Related
I was practicing for an interview and came across this question on a website:
A magical sub-sequence of a string S is a sub-sequence of S that
contains all five vowels in order. Find the length of largest magical sub-sequence of a string S.
For example, if S = aeeiooua, then aeiou and aeeioou are magical sub-sequences
but aeio and aeeioua are not.
I am a beginner in dynamic programming and am finding it hard to come up with a recursive formula for this.
I did it with an iterative approach rather than recursive one. I started building solution similar to LIS (Longest Increasing Subsequence) and then optimised it upto O(n).
#include<iostream>
#include<string>
#include<vector>
using namespace std;
string vowel = "aeiou";
int vpos(char c)
{
for (int i = 0; i < 5; ++i)
if (c == vowel[i])
return i;
return -1;
}
int magical(string s)
{
int l = s.length();
int previndex[5] = {-1, -1, -1, -1, -1}; // for each vowel
vector<int> len (l, 0);
int i = 0, maxlen = 0;
// finding first 'a'
while (s[i] != 'a')
{
++i;
if (i == l)
return 0;
}
previndex[0] = i; //prev index of 'a'
len[i] = 1;
for ( ++i; i < l; ++i)
{
if (vpos(s[i]) >= 0) // a vowel
{
/* Need to append to longest subsequence on its left, only for this vowel (for any vowels) and
* its previous vowel (if it is not 'a')
This important observation makes it O(n) -- differnet from typical LIS
*/
if (previndex[vpos(s[i])] >= 0)
len[i] = 1+len[previndex[vpos(s[i])]];
previndex[vpos(s[i])] = i;
if (s[i] != 'a')
{
if (previndex[vpos(s[i])-1] >= 0)
len[i] = max(len[i], 1+len[previndex[vpos(s[i])-1]]);
}
maxlen = max(maxlen, len[i]);
}
}
return maxlen;
}
int main()
{
string s = "aaejkioou";
cout << magical(s);
return 0;
}
O(input string length) runtime
import java.util.*;
public class Main {
/*
algo:
keep map of runningLongestSubsequence that ends in each letter. loop through String s. for each char, try appending
to runningLongestSubsequence for that char, as well as to runningLongestSubsequence for preceding char.
update map with whichever results in longer subsequence.
for String s = "ieaeiouiaooeeeaaeiou", final map is:
terminal letter in longest running subsequence-> longest running subsequence
a -> aaaa
e -> aeeeee
i -> aeeeeei
o -> aeeeeeio
u -> aeeeeeiou
naming:
precCharMap - precedingCharMap
runningLongestSubMap - runningLongestSubsequenceMap
*/
public static int longestSubsequence(String s) {
if (s.length() <= 0) throw new IllegalArgumentException();
Map<Character, Character> precCharMap = new HashMap<>();
precCharMap.put('u', 'o');
precCharMap.put('o', 'i');
precCharMap.put('i', 'e');
precCharMap.put('e', 'a');
Map<Character, String> runningLongestSubMap = new HashMap<>();
for (char currChar : s.toCharArray()) {
//get longest subs
String currCharLongestSub;
String precCharLongestSub = null;
if (currChar == 'a') {
currCharLongestSub = runningLongestSubMap.getOrDefault(currChar, "");
} else {
currCharLongestSub = runningLongestSubMap.get(currChar);
char precChar = precCharMap.get(currChar);
precCharLongestSub = runningLongestSubMap.get(precChar);
}
//update running longest subsequence map
if (precCharLongestSub == null && currCharLongestSub != null) {
updateRunningLongestSubMap(currCharLongestSub, currChar, runningLongestSubMap);
} else if (currCharLongestSub == null && precCharLongestSub != null) {
updateRunningLongestSubMap(precCharLongestSub, currChar, runningLongestSubMap);
} else if (currCharLongestSub != null && precCharLongestSub != null) {
//pick longer
if (currCharLongestSub.length() < precCharLongestSub.length()) {
updateRunningLongestSubMap(precCharLongestSub, currChar, runningLongestSubMap);
} else {
updateRunningLongestSubMap(currCharLongestSub, currChar, runningLongestSubMap);
}
}
}
if (runningLongestSubMap.get('u') == null) {
return 0;
}
return runningLongestSubMap.get('u').length();
}
private static void updateRunningLongestSubMap(String longestSub, char currChar,
Map<Character, String> runningLongestSubMap) {
String currCharLongestSub = longestSub + currChar;
runningLongestSubMap.put(currChar, currCharLongestSub);
}
public static void main(String[] args) {
//String s = "aeeiooua"; //7
//String s = "aeiaaioooaauuaeiou"; //10
String s = "ieaeiouiaooeeeaaeiou"; //9
//String s = "ieaeou"; //0
//String s = "ieaeoooo"; //0
//String s = "aeiou"; //5
//if u have String s beginning in "ao", it'll do nothing with o and
//continue on to index 2.
System.out.println(longestSubsequence(s));
}
}
#include <iostream>
#include<string>
#include<cstring>
using namespace std;
unsigned int getcount(string a, unsigned int l,unsigned int r );
int main()
{
std::string a("aaaaaeeeeaaaaiiioooeeeeuuuuuuiiiiiaaaaaaoo"
"oooeeeeiiioooouuuu");
//std::string a("aaaaaeeeeaaaaiiioooeeeeuuuuuuiiiiiaaaaaaoooooeeeeiiioooo");
//std::string a("aaaaaeeeeaaaaiiioooeeeeiiiiiaaaaaaoooooeeeeiiioooo"); //sol0
//std::string a{"aeiou"};
unsigned int len = a.length();
unsigned int i=0,cnt =0,countmax =0;
bool newstring = true;
while(i<len)
{
if(a.at(i) == 'a' && newstring == true)
{
newstring = false;
cnt = getcount(a,i,len);
if(cnt > countmax)
{
countmax = cnt;
cnt = 0;
}
}
else if(a.at(i)!='a')
{
newstring = true;
}
i++;
}
cout<<countmax;
return 0;
}
unsigned int getcount(string a, unsigned int l,unsigned int r )
{
std::string b("aeiou");
unsigned int seq=0,cnt =0;
unsigned int current =l;
bool compstr = false;
while(current<r)
{
if(a.at(current) == b.at(seq))
{
cnt++;
}
else if((seq <= (b.size()-2)) && (a.at(current) == b.at(seq+1)))
{
seq++;
cnt++;
if (seq == 4)
compstr =true;
}
current++;
}
if (compstr == true)
return cnt;
return 0;
}
you can use recursive approach here (this should work for string length upto max int (easily memorization can be used)
public class LMV {
static final int NOT_POSSIBLE = -1000000000;
// if out put is this i.e soln not possible
static int longestSubsequence(String s, char[] c) {
//exit conditions
if(s.length() ==0 || c.length ==0){
return 0;
}
if(s.length() < c.length){
return NOT_POSSIBLE;
}
if(s.length() == c.length){
for(int i=0; i<s.length(); i++){
if(s.charAt(i) !=c [i]){
return NOT_POSSIBLE;
}
}
return s.length();
}
if(s.charAt(0) < c[0]){
// ignore, go ahead with next item
return longestSubsequence(s.substring(1), c);
} else if (s.charAt(0) == c[0]){
// <case 1> include item and start search for next item in chars
// <case 2> include but search for same item again in chars
// <case 3> don't include item
return Math.max(
Math.max( ( 1+longestSubsequence(s.substring(1), Arrays.copyOfRange(c, 1, c.length) ) ),
( 1+longestSubsequence(s.substring(1), c ) ) ),
( longestSubsequence(s.substring(1), c )) );
} else {
//ignore
return longestSubsequence(s.substring(1), c);
}
}
public static void main(String[] args) {
char[] chars = {'a', 'e', 'i', 'o', 'u'};
String s1 = "aeio";
String s2 = "aaeeieou";
String s3 = "aaeeeieiioiiouu";
System.out.println(longestSubsequence(s1, chars));
System.out.println(longestSubsequence(s2, chars));
System.out.println(longestSubsequence(s3, chars));
}
}
int func( char *p)
{
char *temp = p;
char ae[] = {'a','e','i','o','u'};
int size = strlen(p), i = 0;
int chari = 0, count_aeiou=0;
for (i=0;i<=size; i++){
if (temp[i] == ae[chari]) {
count_aeiou++;
}
else if ( temp[i] == ae[chari+1]) {
count_aeiou++;
chari++;
}
}
if (chari == 4 ) {
printf ("Final count : %d ", count_aeiou);
} else {
count_aeiou = 0;
}
return count_aeiou;
}
The solution to retrun the VOWELS count as per the hackerrank challenge.
int findsubwithcontinuousvowel(string str){
int curr=0;
int start=0,len=0,maxlen=0,i=0;
for(i=0;i<str.size();i++){
if(str[i]=='u' && (current[curr]=='u' || (curr+1<5 && current[curr+1]=='u'))){
//len++;
maxlen=max(len+1,maxlen);
}
if(str[i]==current[curr]){
len++;
}
else if(curr+1<5 && str[i]==current[curr+1]){
len++;
curr++;
}
else{
len=0;
curr=0;
if(str[i]=='a'){
len=1;
}
}
}
return maxlen;
}
Check if vowels are available in sequence in isInSequence and process the result on processor.
public class one {
private char[] chars = {'a','e','i','o','u'};
private int a = 0;
private boolean isInSequence(char c){
// check if char is repeating
if (c == chars[a]){
return true;
}
// if vowels are in sequence and just passed by 'a' and so on...
if (c == 'e' && a == 0){
a++;
return true;
}
if (c == 'i' && a == 1){
a++;
return true;
}
if (c == 'o' && a == 2){
a++;
return true;
}
if (c == 'u' && a == 3){
a++;
return true;
}
return false;
}
private char[] processor(char[] arr){
int length = arr.length-1;
int start = 0;
// In case if all chars are vowels, keeping length == arr
char array[] = new char[length];
for (char a : arr){
if (isInSequence(a)){
array[start] = a;
start++;
}
}
return array;
}
public static void main(String args[]){
char[] arr = {'m','a','e','l','x','o','i','o','u','a'};
one o = new one();
System.out.print(o.processor(arr));
}
}
#include <bits/stdc++.h>
#define ios ios::sync_with_stdio(NULL);cin.tie(NULL);cout.tie(NULL);
#define ll unsigned long long
using namespace std;
int main() {
// your code goes here
ios
string s;
cin>>s;
int n=s.length();
int dp[n+1][5]={0};
for(int i=1;i<=n;i++)
{
if(s[i-1]=='a')
{
dp[i][0]=1+dp[i-1][0];
dp[i][1]=dp[i-1][1];
dp[i][2]=dp[i-1][2];
dp[i][3]=dp[i-1][3];
dp[i][4]=dp[i-1][4];
}
else if(s[i-1]=='e')
{dp[i][0]=dp[i-1][0];
if(dp[i-1][0]>0)
{dp[i][1]=1+max(dp[i-1][1],dp[i-1][0]);}
else
dp[i-1][1]=0;
dp[i][2]=dp[i-1][2];
dp[i][3]=dp[i-1][3];
dp[i][4]=dp[i-1][4];
}
else if(s[i-1]=='i')
{dp[i][0]=dp[i-1][0];
if(dp[i-1][1]>0)
{dp[i][2]=1+max(dp[i-1][1],dp[i-1][2]);}
else
dp[i-1][2]=0;
dp[i][1]=dp[i-1][1];
dp[i][3]=dp[i-1][3];
dp[i][4]=dp[i-1][4];
}
else if(s[i-1]=='o')
{dp[i][0]=dp[i-1][0];
if(dp[i-1][2]>0)
{dp[i][3]=1+max(dp[i-1][3],dp[i-1][2]);}
else
dp[i-1][3]=0;
dp[i][2]=dp[i-1][2];
dp[i][1]=dp[i-1][1];
dp[i][4]=dp[i-1][4];
}
else if(s[i-1]=='u')
{dp[i][0]=dp[i-1][0];
if(dp[i-1][3]>0)
{dp[i][4]=1+max(dp[i-1][4],dp[i-1][3]);}
else
dp[i-1][4]=0;
dp[i][1]=dp[i-1][1];
dp[i][3]=dp[i-1][3];
dp[i][2]=dp[i-1][2];
}
else
{
dp[i][0]=dp[i-1][0];
dp[i][1]=dp[i-1][1];
dp[i][2]=dp[i-1][2];
dp[i][3]=dp[i-1][3];
dp[i][4]=dp[i-1][4];
}
}
cout<<dp[n][4];
return 0;
}
Assume length of element in table is 1 or 2.
Table: { h, fe, na, o}
input string: nafeo
Output: true
Table: {ab,bc}
input string: abc
Output: false
Please advise my below code will cover all the cases and is this the best solution? Or am I missing anything, any alternate solutions?
import java.util.*;
public class CustomTable {
Set<String> table = new HashSet<String>();
public CustomTable(){
// add your elements here for more test cases
table.add("oh");
table.add("he");
}
public int checkTable( String prev, String curr, String next) {
System.out.print(prev+":"+curr+":"+next);
System.out.println();
if (prev!=null) if (table.contains(prev)) return -1;
if (table.contains(curr)) return 0;
if (table.contains(next)) return 1;
return 2;
}
// ohhe.
public static void main(String args[]) {
CustomTable obj = new CustomTable();
String inputStr = "ohheo"; //Tested ohe,ohhe,ohohe
int result = 0;
String curr, prev, next;
for (int i = 0; i < inputStr.length(); i++) {
// if prev element is found
if (result==-1){
prev = null;
}
else {
if (i > 0) {
prev = inputStr.substring(i - 1, i + 1);
} else {
prev = inputStr.substring(i, i + 1);
}
}
curr = inputStr.substring(i,i+1);
if (i < inputStr.length()-1) {
next = inputStr.substring(i, i+2);
} else {
next = inputStr.substring(i, i+1);
}
result = obj.checkTable(prev, curr, next);
if (result==2) {
System.out.print("false");
return;
}
}
System.out.print("true");
}
}
I think the problem have similarities to well known subset sum problem and you can use its solutions by some customization.
Generate Parentheses Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
For example, given n = 3, a solution set is:
"((()))", "(()())", "(())()", "()(())", "()()()"
Personally, I think
time complexity
= O(n! ) (not including time copying tmpStr), n is the input,
= O(n * n!) ( including time copying tmpStr).
space complexity
= O(n) (stack space usage),
= O(n) (stack + recursion space usage). Code: Java
import java.util.List;
import java.util.ArrayList;
import java.util.Stack;
public class Solution {
public List<String> generateParenthesis(int n) {
List<String> list = new ArrayList<String>();
// Input checking.
if (n <= 0) {
list.add("");
return list;
}
String tmpStr = "";
for (int i = 0; i < n; i ++) tmpStr += "(";
helper(n, tmpStr, 0, list);
return list;
}
private void helper(int n, String tmpStr, int start, List<String> list) {
// Base case.
if (tmpStr.length() == 2 * n) {
if (isValid(tmpStr)) list.add(tmpStr);
return;
}
for (int i = start; i < tmpStr.length(); i ++) {
// Have a try.
tmpStr = tmpStr.substring(0, i + 1) + ")" +
tmpStr.substring(i + 1, tmpStr.length());
// Do recursion.
helper(n, tmpStr, i + 1, list);
// Roll back.
tmpStr = tmpStr.substring(0, i + 1) +
tmpStr.substring(i + 2, tmpStr.length());
}
}
private boolean isValid(String str) {
// Input checking.
if (str == null || str.length() < 2) return false;
Stack<Character> stack = new Stack<Character>();
for (int i = 0; i < str.length(); i ++) {
char curr = str.charAt(i);
if (curr == '(') stack.push(curr);
else {
if (stack.isEmpty()) return false;
stack.pop();
}
}
if (stack.isEmpty()) return true;
else return false;
}
}
If you could do a bit alteration in your validation method you can save a significant execution time for large value of n. You can use a counter instead of a stack to check for validation.
something like this
private boolean isValid(String str)
{
// Input checking.
if (str == null || str.length() < 2)
{
return false;
}
int ct = 0;
for (int i = 0; i < str.length(); i++)
{
char curr = str.charAt(i);
if (ct < 0)
{
return false;
}
if (curr == '(')
{
ct++;
}
else if (curr == ')')
{
ct--;
}
}
if (ct != 0 || ct < 0)
{
return false;
}
else
{
return true;
}
}
I ran it on my machine and the time saving for n=13 is around 2 sec and with this method in place the total execution time for the algorithm is less than 2 sec.
For n=15 time saving is around 12 sec.
This will not only save time but will also save significant amount of memory.
What would be the best approach (performance-wise) in solving this problem?
I was recommended to use suffix trees. Is this the best approach?
Check out this link: http://introcs.cs.princeton.edu/java/42sort/LRS.java.html
/*************************************************************************
* Compilation: javac LRS.java
* Execution: java LRS < file.txt
* Dependencies: StdIn.java
*
* Reads a text corpus from stdin, replaces all consecutive blocks of
* whitespace with a single space, and then computes the longest
* repeated substring in that corpus. Suffix sorts the corpus using
* the system sort, then finds the longest repeated substring among
* consecutive suffixes in the sorted order.
*
* % java LRS < mobydick.txt
* ',- Such a funny, sporty, gamy, jesty, joky, hoky-poky lad, is the Ocean, oh! Th'
*
* % java LRS
* aaaaaaaaa
* 'aaaaaaaa'
*
* % java LRS
* abcdefg
* ''
*
*************************************************************************/
import java.util.Arrays;
public class LRS {
// return the longest common prefix of s and t
public static String lcp(String s, String t) {
int n = Math.min(s.length(), t.length());
for (int i = 0; i < n; i++) {
if (s.charAt(i) != t.charAt(i))
return s.substring(0, i);
}
return s.substring(0, n);
}
// return the longest repeated string in s
public static String lrs(String s) {
// form the N suffixes
int N = s.length();
String[] suffixes = new String[N];
for (int i = 0; i < N; i++) {
suffixes[i] = s.substring(i, N);
}
// sort them
Arrays.sort(suffixes);
// find longest repeated substring by comparing adjacent sorted suffixes
String lrs = "";
for (int i = 0; i < N - 1; i++) {
String x = lcp(suffixes[i], suffixes[i+1]);
if (x.length() > lrs.length())
lrs = x;
}
return lrs;
}
// read in text, replacing all consecutive whitespace with a single space
// then compute longest repeated substring
public static void main(String[] args) {
String s = StdIn.readAll();
s = s.replaceAll("\\s+", " ");
StdOut.println("'" + lrs(s) + "'");
}
}
Have a look at http://en.wikipedia.org/wiki/Suffix_array as well - they are quite space-efficient and have some reasonably programmable algorithms to produce them, such as "Simple Linear Work Suffix Array Construction" by Karkkainen and Sanders
Here is a simple implementation of longest repeated substring using simplest suffix tree. Suffix tree is very easy to implement in this way.
#include <iostream>
#include <vector>
#include <unordered_map>
#include <string>
using namespace std;
class Node
{
public:
char ch;
unordered_map<char, Node*> children;
vector<int> indexes; //store the indexes of the substring from where it starts
Node(char c):ch(c){}
};
int maxLen = 0;
string maxStr = "";
void insertInSuffixTree(Node* root, string str, int index, string originalSuffix, int level=0)
{
root->indexes.push_back(index);
// it is repeated and length is greater than maxLen
// then store the substring
if(root->indexes.size() > 1 && maxLen < level)
{
maxLen = level;
maxStr = originalSuffix.substr(0, level);
}
if(str.empty()) return;
Node* child;
if(root->children.count(str[0]) == 0) {
child = new Node(str[0]);
root->children[str[0]] = child;
} else {
child = root->children[str[0]];
}
insertInSuffixTree(child, str.substr(1), index, originalSuffix, level+1);
}
int main()
{
string str = "banana"; //"abcabcaacb"; //"banana"; //"mississippi";
Node* root = new Node('#');
//insert all substring in suffix tree
for(int i=0; i<str.size(); i++){
string s = str.substr(i);
insertInSuffixTree(root, s, i, s);
}
cout << maxLen << "->" << maxStr << endl;
return 1;
}
/*
s = "mississippi", return "issi"
s = "banana", return "ana"
s = "abcabcaacb", return "abca"
s = "aababa", return "aba"
*/
the LRS problem is one that is best solved using either a suffix tree or a suffix array. Both approaches have a best time complexity of O(n).
Here is an O(nlog(n)) solution to the LRS problem using a suffix array. My solution can be improved to O(n) if you have a linear construction time algorithm for the suffix array (which is quite hard to implement). The code was taken from my library. If you want more information on how suffix arrays work make sure to check out my tutorials
/**
* Finds the longest repeated substring(s) of a string.
*
* Time complexity: O(nlogn), bounded by suffix array construction
*
* #author William Fiset, william.alexandre.fiset#gmail.com
**/
import java.util.*;
public class LongestRepeatedSubstring {
// Example usage
public static void main(String[] args) {
String str = "ABC$BCA$CAB";
SuffixArray sa = new SuffixArray(str);
System.out.printf("LRS(s) of %s is/are: %s\n", str, sa.lrs());
str = "aaaaa";
sa = new SuffixArray(str);
System.out.printf("LRS(s) of %s is/are: %s\n", str, sa.lrs());
str = "abcde";
sa = new SuffixArray(str);
System.out.printf("LRS(s) of %s is/are: %s\n", str, sa.lrs());
}
}
class SuffixArray {
// ALPHABET_SZ is the default alphabet size, this may need to be much larger
int ALPHABET_SZ = 256, N;
int[] T, lcp, sa, sa2, rank, tmp, c;
public SuffixArray(String str) {
this(toIntArray(str));
}
private static int[] toIntArray(String s) {
int[] text = new int[s.length()];
for(int i=0;i<s.length();i++)text[i] = s.charAt(i);
return text;
}
// Designated constructor
public SuffixArray(int[] text) {
T = text;
N = text.length;
sa = new int[N];
sa2 = new int[N];
rank = new int[N];
c = new int[Math.max(ALPHABET_SZ, N)];
construct();
kasai();
}
private void construct() {
int i, p, r;
for (i=0; i<N; ++i) c[rank[i] = T[i]]++;
for (i=1; i<ALPHABET_SZ; ++i) c[i] += c[i-1];
for (i=N-1; i>=0; --i) sa[--c[T[i]]] = i;
for (p=1; p<N; p <<= 1) {
for (r=0, i=N-p; i<N; ++i) sa2[r++] = i;
for (i=0; i<N; ++i) if (sa[i] >= p) sa2[r++] = sa[i] - p;
Arrays.fill(c, 0, ALPHABET_SZ, 0);
for (i=0; i<N; ++i) c[rank[i]]++;
for (i=1; i<ALPHABET_SZ; ++i) c[i] += c[i-1];
for (i=N-1; i>=0; --i) sa[--c[rank[sa2[i]]]] = sa2[i];
for (sa2[sa[0]] = r = 0, i=1; i<N; ++i) {
if (!(rank[sa[i-1]] == rank[sa[i]] &&
sa[i-1]+p < N && sa[i]+p < N &&
rank[sa[i-1]+p] == rank[sa[i]+p])) r++;
sa2[sa[i]] = r;
} tmp = rank; rank = sa2; sa2 = tmp;
if (r == N-1) break; ALPHABET_SZ = r + 1;
}
}
// Use Kasai algorithm to build LCP array
private void kasai() {
lcp = new int[N];
int [] inv = new int[N];
for (int i = 0; i < N; i++) inv[sa[i]] = i;
for (int i = 0, len = 0; i < N; i++) {
if (inv[i] > 0) {
int k = sa[inv[i]-1];
while( (i + len < N) && (k + len < N) && T[i+len] == T[k+len] ) len++;
lcp[inv[i]-1] = len;
if (len > 0) len--;
}
}
}
// Finds the LRS(s) (Longest Repeated Substring) that occurs in a string.
// Traditionally we are only interested in substrings that appear at
// least twice, so this method returns an empty set if this is not the case.
// #return an ordered set of longest repeated substrings
public TreeSet <String> lrs() {
int max_len = 0;
TreeSet <String> lrss = new TreeSet<>();
for (int i = 0; i < N; i++) {
if (lcp[i] > 0 && lcp[i] >= max_len) {
// We found a longer LRS
if ( lcp[i] > max_len )
lrss.clear();
// Append substring to the list and update max
max_len = lcp[i];
lrss.add( new String(T, sa[i], max_len) );
}
}
return lrss;
}
public void display() {
System.out.printf("-----i-----SA-----LCP---Suffix\n");
for(int i = 0; i < N; i++) {
int suffixLen = N - sa[i];
String suffix = new String(T, sa[i], suffixLen);
System.out.printf("% 7d % 7d % 7d %s\n", i, sa[i],lcp[i], suffix );
}
}
}
public class LongestSubString {
public static void main(String[] args) {
String s = findMaxRepeatedString("ssssssssssss this is a ddddddd word with iiiiiiiiiis and loads of these are ppppppppppppps");
System.out.println(s);
}
private static String findMaxRepeatedString(String s) {
Processor p = new Processor();
char[] c = s.toCharArray();
for (char ch : c) {
p.process(ch);
}
System.out.println(p.bigger());
return new String(new char[p.bigger().count]).replace('\0', p.bigger().letter);
}
static class CharSet {
int count;
Character letter;
boolean isLastPush;
boolean assign(char c) {
if (letter == null) {
count++;
letter = c;
isLastPush = true;
return true;
}
return false;
}
void reassign(char c) {
count = 1;
letter = c;
isLastPush = true;
}
boolean push(char c) {
if (isLastPush && letter == c) {
count++;
return true;
}
return false;
}
#Override
public String toString() {
return "CharSet [count=" + count + ", letter=" + letter + "]";
}
}
static class Processor {
Character previousLetter = null;
CharSet set1 = new CharSet();
CharSet set2 = new CharSet();
void process(char c) {
if ((set1.assign(c)) || set1.push(c)) {
set2.isLastPush = false;
} else if ((set2.assign(c)) || set2.push(c)) {
set1.isLastPush = false;
} else {
set1.isLastPush = set2.isLastPush = false;
smaller().reassign(c);
}
}
CharSet smaller() {
return set1.count < set2.count ? set1 : set2;
}
CharSet bigger() {
return set1.count < set2.count ? set2 : set1;
}
}
}
I had an interview and I needed to solve this problem. This is my solution:
public class FindLargestSubstring {
public static void main(String[] args) {
String test = "ATCGATCGA";
System.out.println(hasRepeatedSubString(test));
}
private static String hasRepeatedSubString(String string) {
Hashtable<String, Integer> hashtable = new Hashtable<>();
int length = string.length();
for (int subLength = length - 1; subLength > 1; subLength--) {
for (int i = 0; i <= length - subLength; i++) {
String sub = string.substring(i, subLength + i);
if (hashtable.containsKey(sub)) {
return sub;
} else {
hashtable.put(sub, subLength);
}
}
}
return "No repeated substring!";
}}
There are way too many things that affect performance for us to answer this question with only what you've given us. (Operating System, language, memory issues, the code itself)
If you're just looking for a mathematical analysis of the algorithm's efficiency, you probably want to change the question.
EDIT
When I mentioned "memory issues" and "the code" I didn't provide all the details. The length of the strings you will be analyzing are a BIG factor. Also, the code doesn't operate alone - it must sit inside a program to be useful. What are the characteristics of that program which impact this algorithm's use and performance?
Basically, you can't performance tune until you have a real situation to test. You can make very educated guesses about what is likely to perform best, but until you have real data and real code, you'll never be certain.
I have this string s1 = "My name is X Y Z" and I want to reverse the order of the words so that s1 = "Z Y X is name My".
I can do it using an additional array. I thought hard but is it possible to do it inplace (without using additional data structures) and with the time complexity being O(n)?
Reverse the entire string, then reverse the letters of each individual word.
After the first pass the string will be
s1 = "Z Y X si eman yM"
and after the second pass it will be
s1 = "Z Y X is name My"
reverse the string and then, in a second pass, reverse each word...
in c#, completely in-place without additional arrays:
static char[] ReverseAllWords(char[] in_text)
{
int lindex = 0;
int rindex = in_text.Length - 1;
if (rindex > 1)
{
//reverse complete phrase
in_text = ReverseString(in_text, 0, rindex);
//reverse each word in resultant reversed phrase
for (rindex = 0; rindex <= in_text.Length; rindex++)
{
if (rindex == in_text.Length || in_text[rindex] == ' ')
{
in_text = ReverseString(in_text, lindex, rindex - 1);
lindex = rindex + 1;
}
}
}
return in_text;
}
static char[] ReverseString(char[] intext, int lindex, int rindex)
{
char tempc;
while (lindex < rindex)
{
tempc = intext[lindex];
intext[lindex++] = intext[rindex];
intext[rindex--] = tempc;
}
return intext;
}
Not exactly in place, but anyway: Python:
>>> a = "These pretzels are making me thirsty"
>>> " ".join(a.split()[::-1])
'thirsty me making are pretzels These'
In Smalltalk:
'These pretzels are making me thirsty' subStrings reduce: [:a :b| b, ' ', a]
I know noone cares about Smalltalk, but it's so beautiful to me.
You cannot do the reversal without at least some extra data structure. I think the smallest structure would be a single character as a buffer while you swap letters. It can still be considered "in place", but it's not completely "extra data structure free".
Below is code implementing what Bill the Lizard describes:
string words = "this is a test";
// Reverse the entire string
for(int i = 0; i < strlen(words) / 2; ++i) {
char temp = words[i];
words[i] = words[strlen(words) - i];
words[strlen(words) - i] = temp;
}
// Reverse each word
for(int i = 0; i < strlen(words); ++i) {
int wordstart = -1;
int wordend = -1;
if(words[i] != ' ') {
wordstart = i;
for(int j = wordstart; j < strlen(words); ++j) {
if(words[j] == ' ') {
wordend = j - 1;
break;
}
}
if(wordend == -1)
wordend = strlen(words);
for(int j = wordstart ; j <= (wordend + wordstart) / 2 ; ++j) {
char temp = words[j];
words[j] = words[wordend - (j - wordstart)];
words[wordend - (j - wordstart)] = temp;
}
i = wordend;
}
}
What language?
If PHP, you can explode on space, then pass the result to array_reverse.
If its not PHP, you'll have to do something slightly more complex like:
words = aString.split(" ");
for (i = 0; i < words.length; i++) {
words[i] = words[words.length-i];
}
public static String ReverseString(String str)
{
int word_length = 0;
String result = "";
for (int i=0; i<str.Length; i++)
{
if (str[i] == ' ')
{
result = " " + result;
word_length = 0;
} else
{
result = result.Insert(word_length, str[i].ToString());
word_length++;
}
}
return result;
}
This is C# code.
In Python...
ip = "My name is X Y Z"
words = ip.split()
words.reverse()
print ' '.join(words)
Anyway cookamunga provided good inline solution using python!
This is assuming all words are separated by spaces:
#include <stdio.h>
#include <string.h>
int main()
{
char string[] = "What are you looking at";
int i, n = strlen(string);
int tail = n-1;
for(i=n-1;i>=0;i--)
{
if(string[i] == ' ' || i == 0)
{
int cursor = (i==0? i: i+1);
while(cursor <= tail)
printf("%c", string[cursor++]);
printf(" ");
tail = i-1;
}
}
return 0;
}
class Program
{
static void Main(string[] args)
{
string s1 =" My Name varma:;
string[] arr = s1.Split(' ');
Array.Reverse(arr);
string str = string.Join(" ", arr);
Console.WriteLine(str);
Console.ReadLine();
}
}
This is not perfect but it works for me right now. I don't know if it has O(n) running time btw (still studying it ^^) but it uses one additional array to fulfill the task.
It is probably not the best answer to your problem because i use a dest string to save the reversed version instead of replacing each words in the source string. The problem is that i use a local stack variable named buf to copy all the words in and i can not copy but into the source string as this would lead to a crash if the source string is const char * type.
But it was my first attempt to write s.th. like this :) Ok enough blablub. here is code:
#include <iostream>
using namespace std;
void reverse(char *des, char * const s);
int main (int argc, const char * argv[])
{
char* s = (char*)"reservered. rights All Saints. The 2011 (c) Copyright 11/10/11 on Pfundstein Markus by Created";
char *x = (char*)"Dogfish! White-spotted Shark, Bullhead";
printf("Before: |%s|\n", x);
printf("Before: |%s|\n", s);
char *d = (char*)malloc((strlen(s)+1)*sizeof(char));
char *i = (char*)malloc((strlen(x)+1)*sizeof(char));
reverse(d,s);
reverse(i,x);
printf("After: |%s|\n", i);
printf("After: |%s|\n", d);
free (i);
free (d);
return 0;
}
void reverse(char *dest, char *const s) {
// create a temporary pointer
if (strlen(s)==0) return;
unsigned long offset = strlen(s)+1;
char *buf = (char*)malloc((offset)*sizeof(char));
memset(buf, 0, offset);
char *p;
// iterate from end to begin and count how much words we have
for (unsigned long i = offset; i != 0; i--) {
p = s+i;
// if we discover a whitespace we know that we have a whole word
if (*p == ' ' || *p == '\0') {
// we increment the counter
if (*p != '\0') {
// we write the word into the buffer
++p;
int d = (int)(strlen(p)-strlen(buf));
strncat(buf, p, d);
strcat(buf, " ");
}
}
}
// copy the last word
p -= 1;
int d = (int)(strlen(p)-strlen(buf));
strncat(buf, p, d);
strcat(buf, "\0");
// copy stuff to destination string
for (int i = 0; i < offset; ++i) {
*(dest+i)=*(buf+i);
}
free(buf);
}
We can insert the string in a stack and when we extract the words, they will be in reverse order.
void ReverseWords(char Arr[])
{
std::stack<std::string> s;
char *str;
int length = strlen(Arr);
str = new char[length+1];
std::string ReversedArr;
str = strtok(Arr," ");
while(str!= NULL)
{
s.push(str);
str = strtok(NULL," ");
}
while(!s.empty())
{
ReversedArr = s.top();
cout << " " << ReversedArr;
s.pop();
}
}
This quick program works..not checks the corner cases though.
#include <stdio.h>
#include <stdlib.h>
struct node
{
char word[50];
struct node *next;
};
struct stack
{
struct node *top;
};
void print (struct stack *stk);
void func (struct stack **stk, char *str);
main()
{
struct stack *stk = NULL;
char string[500] = "the sun is yellow and the sky is blue";
printf("\n%s\n", string);
func (&stk, string);
print (stk);
}
void func (struct stack **stk, char *str)
{
char *p1 = str;
struct node *new = NULL, *list = NULL;
int i, j;
if (*stk == NULL)
{
*stk = (struct stack*)malloc(sizeof(struct stack));
if (*stk == NULL)
printf("\n####### stack is not allocated #####\n");
(*stk)->top = NULL;
}
i = 0;
while (*(p1+i) != '\0')
{
if (*(p1+i) != ' ')
{
new = (struct node*)malloc(sizeof(struct node));
if (new == NULL)
printf("\n####### new is not allocated #####\n");
j = 0;
while (*(p1+i) != ' ' && *(p1+i) != '\0')
{
new->word[j] = *(p1 + i);
i++;
j++;
}
new->word[j++] = ' ';
new->word[j] = '\0';
new->next = (*stk)->top;
(*stk)->top = new;
}
i++;
}
}
void print (struct stack *stk)
{
struct node *tmp = stk->top;
int i;
while (tmp != NULL)
{
i = 0;
while (tmp->word[i] != '\0')
{
printf ("%c" , tmp->word[i]);
i++;
}
tmp = tmp->next;
}
printf("\n");
}
Most of these answers fail to account for leading and/or trailing spaces in the input string. Consider the case of str=" Hello world"... The simple algo of reversing the whole string and reversing individual words winds up flipping delimiters resulting in f(str) == "world Hello ".
The OP said "I want to reverse the order of the words" and did not mention that leading and trailing spaces should also be flipped! So, although there are a ton of answers already, I'll provide a [hopefully] more correct one in C++:
#include <string>
#include <algorithm>
void strReverseWords_inPlace(std::string &str)
{
const char delim = ' ';
std::string::iterator w_begin, w_end;
if (str.size() == 0)
return;
w_begin = str.begin();
w_end = str.begin();
while (w_begin != str.end()) {
if (w_end == str.end() || *w_end == delim) {
if (w_begin != w_end)
std::reverse(w_begin, w_end);
if (w_end == str.end())
break;
else
w_begin = ++w_end;
} else {
++w_end;
}
}
// instead of reversing str.begin() to str.end(), use two iterators that
// ...represent the *logical* begin and end, ignoring leading/traling delims
std::string::iterator str_begin = str.begin(), str_end = str.end();
while (str_begin != str_end && *str_begin == delim)
++str_begin;
--str_end;
while (str_end != str_begin && *str_end == delim)
--str_end;
++str_end;
std::reverse(str_begin, str_end);
}
My version of using stack:
public class Solution {
public String reverseWords(String s) {
StringBuilder sb = new StringBuilder();
String ns= s.trim();
Stack<Character> reverse = new Stack<Character>();
boolean hadspace=false;
//first pass
for (int i=0; i< ns.length();i++){
char c = ns.charAt(i);
if (c==' '){
if (!hadspace){
reverse.push(c);
hadspace=true;
}
}else{
hadspace=false;
reverse.push(c);
}
}
Stack<Character> t = new Stack<Character>();
while (!reverse.empty()){
char temp =reverse.pop();
if(temp==' '){
//get the stack content out append to StringBuilder
while (!t.empty()){
char c =t.pop();
sb.append(c);
}
sb.append(' ');
}else{
//push to stack
t.push(temp);
}
}
while (!t.empty()){
char c =t.pop();
sb.append(c);
}
return sb.toString();
}
}
Store Each word as a string in array then print from end
public void rev2() {
String str = "my name is ABCD";
String A[] = str.split(" ");
for (int i = A.length - 1; i >= 0; i--) {
if (i != 0) {
System.out.print(A[i] + " ");
} else {
System.out.print(A[i]);
}
}
}
In Python, if you can't use [::-1] or reversed(), here is the simple way:
def reverse(text):
r_text = text.split(" ")
res = []
for word in range(len(r_text) - 1, -1, -1):
res.append(r_text[word])
return " ".join(res)
print (reverse("Hello World"))
>> World Hello
[Finished in 0.1s]
Printing words in reverse order of a given statement using C#:
void ReverseWords(string str)
{
int j = 0;
for (int i = (str.Length - 1); i >= 0; i--)
{
if (str[i] == ' ' || i == 0)
{
j = i == 0 ? i : i + 1;
while (j < str.Length && str[j] != ' ')
Console.Write(str[j++]);
Console.Write(' ');
}
}
}
Here is the Java Implementation:
public static String reverseAllWords(String given_string)
{
if(given_string == null || given_string.isBlank())
return given_string;
char[] str = given_string.toCharArray();
int start = 0;
// Reverse the entire string
reverseString(str, start, given_string.length() - 1);
// Reverse the letters of each individual word
for(int end = 0; end <= given_string.length(); end++)
{
if(end == given_string.length() || str[end] == ' ')
{
reverseString(str, start, end-1);
start = end + 1;
}
}
return new String(str);
}
// In-place reverse string method
public static void reverseString(char[] str, int start, int end)
{
while(start < end)
{
char temp = str[start];
str[start++] = str[end];
str[end--] = temp;
}
}
Actually, the first answer:
words = aString.split(" ");
for (i = 0; i < words.length; i++) {
words[i] = words[words.length-i];
}
does not work because it undoes in the second half of the loop the work it did in the first half. So, i < words.length/2 would work, but a clearer example is this:
words = aString.split(" "); // make up a list
i = 0; j = words.length - 1; // find the first and last elements
while (i < j) {
temp = words[i]; words[i] = words[j]; words[j] = temp; //i.e. swap the elements
i++;
j--;
}
Note: I am not familiar with the PHP syntax, and I have guessed incrementer and decrementer syntax since it seems to be similar to Perl.
How about ...
var words = "My name is X Y Z";
var wr = String.Join( " ", words.Split(' ').Reverse().ToArray() );
I guess that's not in-line tho.
In c, this is how you might do it, O(N) and only using O(1) data structures (i.e. a char).
#include<stdio.h>
#include<stdlib.h>
main(){
char* a = malloc(1000);
fscanf(stdin, "%[^\0\n]", a);
int x = 0, y;
while(a[x]!='\0')
{
if (a[x]==' ' || a[x]=='\n')
{
x++;
}
else
{
y=x;
while(a[y]!='\0' && a[y]!=' ' && a[y]!='\n')
{
y++;
}
int z=y;
while(x<y)
{
y--;
char c=a[x];a[x]=a[y];a[y]=c;
x++;
}
x=z;
}
}
fprintf(stdout,a);
return 0;
}
It can be done more simple using sscanf:
void revertWords(char *s);
void revertString(char *s, int start, int n);
void revertWordsInString(char *s);
void revertString(char *s, int start, int end)
{
while(start<end)
{
char temp = s[start];
s[start] = s[end];
s[end]=temp;
start++;
end --;
}
}
void revertWords(char *s)
{
int start = 0;
char *temp = (char *)malloc(strlen(s) + 1);
int numCharacters = 0;
while(sscanf(&s[start], "%s", temp) !=EOF)
{
numCharacters = strlen(temp);
revertString(s, start, start+numCharacters -1);
start = start+numCharacters + 1;
if(s[start-1] == 0)
return;
}
free (temp);
}
void revertWordsInString(char *s)
{
revertString(s,0, strlen(s)-1);
revertWords(s);
}
int main()
{
char *s= new char [strlen("abc deff gh1 jkl")+1];
strcpy(s,"abc deff gh1 jkl");
revertWordsInString(s);
printf("%s",s);
return 0;
}
import java.util.Scanner;
public class revString {
static char[] str;
public static void main(String[] args) {
//Initialize string
//str = new char[] { 'h', 'e', 'l', 'l', 'o', ' ', 'a', ' ', 'w', 'o',
//'r', 'l', 'd' };
getInput();
// reverse entire string
reverse(0, str.length - 1);
// reverse the words (delimeted by space) back to normal
int i = 0, j = 0;
while (j < str.length) {
if (str[j] == ' ' || j == str.length - 1) {
int m = i;
int n;
//dont include space in the swap.
//(special case is end of line)
if (j == str.length - 1)
n = j;
else
n = j -1;
//reuse reverse
reverse(m, n);
i = j + 1;
}
j++;
}
displayArray();
}
private static void reverse(int i, int j) {
while (i < j) {
char temp;
temp = str[i];
str[i] = str[j];
str[j] = temp;
i++;
j--;
}
}
private static void getInput() {
System.out.print("Enter string to reverse: ");
Scanner scan = new Scanner(System.in);
str = scan.nextLine().trim().toCharArray();
}
private static void displayArray() {
//Print the array
for (int i = 0; i < str.length; i++) {
System.out.print(str[i]);
}
}
}
In Java using an additional String (with StringBuilder):
public static final String reverseWordsWithAdditionalStorage(String string) {
StringBuilder builder = new StringBuilder();
char c = 0;
int index = 0;
int last = string.length();
int length = string.length()-1;
StringBuilder temp = new StringBuilder();
for (int i=length; i>=0; i--) {
c = string.charAt(i);
if (c == SPACE || i==0) {
index = (i==0)?0:i+1;
temp.append(string.substring(index, last));
if (index!=0) temp.append(c);
builder.append(temp);
temp.delete(0, temp.length());
last = i;
}
}
return builder.toString();
}
In Java in-place:
public static final String reverseWordsInPlace(String string) {
char[] chars = string.toCharArray();
int lengthI = 0;
int lastI = 0;
int lengthJ = 0;
int lastJ = chars.length-1;
int i = 0;
char iChar = 0;
char jChar = 0;
while (i<chars.length && i<=lastJ) {
iChar = chars[i];
if (iChar == SPACE) {
lengthI = i-lastI;
for (int j=lastJ; j>=i; j--) {
jChar = chars[j];
if (jChar == SPACE) {
lengthJ = lastJ-j;
swapWords(lastI, i-1, j+1, lastJ, chars);
lastJ = lastJ-lengthI-1;
break;
}
}
lastI = lastI+lengthJ+1;
i = lastI;
} else {
i++;
}
}
return String.valueOf(chars);
}
private static final void swapWords(int startA, int endA, int startB, int endB, char[] array) {
int lengthA = endA-startA+1;
int lengthB = endB-startB+1;
int length = lengthA;
if (lengthA>lengthB) length = lengthB;
int indexA = 0;
int indexB = 0;
char c = 0;
for (int i=0; i<length; i++) {
indexA = startA+i;
indexB = startB+i;
c = array[indexB];
array[indexB] = array[indexA];
array[indexA] = c;
}
if (lengthB>lengthA) {
length = lengthB-lengthA;
int end = 0;
for (int i=0; i<length; i++) {
end = endB-((length-1)-i);
c = array[end];
shiftRight(endA+i,end,array);
array[endA+1+i] = c;
}
} else if (lengthA>lengthB) {
length = lengthA-lengthB;
for (int i=0; i<length; i++) {
c = array[endA];
shiftLeft(endA,endB,array);
array[endB+i] = c;
}
}
}
private static final void shiftRight(int start, int end, char[] array) {
for (int i=end; i>start; i--) {
array[i] = array[i-1];
}
}
private static final void shiftLeft(int start, int end, char[] array) {
for (int i=start; i<end; i++) {
array[i] = array[i+1];
}
}
Here is a C implementation that is doing the word reversing inlace, and it has O(n) complexity.
char* reverse(char *str, char wordend=0)
{
char c;
size_t len = 0;
if (wordend==0) {
len = strlen(str);
}
else {
for(size_t i=0;str[i]!=wordend && str[i]!=0;i++)
len = i+1;
}
for(size_t i=0;i<len/2;i++) {
c = str[i];
str[i] = str[len-i-1];
str[len-i-1] = c;
}
return str;
}
char* inplace_reverse_words(char *w)
{
reverse(w); // reverse all letters first
bool is_word_start = (w[0]!=0x20);
for(size_t i=0;i<strlen(w);i++){
if(w[i]!=0x20 && is_word_start) {
reverse(&w[i], 0x20); // reverse one word only
is_word_start = false;
}
if (!is_word_start && w[i]==0x20) // found new word
is_word_start = true;
}
return w;
}
c# solution to reverse words in a sentence
using System;
class helloworld {
public void ReverseString(String[] words) {
int end = words.Length-1;
for (int start = 0; start < end; start++) {
String tempc;
if (start < end ) {
tempc = words[start];
words[start] = words[end];
words[end--] = tempc;
}
}
foreach (String s1 in words) {
Console.Write("{0} ",s1);
}
}
}
class reverse {
static void Main() {
string s= "beauty lies in the heart of the peaople";
String[] sent_char=s.Split(' ');
helloworld h1 = new helloworld();
h1.ReverseString(sent_char);
}
}
output:
peaople the of heart the in lies beauty Press any key to continue . . .
Better version
Check my blog http://bamaracoulibaly.blogspot.co.uk/2012/04/19-reverse-order-of-words-in-text.html
public string reverseTheWords(string description)
{
if(!(string.IsNullOrEmpty(description)) && (description.IndexOf(" ") > 1))
{
string[] words= description.Split(' ');
Array.Reverse(words);
foreach (string word in words)
{
string phrase = string.Join(" ", words);
Console.WriteLine(phrase);
}
return phrase;
}
return description;
}
public class manip{
public static char[] rev(char[] a,int left,int right) {
char temp;
for (int i=0;i<(right - left)/2;i++) {
temp = a[i + left];
a[i + left] = a[right -i -1];
a[right -i -1] = temp;
}
return a;
}
public static void main(String[] args) throws IOException {
String s= "i think this works";
char[] str = s.toCharArray();
int i=0;
rev(str,i,s.length());
int j=0;
while(j < str.length) {
if (str[j] != ' ' && j != str.length -1) {
j++;
} else
{
if (j == (str.length -1)) {
j++;
}
rev(str,i,j);
i=j+1;
j=i;
}
}
System.out.println(str);
}
I know there are several correct answers. Here is the one in C that I came up with.
This is an implementation of the excepted answer. Time complexity is O(n) and no extra string is used.
#include<stdio.h>
char * strRev(char *str, char tok)
{
int len = 0, i;
char *temp = str;
char swap;
while(*temp != tok && *temp != '\0') {
len++; temp++;
}
len--;
for(i = 0; i < len/2; i++) {
swap = str[i];
str[i] = str[len - i];
str[len - i] = swap;
}
// Return pointer to the next token.
return str + len + 1;
}
int main(void)
{
char a[] = "Reverse this string.";
char *temp = a;
if (a == NULL)
return -1;
// Reverse whole string character by character.
strRev(a, '\0');
// Reverse every word in the string again.
while(1) {
temp = strRev(temp, ' ');
if (*temp == '\0')
break;
temp++;
}
printf("Reversed string: %s\n", a);
return 0;
}