Uber Interview Question for Software Engineers


Country: United States




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

Your logic is wrong, you switched 3 and 4.
3. If length is odd then all the characters except one should have even occurrences.
4. If even, then all the characters should have even occurrences.

- Shivam Maharshi October 05, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

// ZoomBA
#|select ( mset(s.toCharArray) ) :: {  $.o.value % 2 != 0 }| <= 1

- NoOne October 06, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

1. Count the characters of the string (use a map, dictionary or something like that to map from a character to the num of ocurrences in the string)
2. check len of the string
3. If odd, check all characters has odd number of occurrences. If so, return True.
4. If even, check all characters has odd nomber of ocurrences but one. If so, return True.
5. None of 3, or 4, return False

- Anonymous October 05, 2016 | Flag Reply
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0
of 0 vote

import java.util.HashMap;


public class palindromecanbeformedornotfromgivenstring {
	public static void main(String[] args) {
		boolean result = false;
	  String str ="yakak";
HashMap<Character,Integer>hm= new HashMap<Character,Integer>();
	  int l1=str.length();
	  char[] c1=str.toCharArray();
	  for(char c:c1){
		  if(hm.containsKey(c)){
			  hm.put(c,hm.get(c)+1);
		  }
		  else{
			  hm.put(c,1);
		  }
	  }
	  if(l1%2==0){
		  for(char c2 : c1){
			  int value = hm.get(c2);
			  if(value%2==0)
				  result= true;
			  else
				  result=false;
		  }
	  }
	  else{
		  int oddcount=0;
		  for(char c2 : c1){
			  int value = hm.get(c2);
			  if(value%2==0)
				 continue;
			  else
				  oddcount++;
		  }
		  if(oddcount<2)
			  result= true;
		  else
			  result=false;
	  }
	  
	  if(result==true)
		  System.out.println("possible");
	  else
		  System.out.println("not possible");

	}

}

- priyaa.jass October 08, 2016 | Flag Reply
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0
of 0 vote

import java.util.Scanner;

public class StringPalindrome {

static int[] visited=null;
static char[] array=null;
static boolean Answer;

public static void Palindrome(int index,char[] temp)
{
if(index>=array.length)
{
int m=0,n=temp.length-1;
boolean b=true;
while(m<n)
{
if(temp[m]!=temp[n])
{b=false;break;}

m++;n--;
}
if(b)
Answer=true;
return;
}

for(int p=0;p<array.length;p++)
{
if(visited[index]==0)
{
visited[index]=1;
temp[index]=array[p];
Palindrome(index+1, temp);
visited[index]=0;
}
}
}
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
String s=sc.next();
array=s.toCharArray();
visited=new int[array.length];
char[] tarray=new char[array.length];
Answer=false;
Palindrome(0, tarray);
System.out.println(Answer);
}

}

- Anonymous October 11, 2016 | Flag Reply
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0
of 0 vote

import java.util.Scanner;

/**
 * Created by dushyant.sabharwal on 11/20/16.
 */
public class PalindromicPermutation {
    public static void main(String []args) {
        char [] char_map = new char[122];
        System.out.println("Enter the string for checking");
        Scanner sc = new Scanner(System.in);
        String input = sc.next();
        sc.close();
        boolean even = (0 == input.length() % 2);

        for (int i=0; i < input.length(); i++) {
            char c = input.charAt(i);
            char_map[c]++;
        }

        int i;
        for(i=0; i < char_map.length; i++) {
            if (char_map[i] > 0) {
                if (even && char_map[i] % 2 != 0) {
                    System.out.println("No palindrome");
                    break;
                } else if(!even && char_map[i] % 2 != 0) {
                    even = true;
                }
            }
        }

        if (i == char_map.length) {
            System.out.println("There is a palindrome");
        }
    }
}

- dushyant.sabharwal November 20, 2016 | Flag Reply


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