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I'm trying to write a method that uses recursion to print the string formed by "interleaving" the strings str1 and str2. In other words, it should alternate characters from the two strings: the first character from str1, followed by the first character from str2, followed by the second character from str1, followed by the second character from str2, etc.
How would I go about this?
The general idea of recursion is to have a terminating position that returns a constant value of some sort and then every other case is built on top of that.
For example, the factorial function, f(n) = n * (n-1) * (n-2) * ... * 2 * 1 has the following terminating condition and dependent functions:
f(1) = 1
f(n) = n * f(n-1) for all n > 1
so could be implemented as:
def factorial(n):
if n == 1:
return 1;
return n * factorial(n-1)
In your particular case, the terminating condition is when either string is empty, at which point you just tack the other string onto the end. The dependent function is simply to grab the first character from the first string then call the next level down passing the rest of that string and the other string, but in reverse order so that you alternate.
def mix (s1, s2):
if s1 == "" return s2
if s2 == "" return s1
return s1.firstChar() + mix (s2, s1.allButFirstChar());
In Java, this would translate to the following. Be warned that, if this is homework and you use this, you will almost certainly fail since you would be foolish to think your educators are not monitoring these sites.
public class Demo {
public static String Mix (String s1, String s2) {
if (s1.length() == 0) return s2;
if (s2.length() == 0) return s1;
return s1.substring(0,1) + Mix (s2, s1.substring(1));
}
public static void main(String[] args) {
System.out.println (Mix ("Hello", "There"));
System.out.println (Mix ("Hi", "There"));
System.out.println (Mix ("Hello again", "Pax"));
System.out.println (Mix ("", ""));
System.out.println (Mix ("1111", ""));
System.out.println (Mix ("111", "2"));
System.out.println (Mix ("111", "22"));
System.out.println (Mix ("111", "222"));
System.out.println (Mix ("111", "2222"));
System.out.println (Mix ("11", "2222"));
System.out.println (Mix ("1", "2222"));
System.out.println (Mix ("", "2222"));
}
}
outputs:
HTehlelroe
HTihere
HPealxlo again
1111
1211
12121
121212
1212122
121222
12222
2222
This should work:
public String Interleave( String first, String second )
{
if ( first.length() == 0 )
return second;
if ( second.length() == 0 )
return first;
return first.substring(0,1) + second.substring(0,1) +
Interleave( first.substring(1), second.substring(1) );
}
This is the codingBat/String-2/mixString problem, except you want a recursive solution.
My solution is essentially the same as Kieveli's:
public String mixString(String a, String b) {
return
a.isEmpty() ? b :
b.isEmpty() ? a :
a.substring(0, 1) + b.substring(0, 1)
+ mixString(a.substring(1), b.substring(1));
}
String s1="12345";
String s2="67890";
char ch1[] = s1.toCharArray();
char ch2 [] = s2.toCharArray();
char[] ch3 = new char[10];
// System.out.println("ch1"+ch1);
for(i=0;i<10;i++)
{
ch3[i*2]=ch1[i];
ch3[i*2+1]=ch2[i];
System.out.println("ttttttttttt");
System.out.println(ch3[i]);
}
Related
Let us a have a string "abbashbhqa". We have to remove the duplicate characters in such a manner that the output should be "abshq". One possible solution is to check each character with the others present in the string and then manipulate. But this requires O(n^2) time complexity. Is there any optimised approach to do so ?
O(n):
Define an array L[26] of booleans. Set all to FALSE.
Construct a new empty string
Walk over the string and for each letter check if L [x] is FALSE. If so, append x to the new string and set L [x] to 1.
Copy new string to the old one.
as soon as you iterate string you create a set (or hash set). in case the alphabet is limited (English letters as in your example) you just can create a 256 boolean array and use ASCII code as a key to it. Make all booleans to be false at starting point. Each iteration you check if array[] is false or true. In case it's false, the symbol is not a duplicate, so you mark it into array[] = true, do not remove from the string and go on. in case it's true - the symbol is a duplicate
Probably this will be the implementation of the above problem
import java.util.*;
import java.io.*;
public class String_Duplicate_Removal
{
public static String duplicate_removal(String s)
{
if(s.length()<2)
return s;
else if(s.length()==2)
{
if(s.charAt(0)==s.charAt(1))
s = Character.toString(s.charAt(0));
return s;
}
boolean [] arr = new boolean[26];
for(int i=0;i<s.length();i++)
{
if(arr[s.charAt(i)-'a']==false)
arr[s.charAt(i)-'a']=true;
else
{
s= ((new StringBuilder(s)).deleteCharAt(i)).toString();
i--;
}
}
return s;
}
public static void main(String [] args)
{
String s = "abbashbhqa";
System.out.println(duplicate_removal(s));
}
}
I am solving using Python and it works in O(n) time and O(n) space --
I am using set() as set does not allow duplicates ---
In this case the order of elements gets changed --
If u want the order to remain same then u can use OrderedDict() and it also works in O(n) time --
def remove_duplicates(s , ans_set):
for i in s: # O(n)
ans_set.add(i) # O(1)
ans = ''
for char in ans_set:
ans += char
print ans
s = raw_input()
ans_set = set()
remove_duplicates(s , ans_set)
from collections import OrderedDict
def remove_duplicates_maintain_order(a):
ans_dict = OrderedDict()
for i in a: # O(n)
ans_dict[i] = ans_dict.get(i , 0) + 1 # O(1)
ans = ''
for char in ans_dict:
ans += char
print ans
s = raw_input()
remove_duplicates_maintain_order(s)
I'm trying to find a suitable DP algorithm for simplifying a string. For example I have a string a b a b and a list of rules
a b -> b
a b -> c
b a -> a
c c -> b
The purpose is to get all single chars that can be received from the given string using these rules. For this example it will be b, c. The length of the given string can be up to 200 symbols. Could you please prompt an effective algorithm?
Rules always are 2 -> 1. I've got an idea of creating a tree, root is given string and each child is a string after one transform, but I'm not sure if it's the best way.
If you read those rules from right to left, they look exactly like the rules of a context free grammar, and have basically the same meaning. You could apply a bottom-up parsing algorithm like the Earley algorithm to your data, along with a suitable starting rule; something like
start <- start a
| start b
| start c
and then just examine the parse forest for the shortest chain of starts. The worst case remains O(n^3) of course, but Earley is fairly effective, these days.
You can also produce parse forests when parsing with derivatives. You might be able to efficiently check them for short chains of starts.
For a DP problem, you always need to understand how you can construct the answer for a big problem in terms of smaller sub-problems. Assume you have your function simplify which is called with an input of length n. There are n-1 ways to split the input in a first and a last part. For each of these splits, you should recursively call your simplify function on both the first part and the last part. The final answer for the input of length n is the set of all possible combinations of answers for the first and for the last part, which are allowed by the rules.
In Python, this can be implemented like so:
rules = {'ab': set('bc'), 'ba': set('a'), 'cc': set('b')}
all_chars = set(c for cc in rules.values() for c in cc)
# memoize
def simplify(s):
if len(s) == 1: # base case to end recursion
return set(s)
possible_chars = set()
# iterate over all the possible splits of s
for i in range(1, len(s)):
head = s[:i]
tail = s[i:]
# check all possible combinations of answers of sub-problems
for c1 in simplify(head):
for c2 in simplify(tail):
possible_chars.update(rules.get(c1+c2, set()))
# speed hack
if possible_chars == all_chars: # won't get any bigger
return all_chars
return possible_chars
Quick check:
In [53]: simplify('abab')
Out[53]: {'b', 'c'}
To make this fast enough for large strings (to avoiding exponential behavior), you should use a memoize decorator. This is a critical step in solving DP problems, otherwise you are just doing a brute-force calculation. A further tiny speedup can be obtained by returning from the function as soon as possible_chars == set('abc'), since at that point, you are already sure that you can generate all possible outcomes.
Analysis of running time: for an input of length n, there are 2 substrings of length n-1, 3 substrings of length n-2, ... n substrings of length 1, for a total of O(n^2) subproblems. Due to the memoization, the function is called at most once for every subproblem. Maximum running time for a single sub-problem is O(n) due to the for i in range(len(s)), so the overall running time is at most O(n^3).
Let N - length of given string and R - number of rules.
Expanding a tree in a top down manner yields computational complexity O(NR^N) in the worst case (input string of type aaa... and rules aa -> a).
Proof:
Root of the tree has (N-1)R children, which have (N-1)R^2 children, ..., which have (N-1)R^N children (leafs). So, the total complexity is O((N-1)R + (N-1)R^2 + ... (N-1)R^N) = O(N(1 + R^2 + ... + R^N)) = (using binomial theorem) = O(N(R+1)^N) = O(NR^N).
Recursive Java implementation of this naive approach:
public static void main(String[] args) {
Map<String, Character[]> rules = new HashMap<String, Character[]>() {{
put("ab", new Character[]{'b', 'c'});
put("ba", new Character[]{'a'});
put("cc", new Character[]{'b'});
}};
System.out.println(simplify("abab", rules));
}
public static Set<String> simplify(String in, Map<String, Character[]> rules) {
Set<String> result = new HashSet<String>();
simplify(in, rules, result);
return result;
}
private static void simplify(String in, Map<String, Character[]> rules, Set<String> result) {
if (in.length() == 1) {
result.add(in);
}
for (int i = 0; i < in.length() - 1; i++) {
String two = in.substring(i, i + 2);
Character[] rep = rules.get(two);
if (rep != null) {
for (Character c : rep) {
simplify(in.substring(0, i) + c + in.substring(i + 2, in.length()), rules, result);
}
}
}
}
Bas Swinckels's O(RN^3) Java implementation (with HashMap as a memoization cache):
public static Set<String> simplify2(final String in, Map<String, Character[]> rules) {
Map<String, Set<String>> cache = new HashMap<String, Set<String>>();
return simplify2(in, rules, cache);
}
private static Set<String> simplify2(final String in, Map<String, Character[]> rules, Map<String, Set<String>> cache) {
final Set<String> cached = cache.get(in);
if (cached != null) {
return cached;
}
Set<String> ret = new HashSet<String>();
if (in.length() == 1) {
ret.add(in);
return ret;
}
for (int i = 1; i < in.length(); i++) {
String head = in.substring(0, i);
String tail = in.substring(i, in.length());
for (String c1 : simplify2(head, rules)) {
for (String c2 : simplify2(tail, rules, cache)) {
Character[] rep = rules.get(c1 + c2);
if (rep != null) {
for (Character c : rep) {
ret.add(c.toString());
}
}
}
}
}
cache.put(in, ret);
return ret;
}
Output in both approaches:
[b, c]
Given a string S and a set of n substrings. Remove every instance of those n substrings from S so that S is of the minimum length and output this minimum length.
Example 1
S = ccdaabcdbb
n = 2
substrings = ab, cd
Output
2
Explanation:
ccdaabcdbb -> ccdacdbb -> cabb -> cb (length=2)
Example 2
S = abcd
n = 2
substrings = ab,bcd
Output
1
How do I solve this problem ?
A simple Brute-force search algorithm is:
For each substring, try all possible ways to remove it from the string, then recurse.
In Pseudocode:
def min_final_length (input, substrings):
best = len(input)
for substr in substrings:
beg = 0
// find all occurrences of substr in input and recurse
while (found = find_substring(input, substr, from=beg)):
input_without_substr = input[0:found]+input[found+len(substr):len(input)]
best = min(best, min_final_length(input_without_substr,substrings))
beg = found+1
return best
Let complexity be F(S,n,l) where S is the length of the input string, n is the cardinality of the set substrings and l is the "characteristic length" of substrings. Then
F(S,n,l) ~ n * ( S * l + F(S-l,n,l) )
Looks like it is at most O(S^2*n*l).
The following solution would have an complexity of O(m * n) where m = len(S) and n is the number of substring
def foo(S, sub):
i = 0
while i < len(S):
for e in sub:
if S[i:].startswith(e):
S = S[:i] + S[i+len(e):]
i -= 1
break
else: i += 1
return S, i
If you are for raw performance and your string is very large, you can do better than brute force. Use a suffix trie (E.g, Ukkonnen trie) to store your string. Then find each substring (which us done in O(m) time, m being substring length), and store the offsets to the substrings and length in an array.
Then use the offsets and length info to actually remove the substrings by filling these areas with \0 (in C) or another placeholder character. By counting all non-Null characters you will get the minimal length of the string.
This will als handle overlapping substring, e.g. say your string is "abcd", and you have two substrings "ab" and "abcd".
I solved it using trie+dp.
First insert your substrings in a trie. Then define the state of the dp is some string, walk through that string and consider each i (for i =0 .. s.length()) as the start of some substring. let j=i and increment j as long as you have a suffix in the trie (which will definitely land you to at least one substring and may be more if you have common suffix between some substring, for example "abce" and "abdd"), whenever you encounter an end of some substring, go solve the new sub-problem and find the minimum between all substring reductions.
Here is my code for it. Don't worry about the length of the code. Just read the solve function and forget about the path, I included it to print the string formed.
struct node{
node* c[26];
bool str_end;
node(){
for(int i= 0;i<26;i++){
c[i]=NULL;
}
str_end= false;
}
};
class Trie{
public:
node* root;
Trie(){
root = new node();
}
~Trie(){
delete root;
}
};
class Solution{
public:
typedef pair<int,int>ii;
string get_str(string& s,map<string,ii>&path){
if(!path.count(s)){
return s;
}
int i= path[s].first;
int j= path[s].second;
string new_str =(s.substr(0,i)+s.substr(j+1));
return get_str(new_str,path);
}
int solve(string& s,Trie* &t, map<string,int>&dp,map<string,ii>&path){
if(dp.count(s)){
return dp[s];
}
int mn= (int)s.length();
for(int i =0;i<s.length();i++){
string left = s.substr(0,i);
node* cur = t->root->c[s[i]-97];
int j=i;
while(j<s.length()&&cur!=NULL){
if(cur->str_end){
string new_str =left+s.substr(j+1);
int ret= solve(new_str,t,dp,path);
if(ret<mn){
path[s]={i,j};
}
}
cur = cur->c[s[++j]-97];
}
}
return dp[s]=mn;
}
string removeSubstrings(vector<string>& substrs, string s){
map<string,ii>path;
map<string,int>dp;
Trie*t = new Trie();
for(int i =0;i<substrs.size();i++){
node* cur = t->root;
for(int j=0;j<substrs[i].length();j++){
if(cur->c[substrs[i][j]-97]==NULL){
cur->c[substrs[i][j]-97]= new node();
}
cur = cur->c[substrs[i][j]-97];
if(j==substrs[i].length()-1){
cur->str_end= true;
}
}
}
solve(s,t,dp,path);
return get_str(s, path);
}
};
int main(){
vector<string>substrs;
substrs.push_back("ab");
substrs.push_back("cd");
Solution s;
cout << s.removeSubstrings(substrs,"ccdaabcdbb")<<endl;
return 0;
}
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.
Closed 9 years ago.
*Question (String Permutation) from interviewstreet.com*
Given two strings, write a method to decide if one is a permutation
of
the other. Your solution should consider case sensitivity and
whitespace as significant.
A permutation of a set of objects is an arrangement of those objects
into a particular order. For example, there are six permutations of
the string "abc", namely "abc", "acb", "bac", "bca", "cab", and
"cba".
Output:
Return 1 if the two strings are permutations of each other.
Return 0 if the two strings are not permutations of each other.
abc acb cab cba bac bca
import java.io.*;
import java.util.*;
public class Solution {
private Set permutations;
public static void main(String args[] ) throws Exception {
// Scanner sc = new Scanner(System.in);
//String string1 = sc.nextLine();
//String string2 = sc.nextLine();
String string1 = "str";
String string2 = "str";
Solution solution = new Solution();
int output = solution.permutation(string1, string2);
System.out.println(output);
}
public void stringPermuation(String s1, String s2) {
if (s2.length() > 0) {
for (int i = 0; i < s2.length(); i++) {
System.out.println(s1 + s2.charAt(i)+","+ s2.substring(0, i)+" +"+ s2.substring(i + 1));
stringPermuation(s1 + s2.charAt(i),
s2.substring(0, i) + s2.substring(i + 1));
}}
else{
permutations.add(s1);
System.out.println(s1);
}
}
public Set stringPermuation(String s) {
permutations = new HashSet<String>();
stringPermuation("", s);
return permutations;
}
private int permutation(String string1, String string2) {
int result = 0;
Set<String> setString1 = stringPermuation(string1);
Set<String> setString2 = stringPermuation(string2);
// create an iterator
System.out.println("There are total of " + setString1.size() + " permutations in String1:");
System.out.println("There are total of " + setString2.size() + " permutations in String2:");
if(setString1.size() == setString2.size())
result=IterateSet(setString1,setString2);
//Return 1 if string1 is a permutation of string2
//Return 0 if string1 is not a permutation of string2
return result;
}
public int IterateSet(Set setString1,Set setString2){
int i= 0;
Iterator<String> it = setString1.iterator();
while (it.hasNext()) {
if(setString2.contains(it.next()) && i == 0)
i=1;
}
return i;}}
...Sort the characters of the two strings (lexicographically) and if the two sorted strings are equal, the originals are permutations of each other.
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Closed 7 years ago.
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I saw this in an interview question ,
Given a sorting order string, you are asked to sort the input string based on the given sorting order string.
for example if the sorting order string is dfbcae
and the Input string is abcdeeabc
the output should be dbbccaaee.
any ideas on how to do this , in an efficient way ?
The Counting Sort option is pretty cool, and fast when the string to be sorted is long compared to the sort order string.
create an array where each index corresponds to a letter in the alphabet, this is the count array
for each letter in the sort target, increment the index in the count array which corresponds to that letter
for each letter in the sort order string
add that letter to the end of the output string a number of times equal to it's count in the count array
Algorithmic complexity is O(n) where n is the length of the string to be sorted. As the Wikipedia article explains we're able to beat the lower bound on standard comparison based sorting because this isn't a comparison based sort.
Here's some pseudocode.
char[26] countArray;
foreach(char c in sortTarget)
{
countArray[c - 'a']++;
}
int head = 0;
foreach(char c in sortOrder)
{
while(countArray[c - 'a'] > 0)
{
sortTarget[head] = c;
head++;
countArray[c - 'a']--;
}
}
Note: this implementation requires that both strings contain only lowercase characters.
Here's a nice easy to understand algorithm that has decent algorithmic complexity.
For each character in the sort order string
scan string to be sorted, starting at first non-ordered character (you can keep track of this character with an index or pointer)
when you find an occurrence of the specified character, swap it with the first non-ordered character
increment the index for the first non-ordered character
This is O(n*m), where n is the length of the string to be sorted and m is the length of the sort order string. We're able to beat the lower bound on comparison based sorting because this algorithm doesn't really use comparisons. Like Counting Sort it relies on the fact that you have a predefined finite external ordering set.
Here's some psuedocode:
int head = 0;
foreach(char c in sortOrder)
{
for(int i = head; i < sortTarget.length; i++)
{
if(sortTarget[i] == c)
{
// swap i with head
char temp = sortTarget[head];
sortTarget[head] = sortTarget[i];
sortTarget[i] = temp;
head++;
}
}
}
In Python, you can just create an index and use that in a comparison expression:
order = 'dfbcae'
input = 'abcdeeabc'
index = dict([ (y,x) for (x,y) in enumerate(order) ])
output = sorted(input, cmp=lambda x,y: index[x] - index[y])
print 'input=',''.join(input)
print 'output=',''.join(output)
gives this output:
input= abcdeeabc
output= dbbccaaee
Use binary search to find all the "split points" between different letters, then use the length of each segment directly. This will be asymptotically faster then naive counting sort, but will be harder to implement:
Use an array of size 26*2 to store the begin and end of each letter;
Inspect the middle element, see if it is different from the element left to it. If so, then this is the begin for the middle element and end for the element before it;
Throw away the segment with identical begin and end (if there are any), recursively apply this algorithm.
Since there are at most 25 "split"s, you won't have to do the search for more than 25 segemnts, and for each segment it is O(logn). Since this is constant * O(logn), the algorithm is O(nlogn).
And of course, just use counting sort will be easier to implement:
Use an array of size 26 to record the number of different letters;
Scan the input string;
Output the string in the given sorting order.
This is O(n), n being the length of the string.
Interview questions are generally about thought process and don't usually care too much about language features, but I couldn't resist posting a VB.Net 4.0 version anyway.
"Efficient" can mean two different things. The first is "what's the fastest way to make a computer execute a task" and the second is "what's the fastest that we can get a task done". They might sound the same but the first can mean micro-optimizations like int vs short, running timers to compare execution times and spending a week tweaking every millisecond out of an algorithm. The second definition is about how much human time would it take to create the code that does the task (hopefully in a reasonable amount of time). If code A runs 20 times faster than code B but code B took 1/20th of the time to write, depending on the granularity of the timer (1ms vs 20ms, 1 week vs 20 weeks), each version could be considered "efficient".
Dim input = "abcdeeabc"
Dim sort = "dfbcae"
Dim SortChars = sort.ToList()
Dim output = New String((From c In input.ToList() Select c Order By SortChars.IndexOf(c)).ToArray())
Trace.WriteLine(output)
Here is my solution to the question
import java.util.*;
import java.io.*;
class SortString
{
public static void main(String arg[])throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
// System.out.println("Enter 1st String :");
// System.out.println("Enter 1st String :");
// String s1=br.readLine();
// System.out.println("Enter 2nd String :");
// String s2=br.readLine();
String s1="tracctor";
String s2="car";
String com="";
String uncom="";
for(int i=0;i<s2.length();i++)
{
if(s1.contains(""+s2.charAt(i)))
{
com=com+s2.charAt(i);
}
}
System.out.println("Com :"+com);
for(int i=0;i<s1.length();i++)
if(!com.contains(""+s1.charAt(i)))
uncom=uncom+s1.charAt(i);
System.out.println("Uncom "+uncom);
System.out.println("Combined "+(com+uncom));
HashMap<String,Integer> h1=new HashMap<String,Integer>();
for(int i=0;i<s1.length();i++)
{
String m=""+s1.charAt(i);
if(h1.containsKey(m))
{
int val=(int)h1.get(m);
val=val+1;
h1.put(m,val);
}
else
{
h1.put(m,new Integer(1));
}
}
StringBuilder x=new StringBuilder();
for(int i=0;i<com.length();i++)
{
if(h1.containsKey(""+com.charAt(i)))
{
int count=(int)h1.get(""+com.charAt(i));
while(count!=0)
{x.append(""+com.charAt(i));count--;}
}
}
x.append(uncom);
System.out.println("Sort "+x);
}
}
Here is my version which is O(n) in time. Instead of unordered_map, I could have just used a char array of constant size. i.,e. char char_count[256] (and done ++char_count[ch - 'a'] ) assuming the input strings has all ASCII small characters.
string SortOrder(const string& input, const string& sort_order) {
unordered_map<char, int> char_count;
for (auto ch : input) {
++char_count[ch];
}
string res = "";
for (auto ch : sort_order) {
unordered_map<char, int>::iterator it = char_count.find(ch);
if (it != char_count.end()) {
string s(it->second, it->first);
res += s;
}
}
return res;
}
private static String sort(String target, String reference) {
final Map<Character, Integer> referencesMap = new HashMap<Character, Integer>();
for (int i = 0; i < reference.length(); i++) {
char key = reference.charAt(i);
if (!referencesMap.containsKey(key)) {
referencesMap.put(key, i);
}
}
List<Character> chars = new ArrayList<Character>(target.length());
for (int i = 0; i < target.length(); i++) {
chars.add(target.charAt(i));
}
Collections.sort(chars, new Comparator<Character>() {
#Override
public int compare(Character o1, Character o2) {
return referencesMap.get(o1).compareTo(referencesMap.get(o2));
}
});
StringBuilder sb = new StringBuilder();
for (Character c : chars) {
sb.append(c);
}
return sb.toString();
}
In C# I would just use the IComparer Interface and leave it to Array.Sort
void Main()
{
// we defin the IComparer class to define Sort Order
var sortOrder = new SortOrder("dfbcae");
var testOrder = "abcdeeabc".ToCharArray();
// sort the array using Array.Sort
Array.Sort(testOrder, sortOrder);
Console.WriteLine(testOrder.ToString());
}
public class SortOrder : IComparer
{
string sortOrder;
public SortOrder(string sortOrder)
{
this.sortOrder = sortOrder;
}
public int Compare(object obj1, object obj2)
{
var obj1Index = sortOrder.IndexOf((char)obj1);
var obj2Index = sortOrder.IndexOf((char)obj2);
if(obj1Index == -1 || obj2Index == -1)
{
throw new Exception("character not found");
}
if(obj1Index > obj2Index)
{
return 1;
}
else if (obj1Index == obj2Index)
{
return 0;
}
else
{
return -1;
}
}
}