Amazon Interview Question Software Engineer in Tests




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hi any one attended the test on that day........and gone thru some rounds of interview.....please post the questions here....

Thanks,
Pavan

- Anonymous on October 13, 2009 | Flag Reply
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int a=9999999,sum=10;
while (sum>9)
{
sum=0;
while(a>0)
{
sum=sum+a%10;
a=a/10;
}
a=sum;
}

- Sulabh on October 13, 2009 | Flag Reply
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of 0 votes

sulabh++;

- Anonymous on November 05, 2009 | Flag
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int add_digit(int k){
  while(k>9){
    k = k/10 + k%10
  }

return k;

}

- Wee on October 14, 2009 | Flag Reply
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Excellent!!

- Saurabh Shah on February 22, 2010 | Flag
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//finds the sum of digits of a number
#include<cstdio>

int addDigits(int no) {
	int negFlag=0;
	int sum=0;
	if(no<0) {
		negFlag=1;
		no=-no;
	}
	while(no>0) {
		sum+=no%10;
		no/=10;	
	}
	return sum;
}
int main() {
	printf("%d : %d ",567,addDigits(567));
	printf("%d : %d ",5,addDigits(5));
	printf("%d : %d ",0,addDigits(0));
	printf("%d : %d ",-567,addDigits(-567));
}

- wolverine on October 18, 2009 | Flag Reply
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@wolverine ur solution doesnot return single digit in the end.

- mee on October 24, 2009 | Flag Reply
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Hi, this is simple...

int n = 65; //answer is 2

int sum_till_one_digit = n % 9;

- st0le on October 30, 2009 | Flag Reply
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1
of 1 vote

There is one trick in this solution -- If it returns 0, answer is actually 9. Need to handle n=0 case separately.
e.g. 999 & 9 == 0, but sum_to_1_digit is actually 9.

- Maverick on November 08, 2009 | Flag
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Hi, this is simple...

int n = 65; //answer is 2

int sum_till_one_digit = n % 9;

- st0le on October 30, 2009 | Flag Reply
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Hi, this is simple...

int n = 65; //answer is 2

int sum_till_one_digit = n % 9;

- st0le on October 30, 2009 | Flag Reply
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what abt n = 63?
It should be 9 and not 0

- chandra on January 05, 2010 | Flag
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1
of 1 vote

(n-1)%9+1

- satheesh on January 21, 2010 | Flag
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@ st0le: your solution works. wow

- Anonymous on November 08, 2009 | Flag Reply
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int addnum(int a)
{int temp =a;
int res = 0;
char flag = 1;
do
{
res = res + temp % 10;
temp = temp/10;
if(temp == 0 && res <= 9)
{
flag=0;
}
if(temp == 0 && res > 9)
{
temp=res;
res=0;
}

printf("temp :%d res :%d \n\n",temp,res);

}while(flag);
printf( "final ans :%d \n\n",res);
return res;
}

- umesh on November 15, 2009 | Flag Reply
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int test(int num)
{
int sum=0;
while(num>0)
{
sum+=num%10;
if(sum>9)
sum=sum-9;
num=num/10;
}
return sum;
}

- gkishan on December 02, 2009 | Flag Reply
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// a recursive solution
unsigned int SumDigits(unsigned int number)
{
	if (number >= 0 && number <= 9)
	{
		return number;
	}	
	return (number % 10) + SumDigits(number / 10);
}

unsigned int SumDigitsTillSingleDigit(unsigned int number)
{
	if (number >= 0 && number <= 9)
	{
		return number;
	}
	return SumDigitsTillSingleDigit(SumDigits(number));
}

- ikr on December 17, 2009 | Flag Reply
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of 0 vote

Can anyone please confirm on this code?

public static long addNumber (long n)
{
long sum = 0;
long asum = 0;
while (n>0)
{
sum = sum + n%10;
n/=10;
}
if (sum > 9 )
sum = addNumber(sum);
return sum;
}

- chandra on January 05, 2010 | Flag Reply
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0
of 0 vote

int sum = 0;
int sumDigits( int n)
{
while (n>0){
sum = sum + n%10;
n/=10;
}
return sum;
}

- Mak on January 09, 2010 | Flag Reply
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of 0 votes

This will not return sum as a single digit.
Ex. if n = 364, the above program would return 13; but, it should actually return 4.

Extending the above code,

int sum = 0;
void sumDigits(int n)
{
while (n>0){
sum = sum + n%10;
n/=10;
}
if(sum/10 > 0)
sumDigits(sum);
else
printf("%d", sum);
}

- HR on February 04, 2010 | Flag
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0
of 0 vote

a = 123;
while(a != 0){

num = a%10;
sum = sum + num;
a = a/10;
}
System.out.println("a : "+ sum);

output: 6

- Anonymous on April 04, 2010 | Flag Reply
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0
of 0 vote

#include<stdio.h>

int main(void) {
	int num;
	scanf("%d",&num);
	if(!num) {
		printf("sum of digits : %d\n",num);
		return 0;
	}
	num=num % 9;
	printf("sum of digits : %d\n",num?num:9);
	return 0;
}

- puri on March 02, 2012 | Flag Reply
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0
of 0 vote

int findSum (int n) {
  int sum = 0;
  while (n>0) {
    sum += n%10;
    n /= 10;
  }
  if (sum > 9) 
    return findSum (sum);
  else 
    return sum;
}

- Anonymous on February 20, 2013 | Flag Reply


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