# Issue

I am practising for an upcoming coding interview and here is one of my practice problems and my progress.

How can I improve the program and what is your advice?

Also, are there any cities that could help with improving my coding skills?

Question:

```
A non-empty zero-indexed array A consisting of N integers is given. The array contains an odd number of elements, and each element of the array can be paired with another element that has the same value, except for one element that is left unpaired.
For example, in array A such that:
A[0] = 9 A[1] = 3 A[2] = 9
A[3] = 3 A[4] = 9 A[5] = 7
A[6] = 9
the elements at indexes 0 and 2 have value 9,
the elements at indexes 1 and 3 have value 3,
the elements at indexes 4 and 6 have value 9,
the element at index 5 has value 7 and is unpaired.
Write a function:
class Solution { public int solution(int[] A); }
that, given an array A consisting of N integers fulfilling the above conditions, returns the value of the unpaired element.
For example, given array A such that:
A[0] = 9 A[1] = 3 A[2] = 9
A[3] = 3 A[4] = 9 A[5] = 7
A[6] = 9
the function should return 7, as explained in the example above.
Assume that:
N is an odd integer within the range [1..1,000,000];
each element of array A is an integer within the range [1..1,000,000,000];
all but one of the values in A occur an even number of times.
Complexity:
expected worst-case time complexity is O(N);
expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.
```

Solution:

```
import java.util.*;
public class Solution {
public int solution(int[] A) {
int x;
for(int i = 0; i < 7; i++)
{
//create an integer array containing an odd number of elements of numbers ranging from 1 - 1,000,000
//for(int N = 1; N <= 1,000,000; N++)
int N = 1;
while(N > 1 && N <= 1000000)
{
//check if N is odd then assign to the array
if(N != N/2)
{
A[i] = N;
}
}
//check for any element not paired more than once
if(A[i] != A[i++])
{
x = A[i];
}
else
return 0;
}
//return unpaired elemnent
return x;
}
}
```

# Solution

Something like this should work, here I implemented it in a way where I test all ints it gets against the rest, and only return if there is a solution (note there has to be a default, maybe a better way to handle "no solutions".

```
public class Solution {
public int solution(int[] A) {
boolean possibleSolution = true; // to return and properly break if not possible
for(int i = 0; i < A.length; i++) // run for all ints
{
possibleSolution = true; // set possible true, in case last one failed
for(int j = 0; j < A.length; j++) // take all ints again (to compare to the rest
if(A[i] == A[j] && i != j){ // note i escape comparing to itself
possibleSolution = false; // if there is a math it can't be this one
break; // break to save resources
}
if(possibleSolution) // if it's the solution
return A[i]; // return the current number (from the initial array as that is the reference number and the 2nd is for comparing)
}
return 0; // return default
}
public static void main(String[] args){
Solution solution = new Solution(); // instance
int[] ints = {9,3,9,3,9,7,9}; // new values
System.out.println(solution.solution(ints)); // print the method after it was run
}
}
```

Note that adding the ints, is not included here unsure what types of values is needed

but simply add them and then pass the array, if multiple answers are possible, then instead of return add to a `List<Integers> results = new ArrayList<>();`

, and after all `i`

is run through return the `results`

this would be where the `return 0;`

is at the moment.

Answered By - Mikenno Answer Checked By - Gilberto Lyons (PHPFixing Admin)

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