Adobe Interview Question Testing / Quality Assurances


Country: United States
Interview Type: Phone Interview


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9
of 9 vote

int secondMaxEle(int a[], int len) {
    int max = INT_MIN, secondMax = INT_MIN, i;
    for(i = 0; i < len; i++) {
        if(a[i] > max) {
            secondMax = max;
            max = a[i];
        }
        else if(a[i] > secondMax && a[i] < max)
            secondMax = a[i];
    }
    return secondMax;
}

- cooldaa on January 30, 2012 | Flag Reply
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0
of 0 vote

public int secondMaxElem(int[] ary}
{
if (ary == null || ary.size == 0 || ary.size == 1) return ERROR;

int max[] = new int[2];
max[0] = a[0]; max[1] = a[1];
for (int i = 2; i < ary.size; i++)
{
if (ary[i] > max[0] || ary[i] > max[1]
{
max[0] < max[1] ? max[0] = ary[i] : max[1] = ary[i];
}
}
return max[0] < max[1] ? max[0] : max[1];
}

- sd on January 28, 2012 | Flag Reply
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0
of 0 votes

why not array.size<=1 return ERROR

- anonymous on January 28, 2012 | Flag
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0
of 0 votes

if the interviewer ask u to extend the code to get the Kth of that array,what will u do?

- anonymous on January 28, 2012 | Flag
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0
of 0 votes

use heap to extend..

- abhishekatuw on January 29, 2012 | Flag
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0
of 0 votes

usage of heap will destroy the basic premise of "Single Loop".

- Anonymous on January 29, 2012 | Flag
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0
of 0 votes

This is the stupidest code segment I've ever seen!

int max[] = new int[2];
max[0] = a[0]; max[1] = a[1];
for (int i = 2; i < ary.size; i++)

really??

- codemonkey on February 20, 2012 | Flag
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0
of 0 vote

int secondMaxEle(int a[], int len) {
    int max = a[0], secondMax = a[0];
    for(int i = 1; i < len; i++) {
        if(a[i] > max) {
            secondMax = max;
            max = a[i];
        }
        else if(a[i] > secondMax)
            secondMax = a[i];
    }
    return seconMax;
}

- cooldaa on January 29, 2012 | Flag Reply
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0
of 0 votes

{5,4,3,2,1} return will be 5....

- Anonymous on January 30, 2012 | Flag
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0
of 0 votes

{5,4,3,2,1} return will be 5....

- Anonymous on January 30, 2012 | Flag
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0
of 0 votes

Thanks for pointing the bug. The correction is that I initialize both max and secondMax to the lowest possible integer value and then traverse the loop from the first element.

int secondMaxEle(int a[], int len) {
    int max = INT_MIN, secondMax = INT_MIN, i;
    for(i = 0; i < len; i++) {
        if(a[i] > max) {
            secondMax = max;
            max = a[i];
        }
        else if(a[i] > secondMax && a[i] < max)
            secondMax = a[i];
    }
    return secondMax;
}

- cooldaa on January 30, 2012 | Flag
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0
of 0 votes

Failing in case of arr[] = {1,1};

- Devesh on January 30, 2012 | Flag
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0
of 0 votes

Failing in case of arr[] = {1,1};

- Devesh on January 30, 2012 | Flag
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0
of 0 votes

What is the complexity??

- Anonymous on April 20, 2012 | Flag
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0
of 0 vote

int a[5]={2,5,4,6,7};
    int i,j;
    
    for(i=0;i<5;i++)
    {
                    int j=i+1;
             if(a[i]>a[j]){
                           int temp=a[j];
                           a[j]=a[i];
                           a[i]=temp;       
                    
                    }
                    }
pf("%d",a[1]);

- arun kumar on January 30, 2012 | Flag Reply
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0
of 0 vote

int len=100;
int a[len] = {3,7,8,2,9}
int max=a[0], smax=a[1];

//Getting max & second max between first 2 numbers
if(a[0]<a[1])
{
temp = smax; //maximum number
smax=a[1]; //second max number
max=temp;
}
//Comparing max, smax & swapping it when required in a loop
for(i=2;i<len;i++)
{
    if(a{i}>max)
    {    
           smax=max;   //bcoz max is already greater than smax
           max=a[i];
    }
    else if(a[i]>smax)
    {
     smax=a[i];
     }
}

- rajesh.savvy on February 29, 2012 | Flag Reply
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0
of 0 vote

#include<stdio.h>
#include<stdlib.h>
int select(int a[],int left,int right,int k)
{
if(left>=right) return a[left];
int i=left+1;
int j=right;
int p=a[left];
int t;
while(j>=i)
{
while(a[i]>p)i++;
while(a[j]<p)j--;
if(i>=j)break;
else{
t=a[i];
a[i]=a[j];
a[j]=t;
}
}
if(j-left+1==k)
return p;
t=a[left];
a[left]=a[j];
a[j]=t;
if(j-left+1<k)
return select(a,j+1,right,k-j+left-1);
else
return select(a,left,j-1,k);
}
int main() {
int a[20],k,n,i,d;
printf("Enter no of elements\n");
scanf("%d",&n);
printf("Enter n elements\n");
for(i=0;i<n;i++)
scanf("%d",&a[i]);
printf("Enter the value of k\n");
scanf("%d",&k);
d=select(a,0,n-1,k);
printf("Kth largest element is %d\n",d);
return 0;
}

- narayan kunal on March 05, 2012 | Flag Reply
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0
of 0 vote

public class Second_largest {
public static void main(String[] args) {
int array[]={5,4,3,2,1};
int max1=array[0];
int max2=array[1];
if(array.length<2){
System.out.println("not able to find second largest number");
}
else{
for(int i=1;i<array.length;i++){
if(array[i]>max1){
max2=max1;
max1=array[i];
}

}
System.out.println(max2);}
}

- Anonymous on July 03, 2012 | Flag Reply
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0
of 0 vote

sorry mistake on above code
public class Second_largest {
public static void main(String[] args) {
int array[]={1,2};
if(array.length<2){
System.out.println("not able to find second largest number");
return;
}
int max1=array[0];
int max2=array[1];


for(int i=1;i<array.length;i++){
if(array[i]>max1){
max2=max1;
max1=array[i];
}

}
System.out.println(max2);
}

}

- Anonymous on July 03, 2012 | Flag Reply
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0
of 0 vote

In a given array if you are trying to get the nth largest element.

Than take the nth node from the last as pivot element and than apply the partition logic of quick sort. which will get the nth element from the last in place. Left subarray (unsorted) will be the values less than the element and right subarray(unsorted) will be the values greater than the element.
But you will get the nth largest element from the last in the array. :)

- Reev on September 17, 2012 | Flag Reply
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0
of 0 vote

int len = 12;

int a[] = {2,183,8,2,1,91,23,100,4,2,52,530};
int max=a[0], smax=a[1];
int temp,i;


//Getting max & second max between first 2 numbers
if(a[0]<a[1])
{
temp = smax; //maximum number
smax=a[0]; //second max number
max=temp;
}

printf("%d\n",max);
printf("%d\n",smax);

//Comparing max, smax & swapping it when required in a loop
for(i=2;i<len;i++)
{
if(a[i] > max)
{
smax=max; //bcoz max is already greater than smax
max=a[i];
}
else if(a[i]>smax)
{
smax=a[i];
}
printf("%d\n",max);
printf("%d\n",smax);
}

printf("%d",smax);

- mani on November 17, 2012 | Flag Reply


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