## Amazon Interview Question

Country: United States

Comment hidden because of low score. Click to expand.
1
of 3 vote

``````import java.util.*;
public class CareerCup{
public static void printPairs(int[] a){
HashSet<Integer> hs = new HashSet<Integer>();
for(Integer i:a){
}

for(int i=0; i<a.length - 1; i++){
for(int j=i+1; j<a.length; j++){
int sum = a[i] + a[j];
if(hs.contains(sum)){
System.out.println("{ " + a[i] + "," + a[j] + " }");
}
}
}
}

public static void main(String args[]){
int[] input = {10, 1, 3, 5, 9, 25, 11};
printPairs(input);
}
}``````

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0
of 0 vote

Comment hidden because of low score. Click to expand.
0
of 2 vote

1st Simple solution is:

``````step 1 - sort the array
step 2 - for i=0 to len(arr)
for j=i to len(arr)
num = arr[i]*arr[j]
if(search(num,arr))
store the pair(i,j) //use two dimensional array
return allStoredPair``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

package CareerCup;

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;

public class Solution {
public static void main(String[] args) {
int[] A = { 11, 2, 3, 14, 5 };
Map<Integer, Set> pairs = new HashMap<Integer, Set>();
for (int i : A) {
pairs.put(i, new HashSet<Integer>());
}

for (int i = 0; i < A.length; i++) {
for (int j = A.length - 1; j >= 0; j--) {

if (A[i] <= A[j]){

if (pairs.get(A[j]).contains(A[j] - A[i]) && pairs.containsKey(A[i])
&& pairs.containsKey(A[i] - A[j])) {

System.out.println((A[j] + " " + (-A[j] + A[i])));
}
}
}
}
}

i couln'd remove duplicates though :(

Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void Test()
{
int inputArray[] = { 1, 3, 2, 6, 9, 8, 7 };

System.out.print("Input Array : ");
for(int x : inputArray)
System.out.print(x + " ");

System.out.print("\n");

for(int i=0 ; i<inputArray.length; i++)
{
for(int j = i+1; j<inputArray.length; j++ )
{
for(int k=0; k<inputArray.length; k++)
{
if (k != i && k != j)
{
if( inputArray[i] + inputArray[j] == inputArray[k] )
{
System.out.println( inputArray[i] + "+" + inputArray[j] + "=" + inputArray[k] );
}
}
}
}
}
}

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````public static void Test()
{
int inputArray[] = { 1, 3, 2, 6, 9, 8, 7 };

System.out.print("Input Array : ");
for(int x : inputArray)
System.out.print(x + "  ");

System.out.print("\n");

for(int i=0 ; i<inputArray.length; i++)
{
for(int j = i+1; j<inputArray.length; j++ )
{
for(int k=0; k<inputArray.length; k++)
{
if (k != i && k != j)
{
if( inputArray[i] + inputArray[j] == inputArray[k] )
{
System.out.println( inputArray[i] + "+" + inputArray[j] + "=" + inputArray[k] );
}
}
}
}
}
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

It can be solved in nLogn
Below are the steps
1. Make hashmap
2. sort the array
3. Use 2 pointers approach

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;

/*
* Given an array of numbers, print all pair (2) of numbers in the array
* if sum also present in the array
*/
public class ArrayPairSumInArray {

public static void main(String[] args) {
int[] iArrays = {1, 3, 2, 6, 9, 8, 7 };
Arrays.sort( iArrays );
Set<Integer> set = new HashSet<Integer>();
for( int num: iArrays ){
}
for( int i = 2; i < iArrays.length; i++ ){
for( int j = i - 1; j > 0; j-- ){
if( set.contains(iArrays[i] - iArrays[j]) && (iArrays[i]-iArrays[j]) < iArrays[j]){
System.out.println("[ " + ( iArrays[i] - iArrays[j]) + "+" + iArrays[j] + " ] = " + iArrays[i]);
}
}
}
}

}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

sorting nlogn algo using binary search with two for loops that is n^2logn

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````public class Resolver {
public static IEnumerable <Tuple<int, int>> GetPairs(IEnumerable<int> numbers) {
var array = numbers.ToArray ();
if (array.Length < 2) // we don't have pairs!!
yield break;

Array.Sort (array);

if (array.Length == 3) {
if (array [0] + array [1] == array [2])
yield return new Tuple <int, int> (array [0], array [1]);
// There are no other possibilities due to the fact that the array is sorted!
yield break;
}

var hSet = new HashSet <int> (array);

var lastPosition = array.Length - 1;

for (var i = 0; i < lastPosition; ++i) {
// break condition!
if ((array [i] < 2 ? 2 : array [i] << 1) > array [lastPosition]) {
yield break; // we found a situation where we would not find the sum in our array anylonger!
}

for (var j = i + 1; j < array.Length; ++j) {
var sum = array [i] + array [j];

if (hSet.Contains (sum)) {
yield return new Tuple <int, int> (array [i], array [j]);
}
}
}
}
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````import java.io.BufferedReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
public class PairsWithSumInArray {
public static void main(String[] args) throws IOException{
String [] str = br.readLine().split(" ");
int size = str.length;
int [] numbers = new int[size];
for(int i=0; i<size; i++){
numbers[i] = Integer.parseInt(str[i]);
}
List<Triplet> triplets = getAllTriplets(numbers);
Iterator itr = triplets.iterator();
while (itr.hasNext()){
Triplet triplet = (Triplet) itr.next();
System.out.println(triplet+", value("+numbers[triplet.getA()]+", "+numbers[triplet.getB()]+", "+numbers[triplet.getC()]+")");
}
}
public static List<Triplet>  getAllTriplets(int [] array){
List<Triplet> triplets = new ArrayList<>();
HashMap<Integer, Integer> hashMap = new HashMap<>();
int size = array.length;
for(int i=0; i<size; i++){
hashMap.put(array[i], i);
}
for(int i=0; i<size-2; i++){
for(int j=i+1; j<size; j++){
int a = array[i];
int b = array[j];
if(hashMap.containsKey(a+b)){
triplets.add(new Triplet(i, j, hashMap.get(a+b)));
}
}
}
return triplets;
}
}
class Triplet{
private int a;
private int b;
private int c;

public Triplet(int a, int b, int c) {
this.a = a;
this.b = b;
this.c = c;
}

public int getA() {
return a;
}

public int getB() {
return b;
}

public int getC() {
return c;
}

@Override
public String toString() {
return "Index"+"("+a+","+b+","+c+")";
}``````

}

Comment hidden because of low score. Click to expand.
-1
of 3 vote

This can be solved in O(n^2) time complexity and O(n) space complexity.
1) Create a hashmap with all the elements in the array
2) Use two for loops to get the sum for each pairs and check if the sum is present in the hashmap.

Can anyone provide feedback on how to improve this algorithm?

Name:

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