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# zeroMAx challenge

Posted on 2016-08-14
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Hi,

I am working on below challenge

http://codingbat.com/prob/p187050

My psuedo code of the logic is
1. loop through given array
2. find occurence of zero as long as not a last elemet
3. find the biggest odd after occurence of zero
4 replace i th element with biggest odd.
5 return array

I wrote my code as below
``````public int[] zeroMax(int[] nums) {

int i, smallEven=0, bigOdd=0, countOdd=0, countEven=0;
for(i=0; i<n; i++){
// System.out.print("Enter element "+(i+1));
nums[i]=sc.nextInt();
if(a[i]%2 == 1){
if(countOdd>0 && a[i]>bigOdd) bigOdd=a[i];
else if(countOdd==0) bigOdd=a[i];
countOdd++;
}

for(int i=0;i<nums.length-1;i++){
if(  (nums[i]%2)==1 && i!=nums.len ) {
nums[i]=
}

}
return nums;
}
``````

To find max odd
``````import java.util.Scanner;

public class ZeroMax {

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Enter the number of elements: ");
int n = sc.nextInt();
int[] a = new int[n];
int i, smallEven=0, bigOdd=0, countOdd=0, countEven=0;
for(i=0; i<n; i++){
System.out.print("Enter element "+(i+1));
a[i]=sc.nextInt();
if(a[i]%2 == 1){
if(countOdd>0 && a[i]>bigOdd) bigOdd=a[i];
else if(countOdd==0) bigOdd=a[i];
countOdd++;
}
else{
if(countEven>0 && a[i]<smallEven) smallEven=a[i];
else if(countEven==0)smallEven=a[i];
countEven++;
}
}
System.out.println("The smallest even number is " + smallEven);
System.out.println("The largest odd number is " + bigOdd);
}
}
``````

I am failing tests

0
Question by:gudii9
• 10
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• 2
• +1

LVL 38

Expert Comment

ID: 41756459
What tests are failing?
0

LVL 7

Author Comment

ID: 41756534
none of the tests passing. i still need to refine my code
0

LVL 38

Expert Comment

ID: 41756568
OK
0

LVL 27

Expert Comment

ID: 41756665
You need to work some examples by hand and refine your pseudo code first.
1

LVL 36

Accepted Solution

mccarl earned 1000 total points
ID: 41759331
There's nothing particularly wrong with your pseudo-code it is just point 3 that needs to be expanded, ie. Now you know the i-th element is zero, HOW do you find the biggest odd number after the i-th element.

Hint: to find the max you need to look at each element from the one after the i-th element to the end of the array, ie. you need a second loop. This will be within the first, main loop.
0

LVL 7

Author Comment

ID: 41759940
to find the max you need to look at each element from the one after the i-th element to the end of the array
how to find max element from remaining array?
what happens if i see one other 0? how to handle that scenario?
0

LVL 27

Assisted Solution

d-glitch earned 1000 total points
ID: 41759952
each zero value in the array is replaced by the largest odd value to the right

If the first loop finds a zero, the second loop looks to the right to find the largest odd number to replace the zero.
If you find a zero with the second loop, you ignore it because you are only looking for odd numbers.
If there are more zeros in the array, the first loop will find them eventually.
0

LVL 7

Author Comment

ID: 41759966
``````public int[] zeroMax(int[] nums) {
for (int i = 0; i < nums.length; i++) {
if (nums[i] == 0) {

if (nums[i] == 0) {
int largest=nums[i];
for (int j = i; j < nums.length; j++) {

if(nums[j]>largest){
largest=nums[j];

}
if(largest>0){
nums[i]=largest;
}

}

}

}

}

return nums;

}
``````
Expected      Run
zeroMax([0, 5, 0, 3]) → [5, 5, 3, 3]      [5, 5, 3, 3]      OK
zeroMax([0, 4, 0, 3]) → [3, 4, 3, 3]      [4, 4, 3, 3]      X
zeroMax([0, 1, 0]) → [1, 1, 0]      [1, 1, 0]      OK
zeroMax([0, 1, 5]) → [5, 1, 5]      [5, 1, 5]      OK
zeroMax([0, 2, 0]) → [0, 2, 0]      [2, 2, 0]      X
zeroMax([1]) → [1]      [1]      OK
zeroMax([0]) → [0]      [0]      OK
zeroMax([]) → []      []      OK
zeroMax([7, 0, 4, 3, 0, 2]) → [7, 3, 4, 3, 0, 2]      [7, 4, 4, 3, 2, 2]      X
zeroMax([7, 0, 4, 3, 0, 1]) → [7, 3, 4, 3, 1, 1]      [7, 4, 4, 3, 1, 1]      X
zeroMax([7, 0, 4, 3, 0, 0]) → [7, 3, 4, 3, 0, 0]      [7, 4, 4, 3, 0, 0]      X
zeroMax([7, 0, 1, 0, 0, 7]) → [7, 7, 1, 7, 7, 7]      [7, 7, 1, 7, 7, 7]      OK
other tests
X
i passed all tests with one 0. How pass tests with more than one 0?
``````import java.util.Arrays;
import java.util.Scanner;

/*There's nothing particularly wrong with your pseudo-code it is
*  just point 3 that needs to be expanded, ie. Now you know the
*  i-th element is zero, HOW do you find the biggest odd number
*   after the i-th element.
Hint: to find the max you need to look at each element from the
one after the i-th element to the end of the array, ie. you need
a second loop. This will be within the first, main loop.*/
public class ZeroMax2 {

public static void main(String[] args) {
int[] ar = { 0, 5, 0, 3 };
System.out.println("value is" + Arrays.toString(zeroMax(ar)));

/*
* Return a version of the given array where each zero value in the
* array is replaced by the largest odd value to the right of the zero
* in the array. If there is no odd value to the right of the zero,
* leave the zero as a zero.
*
* zeroMax([0, 5, 0, 3]) → [5, 5, 3, 3] zeroMax([0, 4, 0, 3]) → [3, 4,
* 3, 3] zeroMax([0, 1, 0]) → [1, 1, 0]
*
* Hint: to find the max you need to look at each element from the one
* after the i-th element to the end of the array, ie. you need a second
* loop. This will be within the first, main loop.
*/
}

public static int[] zeroMax(int[] nums) {
for (int i = 0; i < nums.length; i++) {
if (nums[i] == 0) {

if (nums[i] == 0) {
int largest=nums[i];
for (int j = i; j < nums.length; j++) {

if(nums[j]>largest){
largest=nums[j];

}
if(largest>0){
nums[i]=largest;
}

}

}

}

}

return nums;

}

}
``````
value is[5, 5, 3, 3]
0

LVL 7

Author Comment

ID: 41759972
i think i have to tweak my incrementer to go to next 0 somehow like i=i+i like that
0

LVL 7

Author Comment

ID: 41759977
oops i need to find largest odd number not lagest
0

LVL 7

Author Comment

ID: 41759982
``````public int[] zeroMax(int[] nums) {
for (int i = 0; i < nums.length; i++) {
if (nums[i] == 0) {

if (nums[i] == 0) {
int largest=nums[i];
for (int j = i; j < nums.length; j++) {

if(nums[j]>largest&&nums[j]%2==1){
largest=nums[j];

}
if(largest>0){
nums[i]=largest;
}

}

}

}

}

return nums;

}
``````

above passed all tests. any improvements, modifications, optimization, refactoring of my code? please advise
0

LVL 27

Expert Comment

ID: 41760004
You can do the challenge much more efficiently (with one loop) by working right to left, keeping track of the largest odd integer, and replacing 0's as you go.
0

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Author Comment

ID: 41760011
keeping track of the largest odd integer
how?
0

LVL 27

Expert Comment

ID: 41760020
Just like you did here:
``````if(nums[j]>largest&&nums[j]%2==1){
largest=nums[j];
``````
0

LVL 7

Author Comment

ID: 41760088
``````public int[] zeroMax(int[] nums) {

for (int j = nums.length-1; j >=0; j--) {
if(nums[j]==0){
int largest=nums[j];
if(nums[j-1]>largest&&nums[j]%2==1){
largest=nums[j-1];
}
if(largest>0){
nums[0]=largest;
}
}
}
//i=i+i;
return nums;

}
``````
something like above?
failing some tests?
0

LVL 27

Expert Comment

ID: 41760112
Something like that.  But you need to establish the value for largest odd first.
0

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Author Comment

ID: 41760168
But you need to establish the value for largest odd first.
i thought i did as below?
int largest=nums[j];
0

LVL 27

Expert Comment

ID: 41760177
largest has no meaning until you find an odd integer.
But you can give it a clever initial value that allows it to work as a flag and simplify your program.
0

LVL 7

Author Comment

ID: 41764739
not sure on your approach. can you provide the your approach code?
0

LVL 27

Expert Comment

ID: 41765376
Pseudo code:
``````maxodd = 0
loop backwards through the array
if nums[i] is odd  &&  nums[i] > maxodd
then  maxodd = nums[i]
if nums[i] = 0
then nums[i] = maxodd
``````
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