1::unordered_map<char, int> hash_t;
// Returns min substring width in which sentence contains all chars in word
// Returns sentence's length + 1 if not found
size_t get_min_width(const string &sent, const string &word) {
size_t min_size = sent.size() + 1;
hash_t char_set; // char set that word contains
for (size_t i = 0; i < word.size(); i++) {
char_set.insert(hash_t::value_type(word[i], 1));
}
for (size_t i = 0; i < sent.size() - word.size(); i++) {
hash_t s = char_set;
for (size_t j = i; j < min(j + min_size, sent.size()); j++) {
s.erase(sent[j]);
if (s.empty()) {
size_t size = j - i + 1;
if (size < min_size) min_size = size;
break;
}
}
}
return min_size;
}
int main() {
const string x = "coobdafceeaxab";
const string y = "abc";
cout << get_min_width(x, y) << "\n";
}
An implementation of Jack's idea.
public int smallestWindow(String str1, String str2){
if(str1==null || str2==null){
throw new IllegalArgumentException();
}
Map<String, Node> map=new HashMap<String, Node>();
Node head=null, current=null;
for(int i=0;i<str1.length();i++){
char c=str1.charAt(i);
if(head==null){
head=new Node(c);
current=head;
map.put(String.valueOf(c), head);
}
else{
current.next=new Node(c);
current.next.pre=current;
current=current.next;
map.put(String.valueOf(c), current);
}
}
Node end=current;
int min=Integer.MAX_VALUE;
int count=0;
for(int i=0;i<str2.length();i++){
char c = str2.charAt(i);
Node n=map.get(String.valueOf(c));
if(n!=null){
if(n.index==Integer.MAX_VALUE){
count++;
}
n.index=i;
if(n==head){
Node temp=head;
head=head.next;
if(head==null){//one node
return 1;
}
head.pre=null;
temp.pre=end;
end.next=temp;
temp.next=null;
end=temp;
}
else if(end!=n){
n.pre.next=n.next;
n.next.pre=n.pre;
n.pre=end;
n.next=null;
end.next=n;
end=n;
}
if(count==str1.length()){
min=Math.min(end.index-head.index+1, min);
}
}
}
System.out.println(map);
return min;
}
Simple java solution using the sliding window. Extending NitishMD's idea above:
public class StringSearchDemo {
public String getSmallestSubsetOfStringContaingSearchString(String toMatch,
String inputString) {
if (inputString.isEmpty() || toMatch.isEmpty()) {
return null;
}
// List<String> results = new ArrayList<String>(); // optional you can comment this out
String smallestMatch = "";
// String largestMatch = "";
int startPointer = 0, endPointer = 1;
HashMap<Character, Integer> toMatchMap = new HashMap<Character, Integer>();
for (char c : toMatch.toCharArray()) {
if (toMatchMap.containsKey(c)) {
toMatchMap.put(c, (toMatchMap.get(c) + 1));
} else {
toMatchMap.put(c, 1);
}
}
int totalCount = getCountofMatchingString(toMatchMap, toMatch);
for (int i = 0; i < inputString.length();) {
if (!toMatchMap.containsKey(inputString.charAt(i))) {
endPointer++;
i++;
continue;
}
String currentSubString = inputString.substring(startPointer,
endPointer);
if (getCountofMatchingString(toMatchMap, currentSubString) >= totalCount) {
// results.add(currentSubString); // optional you can comment this out
if (smallestMatch.length() > currentSubString.length()) {
smallestMatch = currentSubString;
} else if (smallestMatch.isEmpty()) {
smallestMatch = currentSubString;
}
// if (largestMatch.length() < currentSubString.length()) {
// largestMatch = currentSubString;
// }
startPointer++;
} else {
endPointer++;
i++;
}
}
// System.out.println("all possible combinations = " + results); // optional, you can comment this out
// System.out.println("smallest result = " + smallestMatch);
// System.out.println("largest result = " + largestMatch);
return smallestMatch;
}
public int getCountofMatchingString(HashMap<Character, Integer> toMatchMap,
String toMatch) {
int match = 0;
HashMap<Character, Integer> localMap = new HashMap<Character, Integer>();
for (char c : toMatch.toCharArray()) {
if (toMatchMap.containsKey(c)) {
if (localMap.containsKey(c)) {
if (localMap.get(c) < toMatchMap.get(c)) {
localMap.put(c, (localMap.get(c) + 1));
match++;
}
} else {
localMap.put(c, 1);
match++;
}
}
}
return match;
}
public static void main(String[] args) {
String inputString = "zxaddbddxyy由ccbbwwaay漢字由来";
String matchCriteria = "a由";
System.out.println("input=" + matchCriteria);
System.out.println("matchCriteria=" + inputString);
String result = (new StringSearchDemo())
.getSmallestSubsetOfStringContaingSearchString(matchCriteria, inputString);
System.out.println("smallest possbile match = " + result);
}
}