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Fix45 challenge
Posted on 2016-08-23
Hi,
I am working on below challenge.
http://codingbat.com/prob/p125819
My psedo code of approach is
1. iterate given arry
2. find position of 4
3. set next position to 5 by keeping orignal value in buffere
4. create other for loop from index position where we found 4 plus two and if that value is 5 replace with buffere
5. return modified array
i am failing below tests. please advise
I am working on below challenge.
http://codingbat.com/prob/p125819
My psedo code of approach is
1. iterate given arry
2. find position of 4
3. set next position to 5 by keeping orignal value in buffere
4. create other for loop from index position where we found 4 plus two and if that value is 5 replace with buffere
5. return modified array
public int[] fix45(int[] nums) {
for (int z = 0; z < nums.length; z++)
if (nums[z] == 4) {
int buffer = nums[z + 1];
nums[z + 1] = 5;
for (int j = z + 2; j < nums.length; j++)
if (nums[j] == 5)
nums[j] = buffer;
}
return nums;
}
i am failing below tests. please advise
Expected Run
fix45([5, 4, 9, 4, 9, 5]) → [9, 4, 5, 4, 5, 9] [5, 4, 5, 4, 5, 9] X
fix45([1, 4, 1, 5]) → [1, 4, 5, 1] [1, 4, 5, 1] OK
fix45([1, 4, 1, 5, 5, 4, 1]) → [1, 4, 5, 1, 1, 4, 5] [1, 4, 5, 1, 1, 4, 5] OK
fix45([4, 9, 4, 9, 5, 5, 4, 9, 5]) → [4, 5, 4, 5, 9, 9, 4, 5, 9] [4, 5, 4, 5, 9, 9, 4, 5, 9] OK
fix45([5, 5, 4, 1, 4, 1]) → [1, 1, 4, 5, 4, 5] [5, 5, 4, 5, 4, 5] X
fix45([4, 2, 2, 5]) → [4, 5, 2, 2] [4, 5, 2, 2] OK
fix45([4, 2, 4, 2, 5, 5]) → [4, 5, 4, 5, 2, 2] [4, 5, 4, 5, 2, 2] OK
fix45([4, 2, 4, 5, 5]) → [4, 5, 4, 5, 2] [4, 5, 4, 5, 2] OK
fix45([1, 1, 1]) → [1, 1, 1] [1, 1, 1] OK
fix45([4, 5]) → [4, 5] [4, 5] OK
fix45([5, 4, 1]) → [1, 4, 5] [5, 4, 5] X
fix45([]) → [] [] OK
fix45([5, 4, 5, 4, 1]) → [1, 4, 5, 4, 5] [5, 4, 5, 4, 5] X
fix45([4, 5, 4, 1, 5]) → [4, 5, 4, 5, 1] [4, 5, 4, 5, 1] OK
fix45([3, 4, 5]) → [3, 4, 5] [3, 4, 5] OK
fix45([4, 1, 5]) → [4, 5, 1] [4, 5, 1] OK
fix45([5, 4, 1]) → [1, 4, 5] [5, 4, 5] X
fix45([2, 4, 2, 5]) → [2, 4, 5, 2] [2, 4, 5, 2] OK
other tests
[X]
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15 Comments
Active today
one extra 5That's what's wrong with your result.
But what do the input arrays have in common?
Active today
same number of 4's and 5'sBoth of those must be true for every case according to the problem definition.
also every 4 has a number after it that is not a 4
What do the input arrays that you get wrong have in common that is different from the ones you get correct? <== This is what debugging is all about.
How is this challenge harder than the fix34? What is the difference?
As a hint: Note that the challenge uses the words "this" and "version" twice.
Active today
they have 5 at beginning and failing
Below passes all tests
fix45 -->[1, 4, 5, 1]
any improvement/refinement/alt
Below passes all tests
public int[] fix45(int[] nums) {
for (int x = 0; x < nums.length; x++) {
if (nums[x] == 4) {
for (int y = 0; y < nums.length; y++) {
if (nums[y] == 5) {
if (y > 0 && nums[y-1] != 4) {
int buffer = nums[x+1];
nums[x+1] = 5;
nums[y] = buffer;
}
else if (y == 0) {
int buff = nums[x+1];
nums[x+1] = 5;
nums[y] = buff;
}
}
}
}
}
return nums;
}
Expected Run
fix45([5, 4, 9, 4, 9, 5]) → [9, 4, 5, 4, 5, 9] [9, 4, 5, 4, 5, 9] OK
fix45([1, 4, 1, 5]) → [1, 4, 5, 1] [1, 4, 5, 1] OK
fix45([1, 4, 1, 5, 5, 4, 1]) → [1, 4, 5, 1, 1, 4, 5] [1, 4, 5, 1, 1, 4, 5] OK
fix45([4, 9, 4, 9, 5, 5, 4, 9, 5]) → [4, 5, 4, 5, 9, 9, 4, 5, 9] [4, 5, 4, 5, 9, 9, 4, 5, 9] OK
fix45([5, 5, 4, 1, 4, 1]) → [1, 1, 4, 5, 4, 5] [1, 1, 4, 5, 4, 5] OK
fix45([4, 2, 2, 5]) → [4, 5, 2, 2] [4, 5, 2, 2] OK
fix45([4, 2, 4, 2, 5, 5]) → [4, 5, 4, 5, 2, 2] [4, 5, 4, 5, 2, 2] OK
fix45([4, 2, 4, 5, 5]) → [4, 5, 4, 5, 2] [4, 5, 4, 5, 2] OK
fix45([1, 1, 1]) → [1, 1, 1] [1, 1, 1] OK
fix45([4, 5]) → [4, 5] [4, 5] OK
fix45([5, 4, 1]) → [1, 4, 5] [1, 4, 5] OK
fix45([]) → [] [] OK
fix45([5, 4, 5, 4, 1]) → [1, 4, 5, 4, 5] [1, 4, 5, 4, 5] OK
fix45([4, 5, 4, 1, 5]) → [4, 5, 4, 5, 1] [4, 5, 4, 5, 1] OK
fix45([3, 4, 5]) → [3, 4, 5] [3, 4, 5] OK
fix45([4, 1, 5]) → [4, 5, 1] [4, 5, 1] OK
fix45([5, 4, 1]) → [1, 4, 5] [1, 4, 5] OK
fix45([2, 4, 2, 5]) → [2, 4, 5, 2] [2, 4, 5, 2] OK
other tests
package com.solution;
import java.util.Arrays;
public class Fix45 {
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] ar={1,4,1,5};
System.out.println("fix45 -->"+Arrays.toString(fix45(ar)));
}
public static int[] fix45(int[] nums) {
for (int x = 0; x < nums.length; x++) {
if (nums[x] == 4) {
for (int y = 0; y < nums.length; y++) {
if (nums[y] == 5) {
if (y > 0 && nums[y-1] != 4) {
int buffer = nums[x+1];
nums[x+1] = 5;
nums[y] = buffer;
}
else if (y == 0) {
int buff = nums[x+1];
nums[x+1] = 5;
nums[y] = buff;
}
}
}
}
}
return nums;
}
}
fix45 -->[1, 4, 5, 1]
any improvement/refinement/alt
Active today
I don't know if my solution is any better than yours. I create a new array. I fill it with the values from the input array. I return the new array.
public int[] fix45(int[] nums) {
int[] arr = new int[nums.length];
for (int z = 0; z < nums.length; z++){
if (nums[z] == 4) {
arr[z] = 4;
arr[z + 1] = 5;
}
}
int mark = 0;
for (int j = 0; j < nums.length; j++){
if(arr[j] == 0){
for (int k = mark; k < nums.length; k++){
if(nums[k] != 4 && nums[k] != 5){
arr[j] = nums[k];
mark = k + 1;
break;
}
}
}
}
return arr;
}
Active today
How does your pseudo code compare with this:
b = 0
for i= 0 to length
if (nums[i] == 4)
buff[b] = nums[i+1]
nums[i+1] = 5
b ++
b = 0
for i= 0 to length
if (nums[i] == 5)
if (i == 0)
nums[i] = buff[b]
b++
else if (nums[i-1] != 4)
nums[i] = buff[b]
b++
Active today
what is the purpose of
mark = k + 1;
mark = k + 1;
public int[] fix45(int[] nums) {
int[] arr = new int[nums.length];
for (int z = 0; z < nums.length; z++){
if (nums[z] == 4) {
arr[z] = 4;
arr[z + 1] = 5;
}
}
int mark = 0;
for (int j = 0; j < nums.length; j++){
if(arr[j] == 0){
for (int k = mark; k < nums.length; k++){
if(nums[k] != 4 && nums[k] != 5){
arr[j] = nums[k];
mark = k + 1;
break;
}
}
}
}
return arr;
}
package com.solution;
import java.util.Arrays;
public class Fix46_2 {
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] ar = { 1, 4, 1, 5 };
System.out.println("fix45 -->" + Arrays.toString(fix45(ar)));
}
public static int[] fix45(int[] nums) {
int[] arr = new int[nums.length];
for (int z = 0; z < nums.length; z++){
if (nums[z] == 4) {
arr[z] = 4;
arr[z + 1] = 5;
}
}
int mark = 0;
for (int j = 0; j < nums.length; j++){
if(arr[j] == 0){
for (int k = mark; k < nums.length; k++){
if(nums[k] != 4 && nums[k] != 5){
arr[j] = nums[k];
mark = k + 1;
break;
}
}
}
}
return arr;
}
}
Active today
i see using mark and other for loop you are trying to fill non 4 and non 5 values from imput nums asrray to output new arr array.
I found your solution of filling new array is more straight forward?
why are we starting second for loop from mark rather than 0?
I found your solution of filling new array is more straight forward?
why are we starting second for loop from mark rather than 0?
public int[] fix45(int[] nums) {
int[] arr = new int[nums.length];
for (int z = 0; z < nums.length; z++){
if (nums[z] == 4) {
arr[z] = 4;
arr[z + 1] = 5;
}
}
int mark = 0;
for (int j = 0; j < nums.length; j++){
if(arr[j] == 0){
for (int k = mark; k < nums.length; k++){
if(nums[k] != 4 && nums[k] != 5){
arr[j] = nums[k];
mark = k + 1;
break;
}
}
}
}
return arr;
}
Active today
i see you are using mark to increment inner for loop with k as you are breaking out of it prematurely for k++ to take place later.
Active today
i see you are using mark to increment inner for loop with k as you are breaking out of it prematurely for k++ to take place later.I used the mark variable to know where to begin looking for the next number to use to fill into the arr array. The increment in the loop(k++) is still used.
There probably is room for improvement. But, when you closed this question, I stopped working on this. Anyway, maybe adding some debug statements, will shed some light on my code. See below,
import java.util.Arrays;
public class Fix{
public static void main(String[] args) {
int[] input = {2, 3, 4, 6, 4, 7, 5, 5, 4, 8, 5};
System.out.println(" input -->" + Arrays.toString(input));
fix45(input);
}
public static int[] fix45(int[] nums) {
int[] arr = new int[nums.length];
for (int z = 0; z < nums.length; z++){
if (nums[z] == 4) {
arr[z] = 4;
arr[z + 1] = 5;
}
}
System.out.println(" output -->" + Arrays.toString(arr) + " (after first for loop)");
int mark = 0;
for (int j = 0; j < nums.length; j++){
if(arr[j] == 0){
for (int k = mark; k < nums.length; k++){
if(nums[k] != 4 && nums[k] != 5){
arr[j] = nums[k];
mark = k + 1;
System.out.println(" j is " + j + ", k is " + k + ", mark is " + mark);
System.out.println(" output -->" + Arrays.toString(arr));
break;
}
}
}
}
return arr;
}
}
input -->[2, 3, 4, 6, 4, 7, 5, 5, 4, 8, 5]
output -->[0, 0, 4, 5, 4, 5, 0, 0, 4, 5, 0] (after first for loop)
j is 0, k is 0, mark is 1
output -->[2, 0, 4, 5, 4, 5, 0, 0, 4, 5, 0]
j is 1, k is 1, mark is 2
output -->[2, 3, 4, 5, 4, 5, 0, 0, 4, 5, 0]
j is 6, k is 3, mark is 4
output -->[2, 3, 4, 5, 4, 5, 6, 0, 4, 5, 0]
j is 7, k is 5, mark is 6
output -->[2, 3, 4, 5, 4, 5, 6, 7, 4, 5, 0]
j is 10, k is 9, mark is 10
output -->[2, 3, 4, 5, 4, 5, 6, 7, 4, 5, 8]
Active today
input -->[2, 3, 4, 6, 4, 7, 5, 5, 4, 8, 5]
output -->[0, 0, 4, 5, 4, 5, 0, 0, 4, 5, 0] (after first for loop)
j is 0, k is 0, mark is 1
output -->[2, 0, 4, 5, 4, 5, 0, 0, 4, 5, 0]
j is 1, k is 1, mark is 2
output -->[2, 3, 4, 5, 4, 5, 0, 0, 4, 5, 0]
j is 6, k is 3, mark is 4
output -->[2, 3, 4, 5, 4, 5, 6, 0, 4, 5, 0]
j is 7, k is 5, mark is 6
output -->[2, 3, 4, 5, 4, 5, 6, 7, 4, 5, 0]
j is 10, k is 9, mark is 10
output -->[2, 3, 4, 5, 4, 5, 6, 7, 4, 5, 8]
above output is more clear now. To understand again today taking some time around 1 hour even though i saw this day before yesterday?
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