Booking.com Interview Question for Software Developers


Country: Netherlands
Interview Type: Phone Interview




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1
of 1 vote

Thanks for sharing such a valuable definition :D

- MehrdadAP August 20, 2015 | Flag Reply
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0
of 0 votes

I edited the question to make it more clear ;)

- ersegun August 21, 2015 | Flag
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1
of 1 vote

First store all elements of A (and their counts) in a HashMap. Then traverse B and if the count for current element is greater than 0, then add this element to the result and decrement the count.

static int[] intersection(int[] arr, int[] arr2) {
        ArrayList<Integer> intersection = new ArrayList<Integer>();
        HashMap<Integer, Integer> seen = new HashMap<Integer, Integer>();
        for(int i : arr) {
            if(!seen.containsKey(i))
                seen.put(i, 0);
            seen.put(i, seen.get(i) + 1);
        }
        for(int j : arr2) {
            if(seen.containsKey(j) && seen.get(j) > 0) {
                intersection.add(j);
                seen.put(j, seen.get(j) - 1);
            }
        }
        int[] result = new int[intersection.size()];
        for(int k=0; k<result.length; k++) {
            result[k] = intersection.get(k);
        }
        return result;
    }

- Sunny August 21, 2015 | Flag Reply
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Hi Sunny
once we add it to intersection array, what's the purpose of decrementing with seen.put(j, seen.get(j) - 1);

this can be changed with

seen.put(j, 0);

This will not allow duplicates. Please correct me if I am wrong.

- suresh.garine February 07, 2016 | Flag
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1
of 1 vote

import java.util.*;


public class retainex {

public static void main(String[] args) {
int a[]={1,2,3,4,5,6};
int b[]={4,5,6,7,8,9};
TreeSet<Integer> tr1=new TreeSet<Integer>();
TreeSet<Integer> tr2=new TreeSet<Integer>();
for(int i=0;i<=a.length-1;i++)
{
tr1.add(a[i]);
tr2.add(b[i]);
}
tr1.retainAll(tr2);
Iterator<Integer> it= tr1.iterator();
while(it.hasNext())
{
System.out.println(it.next());
}

}

}

- Anonymous August 23, 2015 | Flag Reply
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1
of 1 vote

In python using a dictionaries

from collections import Counter

def intersect(s1,s2):
	m1 = Counter(s1)
	m2 = Counter(s2)
	result = []
	for i in m1:
		if i in m2:
			result += [i]*min(m1[i],m2[i])
	return result

- rifat April 28, 2016 | Flag Reply
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import java.util.*;
import java.io.*;
/*
* @author Tejashree pc
*/
public class Intersection {
public static void main(String[] args) throws IOException
{
Scanner s=new Scanner(System.in);
int A[], B[];
System.out.println("Number of elements in A should be less than or equal to number of elements in B.");
System.out.println("Enter the number of elements in Multiset A:");
int a=s.nextInt();
System.out.println("Enter the number of elements in Multiset B:");
int b=s.nextInt();
boolean fA[]=new boolean[a];
boolean fB[]=new boolean[b];
A=new int[a];
B=new int[b];
System.out.println("Enter the elements in Multiset A:");
for(int i=0;i<a;i++)
{
System.out.print("A"+i+": ");
A[i]=s.nextInt();
fA[i]=false;
System.out.println();
}
System.out.println("Enter the elements in Multiset B:");
for(int i=0;i<b;i++)
{
System.out.print("B"+i+": ");
B[i]=s.nextInt();
fB[i]=false;
System.out.println();
}
int ii=0;
List<Integer> I=new ArrayList<>();
System.out.print("Intersection of Multisets A and B = C = [ ");
for(int i=0; i<a; i++)
{
for(int j=0; j<b; j++)
{
if(!fB[j] && A[i]==B[j])
{
I.add(A[i]);
fA[i]=true;
fB[j]=true;
System.out.print(I.get(ii++)+" ");
break;
}
}
}
System.out.print("]");
}
}

- Tejashree Gharat August 21, 2015 | Flag Reply
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0
of 0 vote

import java.util.*;
import java.io.*;
/*
* @author Tejashree pc
*/
public class Intersection {
public static void main(String[] args) throws IOException
{
Scanner s=new Scanner(System.in);
int A[], B[];
System.out.println("Number of elements in A should be less than or equal to number of elements in B.");
System.out.println("Enter the number of elements in Multiset A:");
int a=s.nextInt();
System.out.println("Enter the number of elements in Multiset B:");
int b=s.nextInt();
boolean fA[]=new boolean[a];
boolean fB[]=new boolean[b];
A=new int[a];
B=new int[b];
System.out.println("Enter the elements in Multiset A:");
for(int i=0;i<a;i++)
{
System.out.print("A"+i+": ");
A[i]=s.nextInt();
fA[i]=false;
System.out.println();
}
System.out.println("Enter the elements in Multiset B:");
for(int i=0;i<b;i++)
{
System.out.print("B"+i+": ");
B[i]=s.nextInt();
fB[i]=false;
System.out.println();
}
int ii=0;
List<Integer> I=new ArrayList<>();
System.out.print("Intersection of Multisets A and B = C = [ ");
for(int i=0; i<a; i++)
{
for(int j=0; j<b; j++)
{
if(!fB[j] && A[i]==B[j])
{
I.add(A[i]);
fA[i]=true;
fB[j]=true;
System.out.print(I.get(ii++)+" ");
break;
}
}
}
System.out.print("]");
}
}

- Tejashree Gharat August 21, 2015 | Flag Reply
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of 0 vote

# I started writing a Perl script and in the middle of it, realized it's fairly easy to do it in bash.
# Not fancy but gets the job done. If I'm in hurry and not looking for votes :-), I would do the
# bash version any day. Yes, it can be improved and no it's not flawless. Here is the quick
# version. BTW this method can be used for numbers and strings both.

$ cat careercup_5158359730749440.sh
#!/bin/bash

get_intersection () {
  for num in $arrA
  do
    echo $num
  done | sort > /tmp/arrA.$$
  for num in $arrB
  do
    echo $num
  done | sort > /tmp/arrB.$$
  echo "Intersection of ($arrA) and ($arrB) is: ("`comm -12 /tmp/arrA.$$ /tmp/arrB.$$`")"
  rm -f /tmp/arrA.$$ /tmp/arrB.$$
}

arrA="0 1 1 2 2 5"
arrB="0 1 2 2 2 6"
get_intersection

arrA="0 1 1"
arrB="0 1 2 3 4 5 6"
get_intersection

$ careercup_5158359730749440.sh
Intersection of (0 1 1 2 2 5) and (0 1 2 2 2 6) is: (0 1 2 2)
Intersection of (0 1 1) and (0 1 2 3 4 5 6) is: (0 1)

- rajeev@udhar.com August 22, 2015 | Flag Reply
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of 0 vote

import java.util.*;

public class retainex {


public static void main(String[] args) {
int a[]={1,2,3,4,5,6};
int b[]={4,5,6,7,8,9};
TreeSet<Integer> tr1=new TreeSet<Integer>();
TreeSet<Integer> tr2=new TreeSet<Integer>();
for(int i=0;i<=a.length-1;i++)
{
tr1.add(a[i]);
tr2.add(b[i]);
}
tr1.retainAll(tr2);
Iterator<Integer> it= tr1.iterator();
while(it.hasNext())
{
System.out.println(it.next());
}

}

}

- ANAND VERMA August 23, 2015 | Flag Reply
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0
of 0 vote

It's a changed version of longest common string problem. You can read about it in the book by Cormen & Stein, the chapter of dynamical programming.

- artificer August 24, 2015 | Flag Reply
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public static void main(String[] args) {

		
		int[] setOne = new int[6];
		setOne[0] = 1;
		setOne[1] = 1;
		setOne[2] = 2;
		setOne[3] = 4;
		setOne[4] = 5;
		setOne[5] = 6;
		
		int[] setTwo = new int[8];
		setTwo[0] = 1;
		setTwo[1] = 11;
		setTwo[2] = 21;
		setTwo[3] = 41;
		setTwo[4] = 5;
		setTwo[5] = 6;
		setTwo[6] = 6;
		setTwo[7] = 7;

		
		List<Integer> resultArray = new ArrayList<Integer>();
		for (int i=0;i<setOne.length;i++) {
			if(setOne[i]==setTwo[i]){
				resultArray.add(setOne[i]);
			}
		}
		
		System.out.println(resultArray);
		
	}

- Anonymous September 01, 2015 | Flag Reply
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of 0 vote

package com.valli.AlgortithmProblems;

public class CommonElements {

public static void findCommonElements(int a1[],int a2[]){
if((a1.length==0)||(a2.length==0)){
return;
}
for(int i=0,j=0;i<a1.length-1||j<a2.length-1;){

if(a1[i]==a2[j]){
System.out.println(a1[i]);
i++;j++;
}else if(a1[i]<a2[j]){
i++;
}else
{
j++;
}
}
}

public static void main(String[] args) {
int a1[] =new int[]{1,2,3,3,3,4,5,5,6,6};
int a2[] = new int[]{2,3,3,3,3,4,5,6};
findCommonElements(a1, a2);
}

}

- Sri Valli September 03, 2015 | Flag Reply
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of 0 vote

package com.valli.AlgortithmProblems;

public class CommonElements {
	
	public static void findCommonElements(int a1[],int a2[]){
		if((a1.length==0)||(a2.length==0)){
			return;
		}
		for(int i=0,j=0;i<a1.length-1||j<a2.length-1;){
			
			if(a1[i]==a2[j]){
				System.out.println(a1[i]);
				i++;j++;
			}else if(a1[i]<a2[j]){
				i++;
			}else
			{
				j++;
			}
		}
	}

	public static void main(String[] args) {
		int a1[] =new int[]{1,2,3,3,3,4,5,5,6,6};
		int a2[] = new int[]{2,3,3,3,3,4,5,6};
		findCommonElements(a1, a2);
	}

}

- Sri Valli September 03, 2015 | Flag Reply
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package booking;

import java.util.ArrayList;

public class Booking {

    public static void main(String[] args) {
        int a[] = {0,1,1,2,2,5};
        int b[] = {0,1,2,2,2,6};
       
        System.out.println(findUnion(a, b));     
        
        int c[] = {0,1,1};
        int d[] = {0,1,2,3,4,5,6};
        
        System.out.println(findUnion(c,d));  
    }
    
    public static ArrayList<Integer>  findUnion(int first[], int second[]){
        
        int firstLength = first.length;
        int secondLength = second.length;
        ArrayList<Integer> c = new ArrayList<>();
        int min = Math.min(firstLength, secondLength);
        
       for(int i=0; i<min; i++){
           if(first[i] == second[i]){
               c.add(first[i]);
           }
       }
        return c;
    }

- nathanail.manos September 04, 2015 | Flag Reply
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0
of 0 votes

I can also submit this exercise written in javascript!

- nathanail.manos September 04, 2015 | Flag
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0
of 0 votes

It's won't work. Try

int c[] = {0,1,3};
        int d[] = {0,1,2,3,4,5,6};

- Anonymous January 28, 2016 | Flag
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public static void multiIntersection(Integer[] A, Integer[] B){
if(A.length==0||B.length==0){
return;
}
//Sort the arrays
insertionSort(A);insertionSort(B);
int lenA=A.length, lenB=B.length;
for(int i=0,j=0;i<lenA&&j<lenB;){
if(Objects.equals(A[i], B[j])){
System.out.print(A[i++]+" ");
j++;
}
else if(A[i]>B[j]) j++;
else i++;
}
System.out.println("");
}

public static void multiIntersection2(Integer[] A, Integer[] B, int N){
if(A.length==0||B.length==0){
return;
}
int lenA=A.length, lenB=B.length;
//N is the max number value among the arrays such as if A=[0,2,7], B=[0,1,6],
//N=7
Integer[] C=new Integer[N];
Arrays.fill(C, 0);
for(int i=0;i<lenA;i++){
C[A[i]]=C[A[i]]+1;
}

for(int j=0;j<lenB;j++){
if(C[B[j]]>0){
C[B[j]]--;
System.out.print(B[j]+" ");
}
}
}

- Anonymous September 13, 2015 | Flag Reply
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public static void multiIntersection(Integer[] A, Integer[] B){
        if(A.length==0||B.length==0){
            return;
        }
        //Sort the arrays
        insertionSort(A);insertionSort(B);
        int lenA=A.length, lenB=B.length;
        for(int i=0,j=0;i<lenA&&j<lenB;){
            if(Objects.equals(A[i], B[j])){
                System.out.print(A[i++]+" ");
                j++;
            }
            else if(A[i]>B[j]) j++;
            else          i++;      
        }
        System.out.println("");
    }
    
    public static void multiIntersection2(Integer[] A, Integer[] B, int N){
        if(A.length==0||B.length==0){
            return;
        }
        int lenA=A.length, lenB=B.length;
        //N is the max number value among the arrays such as if A=[0,2,7], B=[0,1,6],
        //N=7
        Integer[] C=new Integer[N];
        Arrays.fill(C, 0);
        for(int i=0;i<lenA;i++){
            C[A[i]]=C[A[i]]+1;
        }

        for(int j=0;j<lenB;j++){
            if(C[B[j]]>0){
                C[B[j]]--;
                System.out.print(B[j]+" ");
            }
        }
    }

- Sutapa September 13, 2015 | Flag Reply
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One solution that comes to my mind is, simply sort both the arrays and start comparing from the beginning. Vomit the integer if they are equal else move forward in the array with the samller-integer.

- rajcoool October 24, 2015 | Flag Reply
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another solution is to create a hashtable from the first array (ignore duplicates before hashing). And walk through the second array, lookup each number in hashtable. if it is found, then vomit it out else move forward.

- rajcoool October 24, 2015 | Flag
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Easy to read, simple and effective Java solution using Set:

public List<Integer> intersectArrays(int[] a, int[] b) {
    Map<Integer, Integer> map = new HashMap<>(); // number of appearances
    List<Integer> result = new ArrayList<>();

    for (int i = 0; i < a.length; i++) {
        int curNum = 1;
        if (map.containsKey(a[i]))
            curNum = map.get(a[i]) + 1;

        map.put(a[i], curNum);
    }

    for (int i = 0; i < b.length; i++) {
        if (!map.containsKey(b[i]))
            continue;

        int curNum = map.get(b[i]);
        if (curNum > 0) {
            curNum--;
            result.add(b[i]);
        }

        map.put(b[i], curNum); // update decreased counter
    }

    return result;
}

- megido December 07, 2015 | Flag Reply
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If arrays are not sorted, they should be sorted first

def intersect(a, b)
  a = a.sort
  b = b.sort
  res = []
  while a.any? && b.any?
    if a.first < b.first
      a.shift
    elsif b.first < a.first
      b.shift
    else
      a.shift
      res << b.shift
    end
  end
  res
end

- Anonymous January 08, 2016 | Flag Reply
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public int[] getBagIntersection2(int a[], int b[]) {
        a = Arrays.copyOf(a, a.length);
        b = Arrays.copyOf(b, b.length);
        Arrays.sort(a);
        Arrays.sort(b);

        List<Integer> resultList = new LinkedList<Integer>();
        int aIndex = 0;
        int bIndex = 0;
        while(aIndex <a.length && bIndex < b.length) {
            if(a[aIndex] == b[bIndex]) {
                resultList.add(a[aIndex]);
                aIndex++;
                bIndex++;
            } else if(a[aIndex] > b[bIndex]){
                bIndex++;
            } else {
                aIndex++;
            }
        }
        return toIntArray(resultList);
    }

- Anonymous January 28, 2016 | Flag Reply
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public int[] getBagIntersection2(int a[], int b[]) {
        a = Arrays.copyOf(a, a.length);
        b = Arrays.copyOf(b, b.length);
        Arrays.sort(a);
        Arrays.sort(b);

        List<Integer> resultList = new LinkedList<Integer>();
        int aIndex = 0;
        int bIndex = 0;
        while(aIndex <a.length && bIndex < b.length) {
            if(a[aIndex] == b[bIndex]) {
                resultList.add(a[aIndex]);
                aIndex++;
                bIndex++;
            } else if(a[aIndex] > b[bIndex]){
                bIndex++;
            } else {
                aIndex++;
            }
        }
        return toIntArray(resultList);
    }

- Sam January 28, 2016 | Flag Reply
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from collections import Counter
def intersect(s1,s2):
	m1 = Counter(s1)
	m2 = Counter(s2)
	result = []
	for i in m1:
		if i in m2:
			result+= [i]*min(m1[i],m2[i])
	return result

- rifat April 28, 2016 | Flag Reply
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of 0 vote

from collections import Counter
def intersect(s1,s2):
	m1 = Counter(s1)
	m2 = Counter(s2)
	result = []
	for i in m1:
		if i in m2:
			result+= [i]*min(m1[i],m2[i])
	return result

- rifat April 28, 2016 | Flag Reply
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of 0 vote

from collections import Counter
def intersect(s1,s2):
	m1 = Counter(s1)
	m2 = Counter(s2)
	result = []
	for i in m1:
		if i in m2:
			result+= [i]*min(m1[i],m2[i])
	return result

- rifat April 28, 2016 | Flag Reply
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Max time complexity O(NlogN) {{{ public static void main (String[] args) throws java.lang.Exception { int a[] = {0,1,1}; int b[] = {0,1,2,3,4,5}; //Assuming elements are in sorted form if not then sort both the array in O(NlogN) Arrays.sort(a); Arrays.sort(b); ArrayList<Integer> list = new ArrayList<>(); int i = 0, j = 0; // Max complexity would be O(N) while (i < a.length && j < b.length) { if (a[i] == b[j]) { list.add(a[i]); i++; j++; } else if (a[i] < b[j]){ i++; } else { j++; } } // So time complexity would be O(NlogN) and O(1) space as we need to create one array list for this. System.out.println(list); } }}} Using HashMap it can be done in O(n) {{{{{{HashMap<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < a.length; i++) { if (map.containsKey(a[i])) { map.put(a[i], map.get(a[i]) + 1); } else { map.put(a[i], 1); } } for (int i = 0 ; i < b.length; i ++) { if (map.containsKey(b[i]) && map.get(b[i]) > 0) { System.out.println(b[i]); map.put(b[i], map.get(b[i]) - 1); } }}}}}}} - tasneem December 17, 2016 | Flag Reply
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HashMap<Integer, Integer> map = new HashMap<>();

for (int i = 0; i < a.length; i++) {
if (map.containsKey(a[i])) {
map.put(a[i], map.get(a[i]) + 1);
} else {
map.put(a[i], 1);
}
}


for (int i = 0 ; i < b.length; i ++) {
if (map.containsKey(b[i]) && map.get(b[i]) > 0) {
System.out.println(b[i]);
map.put(b[i], map.get(b[i]) - 1);
}
}

- Tasneem March 15, 2017 | Flag Reply
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HashMap<Integer, Integer> map = new HashMap<>();
        
        for (int i = 0; i < a.length; i++) {
            if (map.containsKey(a[i])) {
                map.put(a[i], map.get(a[i]) + 1);
            } else {
                map.put(a[i], 1);
            }
        }
        
        
        for (int i = 0 ; i < b.length; i ++) {
            if (map.containsKey(b[i]) && map.get(b[i]) > 0) {
                System.out.println(b[i]);
                map.put(b[i], map.get(b[i]) - 1);
            }

}

- Tasneem March 15, 2017 | Flag Reply
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{{{{{{HashMap<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < a.length; i++) { if (map.containsKey(a[i])) { map.put(a[i], map.get(a[i]) + 1); } else { map.put(a[i], 1); } } for (int i = 0 ; i < b.length; i ++) { if (map.containsKey(b[i]) && map.get(b[i]) > 0) { System.out.println(b[i]); map.put(b[i], map.get(b[i]) - 1); } }}}}}}} - Tasneem March 15, 2017 | Flag Reply
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Using HashMap in O(n)

HashMap<Integer, Integer> map = new HashMap<>();
        
        for (int i = 0; i < a.length; i++) {
            if (map.containsKey(a[i])) {
                map.put(a[i], map.get(a[i]) + 1);
            } else {
                map.put(a[i], 1);
            }
        }
        
        
        for (int i = 0 ; i < b.length; i ++) {
            if (map.containsKey(b[i]) && map.get(b[i]) > 0) {
                System.out.println(b[i]);
                map.put(b[i], map.get(b[i]) - 1);
            }

- tasneem March 15, 2017 | Flag Reply
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Objective-C:

-(NSArray *)intersectBetweenTwoArrays:(NSArray *)arr1 array2:(NSArray *)arr2 {
    
    NSMutableArray * intersection = [NSMutableArray array];
    NSCountedSet *set1 = [NSCountedSet setWithArray:arr1];
    NSCountedSet *set2 = [NSCountedSet setWithArray:arr2];
    for (NSNumber *n1 in set1) {
        NSInteger count1 = [set1 countForObject: n1];
        NSInteger count2 = [set2 countForObject: n1];
        NSInteger min = MIN(count1, count2);
        while (min-- != 0) {
            [intersection addObject: n1];
        }
    }
    return intersection;
}

- hossam.ghareb May 18, 2017 | Flag Reply
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import java.util.*;

public class Solution {
	public static void main (String args[]) {
	int [] array1 = new int[] {0, 1, 1, 2, 2, 5};
	int [] array2 = new int[] {0, 1, 2, 2, 2, 6};
		List<Integer> results = new Solution().solve(array1, array2);
		for (Integer i : results){
			System.out.print(i.intValue()+" ");
		}
	}
	private Map<Integer, Integer> map;
	public Solution(){
		this.map = new HashMap<>();
	}
	private void add(int value) {
		if (!this.map.containsKey(value)) {
			this.map.put(value, 1);
		}else{
			this.map.put(value, this.map.get(value)+1);
		}
	}
	public List<Integer> solve(int [] array1, int[] array2) {
		for (int i : array1)
			add(i);
		List<Integer> results = new LinkedList<>();
		for (int i : array2) {
			if (map.containsKey(i)) {
				if (map.get(i) > 1) {
					results.add(i);
					map.put(i, map.get(i)-1);
				}else{
					results.add(i);
					map.remove(i);
				}
				
			}
		}
		return results;
	}
}

- w.kinaan July 27, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public void intersect()
	{
		int a[] = {0,1,1};
		int b[] = {0,1,2,3,4,5,6};
		int max, min;
		
		Map <Integer, Integer> aMap = new HashMap<Integer, Integer>();
		Map <Integer, Integer> bMap = new HashMap<Integer, Integer>();
		
		for(int i=0;i<a.length;i++)
		{
			if(aMap.containsKey(a[i]))
				aMap.put(a[i], aMap.get(a[i])+1);
			else
				aMap.put(a[i], 1);
		}
		
		for(int j=0;j<b.length;j++)
		{
			if(bMap.containsKey(b[j]))
				bMap.put(b[j], bMap.get(b[j])+1);
			else
				bMap.put(b[j], 1);
		}
		
		for(int key : aMap.keySet())
		{
			if(bMap.containsKey(key))
			{
				max = Math.max(aMap.get(key), bMap.get(key));
				min = Math.min(aMap.get(key), bMap.get(key));
				if(max==min)
					System.out.print(key + " ");
				else
					for(int m = 1;m<=min;m++)
						System.out.print(key + " ");			
			}
		}
	}

- Nitesh August 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

function findIntersection(arA, arB) {
    var pivot = arA.length > arB.length ? arB : arA;
    var baseComparator = arA.length > arB.length ? arA : arB;
    var intersection = [];

    for(var i = 0; i < pivot.length; i++) {
        if(pivot[i] == baseComparator[i]) {
            intersection.push(pivot[i]);
        }
    }
    return intersection;
}

- CodeAddict October 15, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Python:

def interscetion(A,B):
	len1 = len(A)
	len2 = len(B)
	C = []

	if len1 == 0 or len2 == 0:
		return C

	dict1 = {}
	dict2 = {}

	for i in range(0,len1):
		if A[i] in dict1:
			dict1[A[i]] += 1
		else:
			dict1[A[i]] = 1

	for i in range(0,len2):
		if B[i] in dict2:
			dict2[B[i]] += 1
		else:
			dict2[B[i]] = 1


	for keys,values in dict1.items():
		if keys in dict2:
			dict2values = dict2[keys]
			if dict2values <= values:
				temp = dict2values
			else:
				temp = values


			for i in range (0,temp):
				C.append(keys)

	return C

C = interscetion(A,B)
print C

- koolkhush February 12, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
-2
of 2 vote

# Python code

A = [0,1,1,2,2,5]
B = [0,1,2,2,2,6]
a = set(A).intersection(B)
print a

- Gautam Pai November 21, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

but set(A) will exclude duplicates

- megido December 07, 2015 | Flag


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