Amazon Interview Question for SDE1s


Team: WebStore
Country: India
Interview Type: Phone Interview




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10
of 10 vote

#include <iostream>
#include <algorithm>

using namespace std;
int multiply(int A[], int lenA, int B[], int lenB) {
	
	int len = max(lenA, lenB);
	
	int sum = 0;
	
	for (int i=0; i<len; i++) {
		sum += A[i%lenA]*B[i%lenB];
	}

	return sum;

}

int main(void) {
	
	int A[] = {1, 2, 3, 4, 5};
	int B[] = {2, 1};
		
	int lenA = sizeof(A)/sizeof(int);
	int lenB = sizeof(B)/sizeof(int);
	
	cout << multiply(A, lenA, B, lenB) << endl;
	
    return 0;
}

- aimjwizards March 20, 2013 | Flag Reply
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2
of 2 vote

public static int sumProduct(int[] a, int[] b) {
if(a == null || b == null)
return 0;
int length = a.length > b.length ? a.length : b.length;
int sumProduct = 0;
int i = 0;

while(i < length) {
sumProduct += a[ i % a.length ] * b[ i % b.length ];
i++ ;
}
return sumProduct;
}

- susantam25 March 20, 2013 | Flag Reply
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0
of 0 votes

public static int sumProduct(int[] a, int[] b) {
		if(a == null || b == null)
			return 0;		
		int length = a.length > b.length ? a.length : b.length;
		int sumProduct = 0;
		int i = 0;
		
		while(i < length) {
			sumProduct += a[ i % a.length ] * b[ i % b.length ];
			i++ ;
		}
		return sumProduct;
	}

- susantam25 March 20, 2013 | Flag
Comment hidden because of low score. Click to expand.
1
of 1 vote

int multiply(int A[], int B[])
{
    int lenA = sizeof(A)/sizeof(int);
    int lenB = sizeof(B)/sizeof(int);
    int sum = 0;
    int i = 0, j = 0;
    if(lenA > lenB)
    {
         for(; i < lenA; i++)
         {
             sum += A[i % lenA] * B[i % lenB];
         }
    }
    else if(lenB > lenA)
    {
        for(; i < lenB; i++)
        {
            sum += B[i % lenB] * A[i % lenA];
        }
    }
    else
    {
        for(; i < lenA; i++)
        {
            sum += A[i] * B[i];
        }
    }
    return sum;
}

- Anonymous March 20, 2013 | Flag Reply
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0
of 0 votes

Above answer is posted by me.

- Nitin Gupta March 20, 2013 | Flag
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0
of 0 votes

The way I see it, it should also work without the else part ( using any of the above two conditions.. i.e. lenA>=lenB or lenA<=lenB )..

- Anonymous March 20, 2013 | Flag
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0
of 0 vote

result = 0 ,j =0;
for(i=0; i<n ;i++)
{
if(sizeof(b)/b[0] == j && i < sizeof(a)/a[0])
j =0;

if(sizeof(b)/b[0] == i )
{
c[i] = a[i] * b[j];
j++;
}
else
{
c[i] = a[i] * b[i];
}
result = result + c[i];
}

- kk March 20, 2013 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Hey Nitin can you tell me more about the interview procedure , what all question they asked?

- kk March 20, 2013 | Flag Reply
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0
of 0 vote

#include <stdio.h>

int main(int argc, char *argv[])
{
    int a[5]={1,2,3,4,5},b[2]={2,1};
	int k=0,i=0,j=0;
    while(i<5)
	{
		k=k+(a[i]*b[j]);
		i++;
		j=i%2;
	
	}
	printf("%d",k);

	return 0;
}

- Satvik Vishnubhatta March 20, 2013 | Flag Reply
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0
of 0 vote

package algo;


public class ArrayMul {

public static double arrayMultiplication(int []arr1, int []arr2){
int len1= arr1.length;
int len2=arr2.length;
System.out.println(""+len1+""+len2);
double sum =0.0;
if(len1>len2){
for(int i=0,j=0;i<len1;i++){
sum = sum + arr1[i]*arr2[j];
j++;
if(j==len2){
j=0;
}
}
}else{
for(int i=0,j=0;i<len2;i++){
sum = sum + arr1[j]*arr2[i];
j++;
if(j==len1){
j=0;
}
}

}
return sum;
}


public static void main(String args[]){
int [] a= {2,1};
int [] b={1,2,3,4,5};
System.out.println("Sum :"+ArrayMul.arrayMultiplication(a, b));
}
}

- Krishna Kumar Tiwari March 20, 2013 | Flag Reply
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0
of 0 vote

package algo;


public class ArrayMul {

	public static double arrayMultiplication(int []arr1, int []arr2){
		int len1= arr1.length;
		int len2=arr2.length;
		System.out.println(""+len1+""+len2);
		double sum =0.0;
		if(len1>len2){
			for(int i=0,j=0;i<len1;i++){
				sum = sum + arr1[i]*arr2[j];
				j++;
				if(j==len2){
					j=0;
				}
			}
		}else{
			for(int i=0,j=0;i<len2;i++){
				sum = sum + arr1[j]*arr2[i];
				j++;
				if(j==len1){
					j=0;
				}
			}
			
		}
		return sum;
	}
	
	
	public static void main(String args[]){
		int [] a= {2,1};
		int [] b={1,2,3,4,5};
		System.out.println("Sum :"+ArrayMul.arrayMultiplication(a, b));
	}
}

- KK March 20, 2013 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

class Product {
    Product(int sum, int[] array) {
        this.sum = sum;
        this.array = array;
    } 

    final int sum;
    final int[] array;
}

Product arrayProduct(int[] a, int[] b) {
    int sum = 0;
    int[] array = new int[Math.max(a.length, b.length)];
    for (int i = 0; i < array.length; i ++) {
        array[i] = a[i % a.length] * b[i % b.length];
        sum += array[i];
    }   
    return new Product(sum, array);
}

- PP March 20, 2013 | Flag Reply
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0
of 0 vote

public static int multiplyAndSumArrays(int[] arrayA, int[] arrayB) {
        int result = 0;
        
        if (arrayA.length > arrayB.length) {
            int arrayBIndex = 0;
            for (int i = 0; i < arrayA.length; i++) {
                arrayBIndex = i % arrayB.length;
                result += arrayA[i] * arrayB[arrayBIndex];
            }
        }
        else {
            result = multiplyAndSumArrays(arrayB, arrayA);
        }
        
        return result;
    }

- Andre March 20, 2013 | Flag Reply
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0
of 0 votes

Correction to the if statement.

if (arrayA.length >= arrayB.length)

- Andre March 20, 2013 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

public int multiplyAndSumArrays()
        {
            List<int> lsA = new List<int>(new int[] { 1, 2, 3, 4, 5 });
            List<int> lsB = new List<int>(new int[] { 2, 1 });
            int k = 0;

            foreach (int i in lsA)
            {
                k = k + i * lsB[(lsA.IndexOf(i) % lsB.Count)];                
            }

            return k;
        }

- Chloe March 22, 2013 | Flag Reply
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0
of 0 vote

int[] arr1 = {1,2,3,4,5};
		int[] arr2 = {2,1};
		
		int sum = 0;
		
		for(int i = 0,j=0; i<arr1.length; i++,j++){
			
			sum = sum + arr1[i]*arr2[j];
			if(j == arr2.length-1)
				j = -1;		
		}
		
		System.out.println("Sum -- " + sum);

- Anonymous March 23, 2013 | Flag Reply
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0
of 0 vote

public static int arraySum(int[] a, int[] b) {
int sum = 0;
for (int i = 0; i < a.length; i++) {
sum += a[i] * b[i%b.length];
}
return sum;
}

- Anson March 24, 2013 | Flag Reply
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0
of 0 vote

public static int getSum(int[] a, int[] b) {
		int sum = 0, len1 = a.length, len2=b.length;
		for(int i=0; i<len1;i++) {
			sum += a[i]*b[i%len2];
		}
		
		return sum;
	}
	public static int prepare(int[] a, int[] b){
		if (a.length > b.length)
			return getSum(a,b);
		else
			return getSum(b,a);
	}

- lin March 28, 2013 | Flag Reply
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0
of 0 vote

class ProdArray {
    private static int prodOfArray( int[] A, int[] B) {
        int i, j, count=0;
        if (A.length < B.length) {
            int sum = prodOfArray(B, A);
            return sum;
        }
        else {
            for (i=0; i<A.length; i++) {
                j = i % B.length;
                count += A[i] * B[j];
                System.out.println(A[i] + " * " + B[j] + "\t= " + A[i] * B[j]);
            }
            return count;
        }
    }
    
    public static void main(String[] args) {
        int[] a = {2, 7, 8, 9};
        int[] b = {2, 13, 10};
        
        System.out.println(prodOfArray(a,b));
    }

}

- Chinna August 19, 2013 | Flag Reply


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