Amazon Interview Question for Quality Assurance Engineers


Team: Amazon Wireless
Country: India
Interview Type: Written Test




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1
of 1 vote

void duplicates(int arr1[]) {
        int i = 0;
        int j = 1;
        int count = 1;
        while (i < arr1.length && j < arr1.length) {
            if (arr1[i] == arr1[j]) {
                count++;
                j++;
            } else {
                if (count > 1) {
                    System.out.println(arr1[i] + "  is  " + count);
                }
                count = 1;
                i = j;
                j++;
            }
        }

        if (count > 1) {
            System.out.println(arr1[i] + "  is  " + count);
        }
    }

- sarbjot singh July 26, 2015 | Flag Reply
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0
of 0 vote

public void test() {

int intArray[] = { 1, 2, 3, 3, 4, 5, 5, 5, 6 };
Set<Integer> setInterger = new HashSet<Integer>();

int countOfInteger = 1;
int repeatIntegerValue = 0;

for (int j : intArray) {
Boolean bool = setInterger.add(j);
if (!bool) {
repeatIntegerValue = j;
countOfInteger++;
} else if (bool && countOfInteger > 1) {
System.out.println(" value " + repeatIntegerValue + " count "
+ countOfInteger);
countOfInteger = 1;
}
}

}

- Pankaj July 25, 2015 | Flag Reply
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0
of 0 vote

public void test() {

		int intArray[] = { 1, 2, 3, 3, 4, 5, 5, 5, 6 };
		Set<Integer> setInterger = new HashSet<Integer>();

		int countOfInteger = 1;
		int repeatIntegerValue = 0;

		for (int j : intArray) {
			Boolean bool = setInterger.add(j);
			if (!bool) {
				repeatIntegerValue = j;
				countOfInteger++;
			} else if (bool && countOfInteger > 1) {
				System.out.println(" value " + repeatIntegerValue + " count "
						+ countOfInteger);
				countOfInteger = 1;
			}
		}

}

- Anonymous July 25, 2015 | Flag Reply
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0
of 0 vote

Code uses BST and uses Log(n) time. Can be done in contant time using hash tables, but will need extra space.

class RepeatedCount{

    private static class BST{

        private static class Node{
            private int value;
            private int count;
            private Node left;
            private Node  right;

            private Node(int val){
                value = val;
                count = 1;
            }

        }
        private Node root = null;
        private void put(int val){
            root = put(root, val);
        }

        private Node put(Node rt, int val){
            if(rt == null){
                Node node = new Node(val);
                return node;
            }
            if(val < rt.value){
               rt.left = put(rt.left, val);
            }
            else if(val > rt.value){
                rt.right = put(rt.right, val);
            }
            else if(val == rt.value){
                rt.count++;
            }
            return rt;
        }
        private void traverse(){
            traverse(root);
        }
        private void traverse(Node rt){
            if(rt == null) return;
            traverse(rt.left);
            System.out.println("Value:" + rt.value + " present " + rt.count + "times");
            traverse(rt.right);
        }
    }

    public static void printRepeats(Integer[] arr){
        System.out.println("printing repeats");
        BST bst = new BST();
        for(int x:arr){
            bst.put(x);
        }
        bst.traverse();
    }
}

- blurred July 26, 2015 | Flag Reply
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1
of 3 votes

How can you do in log(n) time if you are reading all the data linearly (O(n) :)

- AD July 27, 2015 | Flag
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1
of 1 vote

How can you write this algo in logn time when you have to read entire array to create BST in On time.

- AD July 27, 2015 | Flag
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0
of 0 votes

you are reading all values in linear time
+
for each item read, you are going through each node of "linked list" (not a tree - array is sorted already and you are not creating balanced binary tree here)

Worst case is O(n^2)

- kn July 30, 2015 | Flag
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0
of 0 votes

worst case is O(n^2)
avg case O(n log k)

- kn July 30, 2015 | Flag
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0
of 0 vote

public static void findReps(int array[])
	{
		int prev = -1;
		int counter = 1;
		for( int i : array)
		{
			if( i != prev )
			{
				if(counter > 1)
				{
					System.out.println("bla bla " + prev + " bla " + counter);
					counter = 1;
				}
			}
			else
			{
				++counter;
			}
			prev = i; 
		}
		
		//check last item:
		if(counter > 1)
		{
			System.out.println("bla bla " + prev + " bla " + counter);
		}

	}

- ohadr.developer July 26, 2015 | Flag Reply
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0
of 0 vote

int J = 1.//initialize


for i=0; i==lenght of array -1;i++

{

if array[i] == array[i+1]

J = j+1

if i==lenght of array -1 and j>1 // for the last-1 item if true

array[i] is repeated 'j' times

elseif j>1

array[i] is repeated 'j' times

j = 1.

}

- justpseudocoded!! July 27, 2015 | Flag Reply
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0
of 0 vote

int J = 1.//initialize


for i=0; i==lenght of array -1;i++

{

if array[i] == array[i+1]

J = j+1

if i==lenght of array -1 and j>1 // for the last-1 item if true

array[i] is repeated 'j' times

elseif j>1 

array[i] is repeated 'j' times

j = 1.

}

- justpseudocoded!! July 27, 2015 | Flag Reply
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0
of 0 vote

int J = 1.//initialize


for i=0; i==lenght of array -1;i++

{

if array[i] == array[i+1]

J = j+1

if i==lenght of array -1 and j>1 // for the last-1 item if true

array[i] is repeated 'j' times

elseif j>1 

array[i] is repeated 'j' times

j = 1.

}

- Anonymous July 27, 2015 | Flag Reply
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0
of 0 vote

public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
int J = 1 ;//initialize
int i= 0;
int Array[] = new int[]{1,2,2,3,3,3,4,5,6,7,7,8,9,9};

for( i=0; i< (Array.length-1); i++) 
{
if(Array[i] == Array[i+1]) 
   {  J = J+1;
       if( (i+1) == (Array.length-1)) // for the last-1 item if true
  {
         System.out.println(Array[i] + " is repeated" + J + " times " );
  }
       continue;
    }

    if(J>1 ) 
       {  
        	System.out.println(Array[i] + " is repeated" + J + " times " );
         J = 1;
        }
 }
}

- Anonymous July 28, 2015 | Flag Reply
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0
of 0 vote

result is
2 is repeated2 times
3 is repeated3 times
7 is repeated2 times
9 is repeated2 times

- Anonymous July 28, 2015 | Flag Reply
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0
of 0 vote

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Ideone
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
int J = 1 ;//initialize
int i= 0;
int Array[] = new int[]{1,1,1,2,3,4,5,6,7,7,8,9,9,10,11,12,13,14,15,15,15,16,16};

for( i=0; i< (Array.length-1); i++) 
{
if(Array[i] == Array[i+1]) 
   {  
   	   J = J+1;
       if( (i+1) == (Array.length-1)) // for the last-1 item if true
  {
         System.out.println(Array[i] + " is repeated" + J + " times " );
  }
       continue;
    }

    if(J>1 ) 
       {  
        	System.out.println(Array[i] + " is repeated" + J + " times " );
         J = 1;
        }
 }
}
}

- actualcode July 29, 2015 | Flag Reply
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0
of 0 votes

hello dear what if input is {1111,2,3,4,11,5,6,7}
i think this will print result
1 is repeated 4 times
1 is repeated 2 times

- dhiru August 10, 2015 | Flag
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0
of 0 votes

{{hello dear what if input is {1111,2,3,4,11,5,6,7} 
i think this will print result
 1 is repeated 4 times
1is repeated 2 times
plzz correct me if i m wrong .

- dhiru August 10, 2015 | Flag
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0
of 0 vote

result
1 is repeated3 times
7 is repeated2 times
9 is repeated2 times
15 is repeated3 times
16 is repeated2 times

- actualcode July 29, 2015 | Flag Reply
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of 0 vote

function repeatPrint(str) {
  
  var count = 0;
  var arry = str.split(''); 
  for ( var i = 1; i < arry.length; ++i ) {
      
      if ( arry [i-1] == arry [i] ) {
         ++count;
      }
      else if ( count != 0 ) {
          console.log( arry [i-1]+ " repeated " + (count+1) + " times" );
          count = 0;
      }
  }


  if ( count != 0 ) {
     console.log( arry [i-1] + " repeated is " + (count+1) + " times" );
  }
  
}

- stephenpince July 29, 2015 | Flag Reply
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of 0 vote

@Test
	public void testNumberCount() {
		System.out.println("Enter number comma seperated to count ..");
		Scanner sc = new Scanner(System.in);
		String str = sc.nextLine();
		System.out.println("entered is :: " + str);
		String[] strArr = str.split(",");
		List numList = new ArrayList(Arrays.asList(strArr));
		List repeatList = new ArrayList();
		for(int i=0; i < numList.size(); i++) {
			if(Collections.frequency(numList,  numList.get(i)) > 1 && !repeatList.contains(numList.get(i))) {
				repeatList.add(numList.get(i));
				System.out.println("The number " + numList.get(i) + " repeats " + Collections.frequency(numList,  numList.get(i)) + " times..");
			}
		}
	}

- vishyan August 04, 2015 | Flag Reply
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of 0 votes

Array can be sorted or random, For such condition it may be solution like:

var data = [2,9,2,3,4,5,4,5,6,6,7,7,4,7,6,7,5,6,5,7,7,9,8,8,9,2,9,9];
var count = {};
for (var i = 0; i <= data.length-1; ++i) {
var key = data[i];
if(!count.hasOwnProperty(key)){
var num =0;
for(var j=i; j<=data.length-1; j++){
if(data[i]==data[j]){
num += 1;
}
}
count[key]=num;
}
}
console.log(count)

- harikesh Yaadv August 28, 2015 | Flag
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of 0 votes

Array can be sorted or random, For such condition it may be solution like:

var data = [2,9,2,3,4,5,4,5,6,6,7,7,4,7,6,7,5,6,5,7,7,9,8,8,9,2,9,9];
var count = {}; 
for (var i = 0; i <= data.length-1; ++i) {
	var key = data[i];
	if(!count.hasOwnProperty(key)){
		var num =0;
		for(var j=i; j<=data.length-1; j++){
		    if(data[i]==data[j]){
		     	num += 1;
		    }
	    }
        count[key]=num;
    }   
}
console.log(count)

- harikesh Yaadv August 28, 2015 | Flag
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0
of 0 votes

var data = [2,9,2,3,4,5,4,5,6,6,7,7,4,7,6,7,5,6,5,7,7,9,8,8,9,2,9,9];
var count = {}; 
for (var i = 0; i <= data.length-1; ++i) {
	var key = data[i];
	if(!count.hasOwnProperty(key)){
		var num =0;
		for(var j=i; j<=data.length-1; j++){
		    if(data[i]==data[j]){
		     	num += 1;
		    }
	    }
        count[key]=num;
    }   
}
console.log(count)

- harikesh Yaadv August 28, 2015 | Flag
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of 0 votes

var data = [2,2,3,4,5,4,5,6,6,7,7,4,7,6,7,5,6,5,7,7,9,8,8,9,2,9,9];
var count = {}; 
for (var i = 0; i <= data.length-1; ++i) {
	var key = data[i];
	if(!count.hasOwnProperty(key)){
		var num =0;
		for(var j=i; j<=data.length-1; j++){
		    if(data[i]==data[j]){
		     	num += 1;
		    }
	    }
        count[key]=num;
    }   
}
console.log(count)

- harikesh Yaadv August 28, 2015 | Flag
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of 0 vote

This is done in O(n) but uses extra space(HashMap)

public static void main(String[] args) {

int arr[] = {1,2,3,4,5,5,5,6,7,7};
Map<Integer, Integer> map = new HashMap<Integer, Integer>();


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


Iterator<Integer> it = map.keySet().iterator();
while (it.hasNext()) {
Integer integer = (Integer) it.next();
int count = map.get(integer);
if(count > 1){
System.out.println(integer + " is repeated "+ count+" times");
}
}
}

- Infinite Possibilities August 06, 2015 | Flag Reply
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of 0 vote

We can take advantage of the fact that some patterns repeat. So we can check in chunks of say 3 or 5 etc. For example for chunk of 3 if arr[0]==arr[2] we can directly say that the number repeats at least 3 times and skip i to i+3, if not we can go linear. The performance may be a little better than O(N)

- Asif Garhi August 12, 2015 | Flag Reply
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of 0 votes

public class PrintRepeatPattern {
	private static int[] arr = {1,1,1,1,2,3,4,5,5,6,7,7,7,8,8,8};
	public static void main(String[] args) {
		doPrint(3);
	}
	private static void doPrint(int skip) {
		int cnt = 1, i = 0;
		for(; i < arr.length; ) {
			if(i > 0 && arr[i] != arr[i-1]) {
				print(arr[i-1],cnt);
				cnt = 1; 
			}
			if(chunk(i,skip)) {
				cnt = cnt + skip - 1;
				i += skip;
			} else {
				if(arr[i] == arr[i - 1]) ++cnt;
				++i;
			}
		}
		print(arr[i-1],cnt);
	}
	
	private static boolean chunk(int i, int skip) {
		return (i + skip - 1 < arr.length && arr[i] == arr[i + skip - 1]);
	}
	
	private static void print(int number, int cnt) {
		if(cnt > 1) {
			System.out.printf("%d occurs %d times\n",number,cnt);
		}
	}
}

- Asif Garhi August 12, 2015 | Flag
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0
of 0 vote

i think in python we can do like this....

my = [1,2,3,3,4,5,5,5,6]
b = set(my)
for m in b:
    temp = 0
    for k in my:
        if k == m :
            temp = temp + 1
        else:
            pass
    print 'count for {0} is  : {1}'.format(m, temp)

- Sid August 17, 2015 | Flag Reply
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of 0 vote

public class RepeatedElements {
	public static void main(String args[]) {
		int[] array = { 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5 };
		find(array);
	}

	public static void find(int[] array) {
		int i = 0;
		int count = 1;
		while (i < array.length) {
			if (((i+1) != array.length) && array[i] == array[i + 1]) {
				count++;
			} else {
				System.out.println(array[i] + " is repeated " + count
						+ " times");
				count = 1;
			}
			i++;
		}
	}

}

- Srumith August 19, 2015 | Flag Reply
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of 0 vote

public class RepeatedElements {
	public static void main(String args[]) {
		int[] array = { 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5 };
		find(array);
	}

	public static void find(int[] array) {
		int i = 0;
		int count = 1;
		while (i < array.length) {
			if (((i+1) != array.length) && array[i] == array[i + 1]) {
				count++;
			} else {
				System.out.println(array[i] + " is repeated " + count
						+ " times");
				count = 1;
			}
			i++;
		}
	}

}

- Srumith August 19, 2015 | Flag Reply
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of 0 votes

public void findDuplicatesWithoutCollections(int[] intArray)
{
int uniqueIntLocator=0;
int arrayWalker=uniqueIntLocator+1;
int noOfTimesRepeated=1;

while ((uniqueIntLocator<=(intArray.length-1)) && (arrayWalker<=(intArray.length)))
{
if ((intArray.length==1) || (arrayWalker==intArray.length))
{
System.out.println(intArray[uniqueIntLocator] + ":" + "is not repeated");
uniqueIntLocator=arrayWalker;
arrayWalker++;
noOfTimesRepeated=1;
}
else
{
while ((arrayWalker<=(intArray.length-1)) && (intArray[uniqueIntLocator]==intArray[arrayWalker]))
{
noOfTimesRepeated++;
arrayWalker++;
}
if (noOfTimesRepeated>1)
{
System.out.println(intArray[uniqueIntLocator] + ":" + "repeated" + ":" + (noOfTimesRepeated));
uniqueIntLocator=arrayWalker;
arrayWalker++;
noOfTimesRepeated=1;
}
else
{
System.out.println(intArray[uniqueIntLocator] + ":" + "is not repeated");
uniqueIntLocator=arrayWalker;
arrayWalker++;
}
}
}




}

public void findDuplicatesWithoutCollections(String[] stringArray)
{
int uniqueIntLocator=0;
int arrayWalker=uniqueIntLocator+1;
int noOfTimesRepeated=1;

while ((uniqueIntLocator<=(stringArray.length-1)) && (arrayWalker<=(stringArray.length)))
{
if ((stringArray.length==1) || (arrayWalker==stringArray.length))
{
System.out.println(stringArray[uniqueIntLocator] + ":" + "is not repeated");
uniqueIntLocator=arrayWalker;
arrayWalker++;
noOfTimesRepeated=1;
}
else
{
while ((arrayWalker<=(stringArray.length-1)) && (stringArray[uniqueIntLocator]==stringArray[arrayWalker]))
{
noOfTimesRepeated++;
arrayWalker++;
}
if (noOfTimesRepeated>1)
{
System.out.println(stringArray[uniqueIntLocator] + ":" + "repeated" + ":" + (noOfTimesRepeated));
uniqueIntLocator=arrayWalker;
arrayWalker++;
noOfTimesRepeated=1;
}
else
{
System.out.println(stringArray[uniqueIntLocator] + ":" + "is not repeated");
uniqueIntLocator=arrayWalker;
arrayWalker++;
}
}
}

- Khasim Peera Durgam August 22, 2015 | Flag
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of 0 vote

public void findDuplicatesWithoutCollections(int[] intArray)
{
int uniqueIntLocator=0;
int arrayWalker=uniqueIntLocator+1;
int noOfTimesRepeated=1;

while ((uniqueIntLocator<=(intArray.length-1)) && (arrayWalker<=(intArray.length)))
{
if ((intArray.length==1) || (arrayWalker==intArray.length))
{
System.out.println(intArray[uniqueIntLocator] + ":" + "is not repeated");
uniqueIntLocator=arrayWalker;
arrayWalker++;
noOfTimesRepeated=1;
}
else
{
while ((arrayWalker<=(intArray.length-1)) && (intArray[uniqueIntLocator]==intArray[arrayWalker]))
{
noOfTimesRepeated++;
arrayWalker++;
}
if (noOfTimesRepeated>1)
{
System.out.println(intArray[uniqueIntLocator] + ":" + "repeated" + ":" + (noOfTimesRepeated));
uniqueIntLocator=arrayWalker;
arrayWalker++;
noOfTimesRepeated=1;
}
else
{
System.out.println(intArray[uniqueIntLocator] + ":" + "is not repeated");
uniqueIntLocator=arrayWalker;
arrayWalker++;
}
}
}




}

public void findDuplicatesWithoutCollections(String[] stringArray)
{
int uniqueIntLocator=0;
int arrayWalker=uniqueIntLocator+1;
int noOfTimesRepeated=1;

while ((uniqueIntLocator<=(stringArray.length-1)) && (arrayWalker<=(stringArray.length)))
{
if ((stringArray.length==1) || (arrayWalker==stringArray.length))
{
System.out.println(stringArray[uniqueIntLocator] + ":" + "is not repeated");
uniqueIntLocator=arrayWalker;
arrayWalker++;
noOfTimesRepeated=1;
}
else
{
while ((arrayWalker<=(stringArray.length-1)) && (stringArray[uniqueIntLocator]==stringArray[arrayWalker]))
{
noOfTimesRepeated++;
arrayWalker++;
}
if (noOfTimesRepeated>1)
{
System.out.println(stringArray[uniqueIntLocator] + ":" + "repeated" + ":" + (noOfTimesRepeated));
uniqueIntLocator=arrayWalker;
arrayWalker++;
noOfTimesRepeated=1;
}
else
{
System.out.println(stringArray[uniqueIntLocator] + ":" + "is not repeated");
uniqueIntLocator=arrayWalker;
arrayWalker++;
}
}
}

- Khasim Peera Durgam August 22, 2015 | Flag Reply
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of 0 vote

var data = [2,2,3,4,5,4,5,6,6,7,7,4,7,6,7,5,6,5,7,7,9,8,8,9,2,9,9];
var count = {};
for (var i = 0; i <= data.length-1; ++i) {
var key = data[i];
if(!count.hasOwnProperty(key)){
var num =0;
for(var j=i; j<=data.length-1; j++){
if(data[i]==data[j]){
num += 1;
}
}
count[key]=num;
}
}
console.log(count)

- harikesh yadav August 28, 2015 | Flag Reply
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of 0 vote
{{{ var data = [2,2,3,4,5,4,5,6,6,7,7,4,7,6,7,5,6,5,7,7,9,8,8,9,2,9,9]; var count = {}; for (var i = 0; i <= data.length-1; ++i) { var key = data[i]; if(!count.hasOwnProperty(key)){ var num =0; for(var j=i; j<=data.length-1; j++){ if(data[i]==data[j]){ num += 1; } } count[key]=num; } } console.log(count)}} - harikesh yadav August 28, 2015 | Flag Reply
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of 0 vote

var data = [2,2,3,4,5,4,5,6,6,7,7,4,7,6,7,5,6,5,7,7,9,8,8,9,2,9,9];
var count = {};
for (var i = 0; i <= data.length-1; ++i) {
var key = data[i];
if(!count.hasOwnProperty(key)){
var num =0;
for(var j=i; j<=data.length-1; j++){
if(data[i]==data[j]){
num += 1;
}
}
count[key]=num;
}
}
console.log(count)

- harikesh yadav August 28, 2015 | Flag Reply
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0
of 0 vote

var data = [2,2,3,4,5,4,5,6,6,7,7,4,7,6,7,5,6,5,7,7,9,8,8,9,2,9,9];
var count = {};
for (var i = 0; i <= data.length-1; ++i) {
var key = data[i];
if(!count.hasOwnProperty(key)){
var num =0;
for(var j=i; j<=data.length-1; j++){
if(data[i]==data[j]){
num += 1;
}
}
count[key]=num;
}
}
console.log(count)

- harikesh yadav August 28, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

var data = [2,2,3,4,5,4,5,6,6,7,7,4,7,6,7,5,6,5,7,7,9,8,8,9,2,9,9];
var count = {};
for (var i = 0; i <= data.length-1; ++i) {
var key = data[i];
if(!count.hasOwnProperty(key)){
var num =0;
for(var j=i; j<=data.length-1; j++){
if(data[i]==data[j]){
num += 1;
}
}
count[key]=num;
}
}
console.log(count)

- harikesh yadav August 28, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def rep(arr):
    a = dict()
    for el in arr:
        a[el] = a.get(el,0)+1
    for k,v in a.items():
        if v > 1:
            print "%d is repeated %d times"%(k,v)

print rep([1,2,3,3,4,4,5,5,5,6,7,7,8])

- parashar.002 January 14, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace consoleApp
{
class Program
{

static void Main(string[] args)
{
int[] array = new int[] {0,1,2,3,4,5,6,7,8,1,2,3,3,3,3,4,4,4,5,6,77,88,9,0,0,9,9 };
int[] array1 = new int[array.Length];
List<int> list = new List<int> (array.Length) ;
for (int i = 0; i < array.Length; i++ )
{
bool found = false;
for (int j = 0; j < i; j++ )
{
if(array[j] == array[i])
{
found = true;
break;
}
}
if (!found)
{
list.Add(array[i]);
}

}
array1 = list.ToArray();

for (int i = 0; i < array1.Length; i++)
{
Console.WriteLine(array1[i]);
}
RepeatNumebr(array,array1);

Console.ReadKey();
}

static void RepeatNumebr(int [] num, int [] unique)
{
int count = 0;
for (int i = 0; i < unique.Length;i++ )
{
count = 0;
for (int j = 0; j < num.Length;j++ )
{
if(unique[i] == num[j])
{
count++;
}
}
Console.WriteLine("This number {0} has repitition :{1}",unique[i],count);
}

}



}
}

- mike March 17, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

class Program
{

static void Main(string[] args)
{
int[] array = new int[] {0,1,2,3,4,5,6,7,8,1,2,3,3,3,3,4,4,4,5,6,77,88,9,0,0,9,9 };
int[] array1 = new int[array.Length];
List<int> list = new List<int> (array.Length) ;
for (int i = 0; i < array.Length; i++ )
{
bool found = false;
for (int j = 0; j < i; j++ )
{
if(array[j] == array[i])
{
found = true;
break;
}
}
if (!found)
{
list.Add(array[i]);
}

}
array1 = list.ToArray();

for (int i = 0; i < array1.Length; i++)
{
Console.WriteLine(array1[i]);
}
RepeatNumebr(array,array1);

Console.ReadKey();
}

static void RepeatNumebr(int [] num, int [] unique)
{
int count = 0;
for (int i = 0; i < unique.Length;i++ )
{
count = 0;
for (int j = 0; j < num.Length;j++ )
{
if(unique[i] == num[j])
{
count++;
}
}
Console.WriteLine("This number {0} has repitition :{1}",unique[i],count);
}

}

- Anonymous March 17, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#Python Solution
s = [1, 2, 3, 3, 4, 5, 5, 5, 6]

x = []
for i in range (0, len(s)):
    count = 0
    if s[i] not in x:
        x.append(s[i])
        for j in range (0, len(s)):
            if s[i] == s[j] :
                count = count + 1
        #print ("%s is repeated %d times" % (s[i],count))

e = "Salesforce is the best company to work fo"
d=e.lower()
z = []
for i in range (0, len(d)):
    count = 0
    if d[i] not in z:
        z.append(d[i])
        for j in range (0, len(d)):
            if d[i] == d[j] :
                count = count + 1
        if count == 1:
            
            print (d[i])
            break
        else:
            print (d[i], count)

- Prashant April 23, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Below code is working fine.

public class SampleTest {
	public static void main(String[] args) {
		try{
			int[] a  = {1,2,3,3,4};
			Map<Integer,Integer> n = new HashMap<>();
			for(int i=0;i<a.length;i++){
				int b = a[i];
				if(n.containsKey(b)){
					n.put(b, n.get(b)+1);
				} else {
					n.put(b, 1);
				}
			}
			System.out.println(n);
		} catch(Exception e) {
			System.out.println("Exception "+e);
		} 
	}

- Nandhis June 29, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Code in python : O(n)

A=[1,2,3,4,5,5,5,6,7,7]
count=0
A.append(0)

for i in range(0,len(A)):
	if(A[i]==A[i+1]):
		count=count+1
	else:
		if(count>0):
			print("{0} repeated {1} times".format(A[i],count+1))
		count=0

- geek_code July 30, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void main(String[] args) {
		// Sorted array
		int [] array = { 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5 };;
		int counter = 0;
		for( int i = 0; i < array.length; i++ ){
			if( ( counter != 0 && ( i + 1) < array.length && array[i] != array[i+1]) || ( counter != 0 && i == array.length -1 ) ){
				System.out.println( array[i] + " appread " + ++counter + " times");
				counter = 0;
			}else if( (i + 1) < array.length && array[i] == array[i+1]){
				++counter;
			}
		}
	}

- puttappa.raghavendra December 04, 2016 | Flag Reply


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