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Solved
fix34 challenge
Posted on 2016-08-14
Hi,
I am working on below challenge
http://codingbat.com/prob/p159339
I have not understood the description properly
please advise
I am working on below challenge
http://codingbat.com/prob/p159339
I have not understood the description properly
Array-3 > fix34
prev | next | chance
Return an array that contains exactly the same numbers as the given array, but rearranged so that every 3 is immediately followed by a 4. Do not move the 3's, but every other number may move. The array contains the same number of 3's and 4's, every 3 has a number after it that is not a 3 or 4, and a 3 appears in the array before any 4.
fix34([1, 3, 1, 4]) → [1, 3, 4, 1]
fix34([1, 3, 1, 4, 4, 3, 1]) → [1, 3, 4, 1, 1, 3, 4]
fix34([3, 2, 2, 4]) → [3, 4, 2, 2]
please advise
[X]
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9 Comments
Active 5 days ago
You need to put the numbers 4 that exist in the given array after the numbers 3.
In the given array, numbers 3 are not followed by a 4.
So:
1,3,5,4 will become 1,3,4,5
In the given array, numbers 3 are not followed by a 4.
So:
1,3,5,4 will become 1,3,4,5
Active today
how
fix34([1, 3, 1, 4, 4, 3, 1]) becomes [1, 3, 4, 1, 1, 3, 4] ?
or just move 4 next 3 without moving 3?
what happens to non 4 and non 3 numbers? where we have to move them?
fix34([1, 3, 1, 4, 4, 3, 1]) becomes [1, 3, 4, 1, 1, 3, 4] ?
or just move 4 next 3 without moving 3?
what happens to non 4 and non 3 numbers? where we have to move them?
Active 1 day ago
Moving does include exchanging numbers.... Only 3's are fixed.... and there is a hint in the question as well.
There will be a 3 before any 4 anyway..., Also for every three there will be a 4....
so it can be done in a single pass through the array.
so
There will be a 3 before any 4 anyway..., Also for every three there will be a 4....
so it can be done in a single pass through the array.
so
fix34([1, 3, 1, 4, 4, 3, 1])
becomes [ 1, 3, 4, 1, 4, 3, 1]
-----
becomes [1, 3, 4, 1, 1, 3, 4] ?
-- --
Active today
public int[] fix34(int[] nums) {
for (int z = 0; z < nums.length; z++)
if (nums[z] == 3) {
int buffer = nums[z + 1];
nums[z + 1] = 4;
for (int j = z + 2; j < nums.length; j++)
if (nums[j] == 4)
nums[j] = buffer;
}
return nums;
}
Expected Run
fix34([1, 3, 1, 4]) → [1, 3, 4, 1] [1, 3, 4, 1] OK
fix34([1, 3, 1, 4, 4, 3, 1]) → [1, 3, 4, 1, 1, 3, 4] [1, 3, 4, 1, 1, 3, 4] OK
fix34([3, 2, 2, 4]) → [3, 4, 2, 2] [3, 4, 2, 2] OK
fix34([3, 2, 3, 2, 4, 4]) → [3, 4, 3, 4, 2, 2] [3, 4, 3, 4, 2, 2] OK
fix34([2, 3, 2, 3, 2, 4, 4]) → [2, 3, 4, 3, 4, 2, 2] [2, 3, 4, 3, 4, 2, 2] OK
fix34([3, 1, 4]) → [3, 4, 1] [3, 4, 1] OK
fix34([3, 4, 1]) → [3, 4, 1] [3, 4, 1] OK
fix34([1, 1, 1]) → [1, 1, 1] [1, 1, 1] OK
fix34([1]) → [1] [1] OK
fix34([]) → [] [] OK
fix34([7, 3, 7, 7, 4]) → [7, 3, 4, 7, 7] [7, 3, 4, 7, 7] OK
fix34([3, 1, 4, 3, 1, 4]) → [3, 4, 1, 3, 4, 1] [3, 4, 1, 3, 4, 1] OK
fix34([3, 1, 1, 3, 4, 4]) → [3, 4, 1, 3, 4, 1] [3, 4, 1, 3, 4, 1] OK
other tests
OK
above passes all tests
package com.solution;
import java.util.Arrays;
public class Fix34 {
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] ar={1, 3, 1, 4, 4, 3, 1};
System.out.println("value-->"+Arrays.toString(fix34(ar)));
}
public static int[] fix34(int[] nums) {
for (int z = 0; z < nums.length; z++)
if (nums[z] == 3) {
int buffer = nums[z + 1];
nums[z + 1] = 4;
for (int j = z + 2; j < nums.length; j++)
if (nums[j] == 4)
nums[j] = buffer;
}
return nums;
}
}
Active today
public int[] fix34(int[] nums) {
int[] arr = new int[nums.length];
for (int z = 0; z < nums.length; z++){
if (nums[z] == 3) {
arr[z] = 3;
arr[z + 1] = 4;
}
}
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] != 3 && nums[k] != 4){
arr[j] = nums[k];
mark = k + 1;
break;
}
}
}
}
return arr;
}
Expected Runabove similar to fix45 challenge passed all tests where fill new array simply with values which seems to me more efficient and easy way
fix34([1, 3, 1, 4]) → [1, 3, 4, 1] [1, 3, 4, 1] OK
fix34([1, 3, 1, 4, 4, 3, 1]) → [1, 3, 4, 1, 1, 3, 4] [1, 3, 4, 1, 1, 3, 4] OK
fix34([3, 2, 2, 4]) → [3, 4, 2, 2] [3, 4, 2, 2] OK
fix34([3, 2, 3, 2, 4, 4]) → [3, 4, 3, 4, 2, 2] [3, 4, 3, 4, 2, 2] OK
fix34([2, 3, 2, 3, 2, 4, 4]) → [2, 3, 4, 3, 4, 2, 2] [2, 3, 4, 3, 4, 2, 2] OK
fix34([3, 1, 4]) → [3, 4, 1] [3, 4, 1] OK
fix34([3, 4, 1]) → [3, 4, 1] [3, 4, 1] OK
fix34([1, 1, 1]) → [1, 1, 1] [1, 1, 1] OK
fix34([1]) → [1] [1] OK
fix34([]) → [] [] OK
fix34([7, 3, 7, 7, 4]) → [7, 3, 4, 7, 7] [7, 3, 4, 7, 7] OK
fix34([3, 1, 4, 3, 1, 4]) → [3, 4, 1, 3, 4, 1] [3, 4, 1, 3, 4, 1] OK
fix34([3, 1, 1, 3, 4, 4]) → [3, 4, 1, 3, 4, 1] [3, 4, 1, 3, 4, 1] OK
other tests
OK
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