Arrays in Java - Core Java

Question

In This Question A value is "everywhere" in an array if for every pair of adjacent elements in the array, at least one of the pair is that value. Return true if the given value is everywhere in the array.

isEverywhere({1, 2, 1, 3}, 1) → true,coz 1 is present in(1,2) and (1,3) isEverywhere({1, 2, 1, 3}, 2) → false,coz 2 is in (1,2) but not in (1,3) isEverywhere({1, 2, 1, 3, 4}, 1) → false,coz 1 is present in 2 pairs (1,2) and (1,3) but 4 is not having a pair of 1

My Partial working code is below,could you help me out with this problem,stuck up for a long time.

::Code::

public boolean isEverywhere(int[] nums, int val) {    
    boolean flag = false;    
    for(int i=0;i<nums.length;i++){    
      for(int j=i+2;j<nums.length;j++){    
            if(nums[i] == nums[j]){    
                 flag = true;    
            }    
      }    
    }      
  return flag;     
}

Result expected:

                           `Expected` `This` `Run`

isEverywhere({1, 2, 1, 3}, 1) → true true OK
isEverywhere({1, 2, 1, 3}, 2) → false true X
isEverywhere({1, 2, 1, 3, 4}, 1) → false true X

You don't need to nest for loops for this. Break down the problem. First try to enumerate through the pairs you are considering and doing a System.out.println to see that you are considering the right pairs. Even tiny problems can be taken step by step. — Dan Rosenstark
– Dan Rosenstark, Commented Dec 21, 2010 at 19:00

codaddict · Accepted Answer · 2010-12-21 19:12:45Z

3

Try this:

boolean isEverywhere(int[] nums, int val) {    

        // use i+=2 to get start index of pair.
        for(int i=0;i<nums.length;i+=2) {

                // other index in the pair.
                int j = i + 1;

                // make sure the other index really exists.
                if(j < nums.length) { 

                        // if it exists..and val is not equal to 
                        // either in the pair..return false.
                        if(nums[i] != val && nums[j] != val) {
                                return false;
                        }       
                } else {
                        // no pair..one element case.
                        // return true if that element is val..else return false.
                        return nums[i] == val; 
                }       
        }

        // array has no unpaired element..and all pairs have val.
        return true;     
}

edited Dec 21, 2010 at 19:12

answered Dec 21, 2010 at 19:05

codaddict

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9 Comments

Deepak Over a year ago

@codaddict,i have one question,what the exception scenario isEverywhere({3}, 1) → true false X

codaddict Over a year ago

@Deepak: ` isEverywhere({3}, 1)` returns false as expected as there is no 1. Is that not you want?

Deepak Over a year ago

yes,correct,probably i need to check my test cases once,Thanks once again,sorry for the trouble,@coaddict

Shaded Over a year ago

@Deepak: what about a case such as isEverywhere({1, 3, 4, 1}, 1) If I'm not mistaken this would be expected to return false, but this code would return true. This is because it would only check (1, 3) and (4, 1) isn't the pair (3, 4) considered?

Deepak Over a year ago

@shaded,only the adjacency pairs are considered and will return true.@codaddict is correct,i just checked the algorithm i have last clarification,why have u return nums[i] == val;

|

Eugene Yokota · Accepted Answer · 2010-12-21 19:02:55Z

2

The key is

for every pair of adjacent elements in the array

Print out i and j before if(nums[i] == nums[j]) and you'd see what's going on.

For an array of length 2, you need (0, 1), for array of length 3, you need (0, 1), (1, 2), and so on.

answered Dec 21, 2010 at 19:02

Eugene Yokota

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Comments

Jean-Bernard Pellerin · Accepted Answer · 2010-12-21 18:59:19Z

1

You never use val in your code
Try figuring out the logic for that first

And this can be done with a single loop, try to figure out how you would as a person apply an algorithm, then code it

answered Dec 21, 2010 at 18:59

Jean-Bernard Pellerin

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The Scrum Meister · Accepted Answer · 2010-12-21 18:59:46Z

0

Your code is failing because the flag does not turn off in the case that the second loop goes beyond the array length.

answered Dec 21, 2010 at 18:59

The Scrum Meister

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brian_d · Accepted Answer · 2010-12-22 15:36:24Z

0

Not a full solution, but if the length of the array is odd, check that the last value equals the target val

...
    if( !(nums.length % 2) == 0 ){
       if( nums[length-1] != val){
          return false;
       }
    }
...

edited Dec 22, 2010 at 15:36

answered Dec 21, 2010 at 19:04

brian_d

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Comments

LinuxHaxxor · Accepted Answer · 2016-12-26 18:54:50Z

I did mine by simply counting if there are spaces: Here it is in it's simple glory:

public boolean isEverywhere(int[] nums, int val) {
    //Create an int to count the number of spaces between each instance of val.
    int counter = 0;
    //Create a for loop that checks val;
    for (int i = 0; i < nums.length; i++)
    {
      //If the number at iteration i != val;
      if (nums[i] != val)
      {
        //We increment the counter by one.
        counter++;
      }
      //If the number does equal val, we put the counter back to 0;
      else
      {
        counter = 0;
      }
      //If the counter is greater than or equal to two,
      //We know that there is a space of more than 2, and so
      //We return false;
      if (counter >= 2)
      {
        return false;
      }
    }
    //Return true if not false;
    return true;
  }

ritesh_ratn · Accepted Answer · 2020-06-28 08:15:45Z

0

This code works fine:-

public boolean isEverywhere(int[] nums, int val) {
  for(int i=0;i<=nums.length-2;i++){
    if(nums[i]!=val&&nums[i+1]!=val){
      return false;
    }
  }
  return true;
}

answered Jun 28, 2020 at 8:15

ritesh_ratn

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Comments

Nimantha · Accepted Answer · 2021-07-14 12:51:33Z

0

public class IE {
    static boolean isEverywhere(int[] a, int val) {   
      for (int i = 0; i < a.length; i++) {
        System.out.println("*** iteration: " + i);
        System.out.println("a[" + (i-1) + "]=" + (i > 0 ? a[i-1] : "n/a"));
        System.out.println("a[" + i + "]=" + a[i]);
        System.out.println("a[" + (i+1) + "]=" + (i < a.length -1 ? a[i+1] : "n/a"));
        if (a[i] != val && (i > 0 ? a[i-1] != val : true) && (i < a.length-1 ? a[i+1] != val : true)) {
          return false;
        }   
      }   
      return true; 
    }
    
    public static void main(String[] args) {
        int[] a = new int[] {1, 2, 1, 3};
        System.out.println(isEverywhere(a, 1));
        a = new int[] {1, 2, 1, 3};
        System.out.println(isEverywhere(a, 2));
        a = new int[] {1, 2, 1, 3, 4};
        System.out.println(isEverywhere(a, 1));
    }
}

Basically, for a given element in the array, the test fails if the value isn't adjacent, either before or after it.

edited Jul 14, 2021 at 12:51

Nimantha

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answered Dec 21, 2010 at 19:00

Jeffrey Blattman

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2 Comments

Mark Peters Over a year ago

This doesn't solve the problem at all, and you shouldn't just be doing his homework for him.

Jeffrey Blattman Over a year ago

he didn't say it was homework did he? and yes it does work for the given input (now that i edited it).

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