Find out n numbers of missing elements from an array in java
I have an array which carry some integer numbers. Say,numbers={3,0,1}
or say, numbers={9,6,4,2,3,5,7,0,1}
. Now I have to find out the missing numbers from the array. As per this example there is only one missing number in each set. 1st one miss 2 and the 2nd one miss 8.
I have already code it. My code not only find out one missing number from specified set it can also find out more than 1 missing numbers from a given set.
But if two consecutive numbers are missing from same set it fail to find out.
My code
import java.util.Arrays;
public class Missing_number
{
public static void main( String args )
{
int numbers={9,6,4,5,7,0,1};
Arrays.sort(numbers);
int i=1;
while ( i < numbers.length )
{
if ( numbers[i] - numbers[i-1] == 1 )
{
}
else
{
System.out.println( "Missing number is " + ( numbers[i-1] + 1 ) );
}
i++;
}
}
}
I am thinking like that if I am able to append the 1st missing number in the array and then start searching then how's the code look like? numbers={9,6,4,5,7,0,1}
Now, 8 is already missing from this set. Now I have terminated two more element (2,3) from the list. Output: as per my code: 2,8 But 3 is also missing but that does not display.
I am thinking like if I am able to append 2 in number array then it may be little bit easier. But as we all know Java array is immutable so we cannot increase it's length.
So, maybe I will use List. But in list this type of indexing number[0]=something
not supported. So how could I proceed then. Am I using list or still stuck into array?
So I take a attempt to create it with an arraylist.
Mycode(modified version from array)
public class T1 {
public static void main(String args){
List<Integer> numbers=new ArrayList<>();
numbers.add(9);
numbers.add(6);
numbers.add(4);
numbers.add(5);
numbers.add(7);
numbers.add(0);
numbers.add(1);
Collections.sort(numbers);
int i=1;
while(i< numbers.size()) {
if (numbers.get(i) - numbers.get(i-1) == 1) {
} else {
System.out.println("Missing number is " + (numbers.get(i-1) + 1));
numbers.add((numbers.get(i-1)+1));
Collections.sort(numbers);
}
i++;
}
}
}
Arraylist can solve my problem. But is there any possibility that a simple array can solve this problem?
java arrays algorithm arraylist
add a comment |
I have an array which carry some integer numbers. Say,numbers={3,0,1}
or say, numbers={9,6,4,2,3,5,7,0,1}
. Now I have to find out the missing numbers from the array. As per this example there is only one missing number in each set. 1st one miss 2 and the 2nd one miss 8.
I have already code it. My code not only find out one missing number from specified set it can also find out more than 1 missing numbers from a given set.
But if two consecutive numbers are missing from same set it fail to find out.
My code
import java.util.Arrays;
public class Missing_number
{
public static void main( String args )
{
int numbers={9,6,4,5,7,0,1};
Arrays.sort(numbers);
int i=1;
while ( i < numbers.length )
{
if ( numbers[i] - numbers[i-1] == 1 )
{
}
else
{
System.out.println( "Missing number is " + ( numbers[i-1] + 1 ) );
}
i++;
}
}
}
I am thinking like that if I am able to append the 1st missing number in the array and then start searching then how's the code look like? numbers={9,6,4,5,7,0,1}
Now, 8 is already missing from this set. Now I have terminated two more element (2,3) from the list. Output: as per my code: 2,8 But 3 is also missing but that does not display.
I am thinking like if I am able to append 2 in number array then it may be little bit easier. But as we all know Java array is immutable so we cannot increase it's length.
So, maybe I will use List. But in list this type of indexing number[0]=something
not supported. So how could I proceed then. Am I using list or still stuck into array?
So I take a attempt to create it with an arraylist.
Mycode(modified version from array)
public class T1 {
public static void main(String args){
List<Integer> numbers=new ArrayList<>();
numbers.add(9);
numbers.add(6);
numbers.add(4);
numbers.add(5);
numbers.add(7);
numbers.add(0);
numbers.add(1);
Collections.sort(numbers);
int i=1;
while(i< numbers.size()) {
if (numbers.get(i) - numbers.get(i-1) == 1) {
} else {
System.out.println("Missing number is " + (numbers.get(i-1) + 1));
numbers.add((numbers.get(i-1)+1));
Collections.sort(numbers);
}
i++;
}
}
}
Arraylist can solve my problem. But is there any possibility that a simple array can solve this problem?
java arrays algorithm arraylist
ButArrayList
supports indexing, right?
– Kingsley
Dec 17 '18 at 22:04
1
With a list, instead of subscripting, you use theget(int)
method, with the index as the method argument.
– Ted Hopp
Dec 17 '18 at 22:07
"But as we all know Java array is immutable" - no, Java arrays are mutable. They're not resizable, but they're mutable.
– user2357112
Dec 18 '18 at 1:03
add a comment |
I have an array which carry some integer numbers. Say,numbers={3,0,1}
or say, numbers={9,6,4,2,3,5,7,0,1}
. Now I have to find out the missing numbers from the array. As per this example there is only one missing number in each set. 1st one miss 2 and the 2nd one miss 8.
I have already code it. My code not only find out one missing number from specified set it can also find out more than 1 missing numbers from a given set.
But if two consecutive numbers are missing from same set it fail to find out.
My code
import java.util.Arrays;
public class Missing_number
{
public static void main( String args )
{
int numbers={9,6,4,5,7,0,1};
Arrays.sort(numbers);
int i=1;
while ( i < numbers.length )
{
if ( numbers[i] - numbers[i-1] == 1 )
{
}
else
{
System.out.println( "Missing number is " + ( numbers[i-1] + 1 ) );
}
i++;
}
}
}
I am thinking like that if I am able to append the 1st missing number in the array and then start searching then how's the code look like? numbers={9,6,4,5,7,0,1}
Now, 8 is already missing from this set. Now I have terminated two more element (2,3) from the list. Output: as per my code: 2,8 But 3 is also missing but that does not display.
I am thinking like if I am able to append 2 in number array then it may be little bit easier. But as we all know Java array is immutable so we cannot increase it's length.
So, maybe I will use List. But in list this type of indexing number[0]=something
not supported. So how could I proceed then. Am I using list or still stuck into array?
So I take a attempt to create it with an arraylist.
Mycode(modified version from array)
public class T1 {
public static void main(String args){
List<Integer> numbers=new ArrayList<>();
numbers.add(9);
numbers.add(6);
numbers.add(4);
numbers.add(5);
numbers.add(7);
numbers.add(0);
numbers.add(1);
Collections.sort(numbers);
int i=1;
while(i< numbers.size()) {
if (numbers.get(i) - numbers.get(i-1) == 1) {
} else {
System.out.println("Missing number is " + (numbers.get(i-1) + 1));
numbers.add((numbers.get(i-1)+1));
Collections.sort(numbers);
}
i++;
}
}
}
Arraylist can solve my problem. But is there any possibility that a simple array can solve this problem?
java arrays algorithm arraylist
I have an array which carry some integer numbers. Say,numbers={3,0,1}
or say, numbers={9,6,4,2,3,5,7,0,1}
. Now I have to find out the missing numbers from the array. As per this example there is only one missing number in each set. 1st one miss 2 and the 2nd one miss 8.
I have already code it. My code not only find out one missing number from specified set it can also find out more than 1 missing numbers from a given set.
But if two consecutive numbers are missing from same set it fail to find out.
My code
import java.util.Arrays;
public class Missing_number
{
public static void main( String args )
{
int numbers={9,6,4,5,7,0,1};
Arrays.sort(numbers);
int i=1;
while ( i < numbers.length )
{
if ( numbers[i] - numbers[i-1] == 1 )
{
}
else
{
System.out.println( "Missing number is " + ( numbers[i-1] + 1 ) );
}
i++;
}
}
}
I am thinking like that if I am able to append the 1st missing number in the array and then start searching then how's the code look like? numbers={9,6,4,5,7,0,1}
Now, 8 is already missing from this set. Now I have terminated two more element (2,3) from the list. Output: as per my code: 2,8 But 3 is also missing but that does not display.
I am thinking like if I am able to append 2 in number array then it may be little bit easier. But as we all know Java array is immutable so we cannot increase it's length.
So, maybe I will use List. But in list this type of indexing number[0]=something
not supported. So how could I proceed then. Am I using list or still stuck into array?
So I take a attempt to create it with an arraylist.
Mycode(modified version from array)
public class T1 {
public static void main(String args){
List<Integer> numbers=new ArrayList<>();
numbers.add(9);
numbers.add(6);
numbers.add(4);
numbers.add(5);
numbers.add(7);
numbers.add(0);
numbers.add(1);
Collections.sort(numbers);
int i=1;
while(i< numbers.size()) {
if (numbers.get(i) - numbers.get(i-1) == 1) {
} else {
System.out.println("Missing number is " + (numbers.get(i-1) + 1));
numbers.add((numbers.get(i-1)+1));
Collections.sort(numbers);
}
i++;
}
}
}
Arraylist can solve my problem. But is there any possibility that a simple array can solve this problem?
java arrays algorithm arraylist
java arrays algorithm arraylist
edited Dec 18 '18 at 1:53
Mis94
1,0601921
1,0601921
asked Dec 17 '18 at 22:01
EncipherEncipher
31712
31712
ButArrayList
supports indexing, right?
– Kingsley
Dec 17 '18 at 22:04
1
With a list, instead of subscripting, you use theget(int)
method, with the index as the method argument.
– Ted Hopp
Dec 17 '18 at 22:07
"But as we all know Java array is immutable" - no, Java arrays are mutable. They're not resizable, but they're mutable.
– user2357112
Dec 18 '18 at 1:03
add a comment |
ButArrayList
supports indexing, right?
– Kingsley
Dec 17 '18 at 22:04
1
With a list, instead of subscripting, you use theget(int)
method, with the index as the method argument.
– Ted Hopp
Dec 17 '18 at 22:07
"But as we all know Java array is immutable" - no, Java arrays are mutable. They're not resizable, but they're mutable.
– user2357112
Dec 18 '18 at 1:03
But
ArrayList
supports indexing, right?– Kingsley
Dec 17 '18 at 22:04
But
ArrayList
supports indexing, right?– Kingsley
Dec 17 '18 at 22:04
1
1
With a list, instead of subscripting, you use the
get(int)
method, with the index as the method argument.– Ted Hopp
Dec 17 '18 at 22:07
With a list, instead of subscripting, you use the
get(int)
method, with the index as the method argument.– Ted Hopp
Dec 17 '18 at 22:07
"But as we all know Java array is immutable" - no, Java arrays are mutable. They're not resizable, but they're mutable.
– user2357112
Dec 18 '18 at 1:03
"But as we all know Java array is immutable" - no, Java arrays are mutable. They're not resizable, but they're mutable.
– user2357112
Dec 18 '18 at 1:03
add a comment |
4 Answers
4
active
oldest
votes
This code uses a HashSet
:
public static void main(String args) {
int numbers = {9, 6, 4, 5, 7, 0, 1};
Arrays.sort(numbers);
HashSet<Integer> set = new HashSet<>();
for (int i = numbers[0]; i < numbers[numbers.length - 1]; i++) {
set.add(i);
}
for (int i = 0; i < numbers.length; i++) {
set.remove(numbers[i]);
}
for (int x : set) {
System.out.print(x + " ");
}
}
will print:
2 3 8
Here is how it works:
1. Adds all numbers from the minimum number of the array to the maximum number of the array to the set.
2. Iterates through the array and removes every item of the array from the set.
3. Prints the remaining items in the set, which are all the missing items of the array.
add a comment |
replace the else
clause to be:
for(int j=numbers[i-1] + 1; j <= numbers[i] - 1; j++) {
System.out.println( "Missing number is " + ( j ) );
}
let's examine the case: {9 ,6 ,4 ,5 ,7 ,0 , 1}
after sorting it will be: {0, 1, 4, 5, 6, 7, 9}
now if i
is at index 2 it finds the difference between numbers[i]
and numbers[i-1]
to be not equal 1 (4 - 1 = 3), now you need ALL the numbers between 1 and 4 which are 2, 3 and thus you must loop from numbers[i-1]
to numbers[i]
(exclusive) in order to achieve this.
The complexity of this code is big O of N
(O(N))
, where N
is the biggest element in your array.
add a comment |
There are a lot of questions which are unanswered here. For Example, Does an array always start with zero?, What is the maximum possible size? etc.
Here is a simple way to approach this problem,
- Find the maximum number in your set.
- Create an empty
boolean
array of the length as that of the max number you found in the last step plus one. - Scan your original set and set the value of your new boolean array at the index equal to the number in your original set as
true
. - Finally scan your
boolean
array to find and pront all the indices with valuefalse
.
Example:
Original Set: {1,0,3}
Step 1: Maximum number in the set = 3
Step 2: Boolean Array with Length = 3+1 --> {false, false, false, false}
Step 3: While scanning original set and setting values in theboolean
array this will be the final state --> {true, true, false, true}
Step 4: You'll finally scan theboolean
array to print 2 since this it is only index which has value =false
My code shown that it is not mandatory to start an array from zero. As the elements depends upon array index. if(numbers[0]=0 ) then its ok if (numbers[0]=5) then it is also ok. All it is matter that the array is a sorted array.
– Encipher
Dec 17 '18 at 22:13
add a comment |
int numbers = { 11, 6, 4, 5, 7, 1 };
Arrays.sort(numbers);
int numbersArrayIndex = 0;
for (int i = 0; i < numbers[numbers.length - 1]; i++) {
if (i == numbers[numbersArrayIndex]) {
numbersArrayIndex++;
}
else {
System.out.println(i);
}
}
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
This code uses a HashSet
:
public static void main(String args) {
int numbers = {9, 6, 4, 5, 7, 0, 1};
Arrays.sort(numbers);
HashSet<Integer> set = new HashSet<>();
for (int i = numbers[0]; i < numbers[numbers.length - 1]; i++) {
set.add(i);
}
for (int i = 0; i < numbers.length; i++) {
set.remove(numbers[i]);
}
for (int x : set) {
System.out.print(x + " ");
}
}
will print:
2 3 8
Here is how it works:
1. Adds all numbers from the minimum number of the array to the maximum number of the array to the set.
2. Iterates through the array and removes every item of the array from the set.
3. Prints the remaining items in the set, which are all the missing items of the array.
add a comment |
This code uses a HashSet
:
public static void main(String args) {
int numbers = {9, 6, 4, 5, 7, 0, 1};
Arrays.sort(numbers);
HashSet<Integer> set = new HashSet<>();
for (int i = numbers[0]; i < numbers[numbers.length - 1]; i++) {
set.add(i);
}
for (int i = 0; i < numbers.length; i++) {
set.remove(numbers[i]);
}
for (int x : set) {
System.out.print(x + " ");
}
}
will print:
2 3 8
Here is how it works:
1. Adds all numbers from the minimum number of the array to the maximum number of the array to the set.
2. Iterates through the array and removes every item of the array from the set.
3. Prints the remaining items in the set, which are all the missing items of the array.
add a comment |
This code uses a HashSet
:
public static void main(String args) {
int numbers = {9, 6, 4, 5, 7, 0, 1};
Arrays.sort(numbers);
HashSet<Integer> set = new HashSet<>();
for (int i = numbers[0]; i < numbers[numbers.length - 1]; i++) {
set.add(i);
}
for (int i = 0; i < numbers.length; i++) {
set.remove(numbers[i]);
}
for (int x : set) {
System.out.print(x + " ");
}
}
will print:
2 3 8
Here is how it works:
1. Adds all numbers from the minimum number of the array to the maximum number of the array to the set.
2. Iterates through the array and removes every item of the array from the set.
3. Prints the remaining items in the set, which are all the missing items of the array.
This code uses a HashSet
:
public static void main(String args) {
int numbers = {9, 6, 4, 5, 7, 0, 1};
Arrays.sort(numbers);
HashSet<Integer> set = new HashSet<>();
for (int i = numbers[0]; i < numbers[numbers.length - 1]; i++) {
set.add(i);
}
for (int i = 0; i < numbers.length; i++) {
set.remove(numbers[i]);
}
for (int x : set) {
System.out.print(x + " ");
}
}
will print:
2 3 8
Here is how it works:
1. Adds all numbers from the minimum number of the array to the maximum number of the array to the set.
2. Iterates through the array and removes every item of the array from the set.
3. Prints the remaining items in the set, which are all the missing items of the array.
answered Dec 17 '18 at 22:28
forpasforpas
9,9271421
9,9271421
add a comment |
add a comment |
replace the else
clause to be:
for(int j=numbers[i-1] + 1; j <= numbers[i] - 1; j++) {
System.out.println( "Missing number is " + ( j ) );
}
let's examine the case: {9 ,6 ,4 ,5 ,7 ,0 , 1}
after sorting it will be: {0, 1, 4, 5, 6, 7, 9}
now if i
is at index 2 it finds the difference between numbers[i]
and numbers[i-1]
to be not equal 1 (4 - 1 = 3), now you need ALL the numbers between 1 and 4 which are 2, 3 and thus you must loop from numbers[i-1]
to numbers[i]
(exclusive) in order to achieve this.
The complexity of this code is big O of N
(O(N))
, where N
is the biggest element in your array.
add a comment |
replace the else
clause to be:
for(int j=numbers[i-1] + 1; j <= numbers[i] - 1; j++) {
System.out.println( "Missing number is " + ( j ) );
}
let's examine the case: {9 ,6 ,4 ,5 ,7 ,0 , 1}
after sorting it will be: {0, 1, 4, 5, 6, 7, 9}
now if i
is at index 2 it finds the difference between numbers[i]
and numbers[i-1]
to be not equal 1 (4 - 1 = 3), now you need ALL the numbers between 1 and 4 which are 2, 3 and thus you must loop from numbers[i-1]
to numbers[i]
(exclusive) in order to achieve this.
The complexity of this code is big O of N
(O(N))
, where N
is the biggest element in your array.
add a comment |
replace the else
clause to be:
for(int j=numbers[i-1] + 1; j <= numbers[i] - 1; j++) {
System.out.println( "Missing number is " + ( j ) );
}
let's examine the case: {9 ,6 ,4 ,5 ,7 ,0 , 1}
after sorting it will be: {0, 1, 4, 5, 6, 7, 9}
now if i
is at index 2 it finds the difference between numbers[i]
and numbers[i-1]
to be not equal 1 (4 - 1 = 3), now you need ALL the numbers between 1 and 4 which are 2, 3 and thus you must loop from numbers[i-1]
to numbers[i]
(exclusive) in order to achieve this.
The complexity of this code is big O of N
(O(N))
, where N
is the biggest element in your array.
replace the else
clause to be:
for(int j=numbers[i-1] + 1; j <= numbers[i] - 1; j++) {
System.out.println( "Missing number is " + ( j ) );
}
let's examine the case: {9 ,6 ,4 ,5 ,7 ,0 , 1}
after sorting it will be: {0, 1, 4, 5, 6, 7, 9}
now if i
is at index 2 it finds the difference between numbers[i]
and numbers[i-1]
to be not equal 1 (4 - 1 = 3), now you need ALL the numbers between 1 and 4 which are 2, 3 and thus you must loop from numbers[i-1]
to numbers[i]
(exclusive) in order to achieve this.
The complexity of this code is big O of N
(O(N))
, where N
is the biggest element in your array.
edited Dec 17 '18 at 22:47
answered Dec 17 '18 at 22:10
Mis94Mis94
1,0601921
1,0601921
add a comment |
add a comment |
There are a lot of questions which are unanswered here. For Example, Does an array always start with zero?, What is the maximum possible size? etc.
Here is a simple way to approach this problem,
- Find the maximum number in your set.
- Create an empty
boolean
array of the length as that of the max number you found in the last step plus one. - Scan your original set and set the value of your new boolean array at the index equal to the number in your original set as
true
. - Finally scan your
boolean
array to find and pront all the indices with valuefalse
.
Example:
Original Set: {1,0,3}
Step 1: Maximum number in the set = 3
Step 2: Boolean Array with Length = 3+1 --> {false, false, false, false}
Step 3: While scanning original set and setting values in theboolean
array this will be the final state --> {true, true, false, true}
Step 4: You'll finally scan theboolean
array to print 2 since this it is only index which has value =false
My code shown that it is not mandatory to start an array from zero. As the elements depends upon array index. if(numbers[0]=0 ) then its ok if (numbers[0]=5) then it is also ok. All it is matter that the array is a sorted array.
– Encipher
Dec 17 '18 at 22:13
add a comment |
There are a lot of questions which are unanswered here. For Example, Does an array always start with zero?, What is the maximum possible size? etc.
Here is a simple way to approach this problem,
- Find the maximum number in your set.
- Create an empty
boolean
array of the length as that of the max number you found in the last step plus one. - Scan your original set and set the value of your new boolean array at the index equal to the number in your original set as
true
. - Finally scan your
boolean
array to find and pront all the indices with valuefalse
.
Example:
Original Set: {1,0,3}
Step 1: Maximum number in the set = 3
Step 2: Boolean Array with Length = 3+1 --> {false, false, false, false}
Step 3: While scanning original set and setting values in theboolean
array this will be the final state --> {true, true, false, true}
Step 4: You'll finally scan theboolean
array to print 2 since this it is only index which has value =false
My code shown that it is not mandatory to start an array from zero. As the elements depends upon array index. if(numbers[0]=0 ) then its ok if (numbers[0]=5) then it is also ok. All it is matter that the array is a sorted array.
– Encipher
Dec 17 '18 at 22:13
add a comment |
There are a lot of questions which are unanswered here. For Example, Does an array always start with zero?, What is the maximum possible size? etc.
Here is a simple way to approach this problem,
- Find the maximum number in your set.
- Create an empty
boolean
array of the length as that of the max number you found in the last step plus one. - Scan your original set and set the value of your new boolean array at the index equal to the number in your original set as
true
. - Finally scan your
boolean
array to find and pront all the indices with valuefalse
.
Example:
Original Set: {1,0,3}
Step 1: Maximum number in the set = 3
Step 2: Boolean Array with Length = 3+1 --> {false, false, false, false}
Step 3: While scanning original set and setting values in theboolean
array this will be the final state --> {true, true, false, true}
Step 4: You'll finally scan theboolean
array to print 2 since this it is only index which has value =false
There are a lot of questions which are unanswered here. For Example, Does an array always start with zero?, What is the maximum possible size? etc.
Here is a simple way to approach this problem,
- Find the maximum number in your set.
- Create an empty
boolean
array of the length as that of the max number you found in the last step plus one. - Scan your original set and set the value of your new boolean array at the index equal to the number in your original set as
true
. - Finally scan your
boolean
array to find and pront all the indices with valuefalse
.
Example:
Original Set: {1,0,3}
Step 1: Maximum number in the set = 3
Step 2: Boolean Array with Length = 3+1 --> {false, false, false, false}
Step 3: While scanning original set and setting values in theboolean
array this will be the final state --> {true, true, false, true}
Step 4: You'll finally scan theboolean
array to print 2 since this it is only index which has value =false
answered Dec 17 '18 at 22:08
user2004685user2004685
7,35642143
7,35642143
My code shown that it is not mandatory to start an array from zero. As the elements depends upon array index. if(numbers[0]=0 ) then its ok if (numbers[0]=5) then it is also ok. All it is matter that the array is a sorted array.
– Encipher
Dec 17 '18 at 22:13
add a comment |
My code shown that it is not mandatory to start an array from zero. As the elements depends upon array index. if(numbers[0]=0 ) then its ok if (numbers[0]=5) then it is also ok. All it is matter that the array is a sorted array.
– Encipher
Dec 17 '18 at 22:13
My code shown that it is not mandatory to start an array from zero. As the elements depends upon array index. if(numbers[0]=0 ) then its ok if (numbers[0]=5) then it is also ok. All it is matter that the array is a sorted array.
– Encipher
Dec 17 '18 at 22:13
My code shown that it is not mandatory to start an array from zero. As the elements depends upon array index. if(numbers[0]=0 ) then its ok if (numbers[0]=5) then it is also ok. All it is matter that the array is a sorted array.
– Encipher
Dec 17 '18 at 22:13
add a comment |
int numbers = { 11, 6, 4, 5, 7, 1 };
Arrays.sort(numbers);
int numbersArrayIndex = 0;
for (int i = 0; i < numbers[numbers.length - 1]; i++) {
if (i == numbers[numbersArrayIndex]) {
numbersArrayIndex++;
}
else {
System.out.println(i);
}
}
add a comment |
int numbers = { 11, 6, 4, 5, 7, 1 };
Arrays.sort(numbers);
int numbersArrayIndex = 0;
for (int i = 0; i < numbers[numbers.length - 1]; i++) {
if (i == numbers[numbersArrayIndex]) {
numbersArrayIndex++;
}
else {
System.out.println(i);
}
}
add a comment |
int numbers = { 11, 6, 4, 5, 7, 1 };
Arrays.sort(numbers);
int numbersArrayIndex = 0;
for (int i = 0; i < numbers[numbers.length - 1]; i++) {
if (i == numbers[numbersArrayIndex]) {
numbersArrayIndex++;
}
else {
System.out.println(i);
}
}
int numbers = { 11, 6, 4, 5, 7, 1 };
Arrays.sort(numbers);
int numbersArrayIndex = 0;
for (int i = 0; i < numbers[numbers.length - 1]; i++) {
if (i == numbers[numbersArrayIndex]) {
numbersArrayIndex++;
}
else {
System.out.println(i);
}
}
answered Dec 20 '18 at 12:56
techtech
364
364
add a comment |
add a comment |
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But
ArrayList
supports indexing, right?– Kingsley
Dec 17 '18 at 22:04
1
With a list, instead of subscripting, you use the
get(int)
method, with the index as the method argument.– Ted Hopp
Dec 17 '18 at 22:07
"But as we all know Java array is immutable" - no, Java arrays are mutable. They're not resizable, but they're mutable.
– user2357112
Dec 18 '18 at 1:03