Interview Questions (87): Two arrays containing questions (from Microsoft interview questions)

Question:

You have given two arrays, say

A: 4, 1, 6, 2, 8, 9, 5, 3, 2, 9, 8, 4, 6
B: 6, 1, 2, 9, 8

Where B contains elements which are in a consecutive locations but may be in any order.
Find their starting and ending indexes in A. (Be careful of duplicate numbers ).

 

Answer is (1, 5)

 

First, give the code, and then explain it with the Code as follows:

# Include <iostream> <br/> # include <map> </P> <p> using namespace STD; </P> <p> void findconsecutivesubarrlocation (int A [], int Lena, int B [], int lenb) <br/>{< br/> Map <int, int> bmap; <br/> Map <int, int> windowmap; <br/> Map <int, int> diffmap; <br/> for (INT I = 0; I <lenb; I ++) <br/>{< br/> If (bmap. count (B [I]) = 0) <br/> bmap [B [I] = 1; <br/> else <br/> ++ bmap [B [I]; <br/> If (windowmap. count (A [I]) = 0) <br/> windowmap [A [I] = 1; <br/> else <br/> ++ windowmap [A [I]; <br/>}</P> <p> Map <int, int >:: iterator it = bmap. begin (); <br/> int sameelement = 0; <br/> while (it! = Bmap. End () <br/>{< br/> If (windowmap. Count (* It). First )! = 0) <br/>{< br/> diffmap [(* it ). first] = windowmap [(* it ). first]-(* it ). second; <br/> If (diffmap [(* it ). first] = 0) <br/> sameelement ++; <br/>}< br/> else <br/> diffmap [(* it ). first] =-(* it ). second; <br/> it ++; <br/>}</P> <p> If (sameelement = lenb) <br/>{ <br/> cout <"---------- find one ---------" <Endl; <br/> cout <"Start index: "<0 <Endl; <br/> cout <" End index: "<lenB-1 <Endl; <br/>}</P> <p> For (INT I = lenb; I <Lena; I ++) <br/> {<br/> If (diffmap. Count (A [I-lenb])! = 0) <br/>{< br/> diffmap [A [I-lenb] --; <br/> If (diffmap [A [I-lenb] = 0) <br/> sameelement ++; <br/> else if (diffmap [A [I-lenb] =-1) <br/> sameelement --; <br/>}< br/> If (diffmap. count (A [I])! = 0) <br/>{< br/> diffmap [A [I] ++; <br/> If (diffmap [A [I] = 0) <br/> sameelement ++; <br/> else if (diffmap [A [I] = 1) <br/> sameelement --; <br/>}< br/> If (sameelement = diffmap. size () <br/>{< br/> cout <"---------- find one ---------" <Endl; <br/> cout <"Start index: "<I-lenb + 1 <Endl; <br/> cout <" End index: "<I + 1 <Endl; <br/>}</P> <p> int main () <br/> {<br/> int A [] = {4, 1, 2, 1, 8, 9, 2, 1, 2, 9, 8, 4, 6 }; <br/> int B [] = {1, 1, 2, 8, 9}; <br/> int Lena = sizeof () /sizeof (INT); <br/> int lenb = sizeof (B)/sizeof (INT); <br/> findconsecutivesubarrlocation (A, Lena, B, lenb ); <br/> return 0; <br/>}< br/>

 

 

Ideas:

Ex,

Int A [] = {4, 1, 2, 1, 8, 9, 5, 3, 2, 9, 8, 4, 6 };
Int B [] = {1, 1, 2, 8, 9 };
Int Lena = 13;
Int lenb = 5;
Map <int, int> bmap;
Map <int, int> windowmap;
Map <int, int> diffmap;

Initialize map first:
Bmap =}
Windowmap =}
Diffmap = {1:0, 9:-1}
"1: 0" means that the current Sliding Window of array a exactly has the same number of "1" as array B, while "9: -1 indicates that the current Sliding Window of A is missing 1 "9" compared with that of B ". In addition, we noticed that diffmap and bmap have the same key.

The variable "sameelement" in the Code indicates the number of pair pairs in diffmap matching bmap. Obviously, the value of the initialized "sameelement" variable will be 3 (, in diffmap ).

Next
In array A, move the window with a size of lenb without moving forward. You only need to check the elements El drawn on the left of the sliding window and the elements er slide on the right to update diffmap and sameelement, as shown below:

If (diffmap. Count (A [I-lenb])! = 0)
{
Diffmap [A [I-lenb] --;
If (diffmap [A [I-lenb] = 0)
Sameelement ++;
Else if (diffmap [A [I-lenb] =-1)
Sameelement --;
}
If (diffmap. Count (A [I])! = 0)
{
Diffmap [A [I] ++;
If (diffmap [A [I] = 0)
Sameelement ++;
Else if (diffmap [A [I] = 1)
Sameelement --;
}
If (sameelement = diffmap. Size ())
{
Cout <"---------- find one ---------" <Endl;
Cout <"Start index:" <I-lenb + 1 <Endl;
Cout <"End index:" <I + 1 <Endl;
}