gudii9
asked on
zeroMAx challenge
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
To find max odd
I am failing tests
How to improve my design, approach, code? please advise
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
How to improve my design, approach, code? please advise
What tests are failing?
ASKER
none of the tests passing. i still need to refine my code
OK
You need to work some examples by hand and refine your pseudo code first.
ASKER CERTIFIED SOLUTION
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ASKER
to find the max you need to look at each element from the one after the i-th element to the end of the arrayhow to find max element from remaining array?
what happens if i see one other 0? how to handle that scenario?
SOLUTION
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ASKER
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 Runi passed all tests with one 0. How pass tests with more than one 0?
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
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]
ASKER
i think i have to tweak my incrementer to go to next 0 somehow like i=i+i like that
ASKER
oops i need to find largest odd number not lagest
ASKER
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
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.
ASKER
keeping track of the largest odd integerhow?
Just like you did here:
if(nums[j]>largest&&nums[j]%2==1){
largest=nums[j];
ASKER
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?
Something like that. But you need to establish the value for largest odd first.
ASKER
But you need to establish the value for largest odd first.i thought i did as below?
int largest=nums[j];
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.
But you can give it a clever initial value that allows it to work as a flag and simplify your program.
ASKER
not sure on your approach. can you provide the your approach code?
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