if you don`t no how elements are given you will follow this code also ---> public static void main(String[] args) { int a[]= {10,11,13,14,15,16,17}; int sum1=0,sum2=0; for(int i=0;i
And I think instead of doing -->i++ in sum2,first we do like int x=a[1]-a[0]; Then in sum2 we can do like -->i+x; It can help when sorted order is like [2,4,6,8]
For the missing Element of Array, I have the following code which does not give me the missing number. I think there is something wrong with the condition you stated: public static void main(String[] args) {
The above solution is only for any one value is missing in an array what if multiple values are missing in an array, If any one has the solution for my quiry please respond
Obviously it will come bro but while iteration 2nd for loop we have to give I value as 8 and then in the condition section we have to pass stopping range called 13. So final result we will get 10
hello sir if it is depends on user if they type some other numbers. then how can we find the missing numbers? Your logic is only when input is given in the program. please clarify
package javaapplication23; import java.util.ArrayList; import java.util.Scanner; public class JavaApplication23 { public static void main(String[] args) { int sum = 0, sum2 = 0; int[] a = {1,2,3}; int[] b = {1,2,3,4};
for (int i = 0; i < a.length; i++){ sum += a[i]; }
for (int i = 0; i < b.length; i++){ sum2 += b[i]; }
We can have below solution as well in more simpler public class FindMissingNumberInArray { public static void main(String[] args) { int arr[] = { 1, 2, 3, 4, 5, 6, 8 }; for (int i = 0; i < arr.length; i++) { for (int j = i + 1; j
//Secound for loop modified public class MissingNumberFromArray { public static void main(String[] args) { //Array should not have duplicates //Array no need to be sorted order //value should be in range int a[]= {11,12,14,15}; int sum=0; for(int value:a) { sum=sum+value; } System.out.println("Sum of element in array: "+sum); int sum2=0; for(int i=a[0];i
package FA_Java_Programs; public class ArrayMissingNumber { public static void main(String[] args) { int[] a= {6,3,4,7}; int sum=0; for(int x:a) sum+=x;
if you don`t no how elements are given you will follow this code also --->
public static void main(String[] args) {
int a[]= {10,11,13,14,15,16,17};
int sum1=0,sum2=0;
for(int i=0;i
to get correct result of sum2, array needs to be in sorted order
@@harshadagothankar965 yes
I think here we should know the last element also
And I think instead of doing -->i++ in sum2,first we do like int x=a[1]-a[0];
Then in sum2 we can do like -->i+x;
It can help when sorted order is like [2,4,6,8]
Is it write or wrong bro because I am so confusing here
line 27
for(int i=1;i
array index start from zero
yes we can use
This is possible only when 1st value is 1
For the missing Element of Array, I have the following code which does not give me the missing number. I think there is something wrong with the condition you stated: public static void main(String[] args) {
int arr[]= {1,2,3,4,5,7,8,9};
// steps - sum1 is based on the Index
int sum1=0;
for (int i=0; i
For step2 as per your case range will 1 to 9 rather than 5
You've given range till 9 in array but in sum2 for loop, you've given range till 5. So missing number will give you -24
The range is wrong
How can we calculate if two numbers are missing?
I have also the same dought. Could you possible to write the code for that.
int arr[] = { 15 ,6 , 7, 10, 11, 13 };
System.out.println("array without sort: "+Arrays.toString(arr));
//sort array ascending
Arrays.sort(arr);
System.out.println("array with ascending:"+Arrays.toString(arr));
//sort array ascending without Arrays.sort(arr)
// int t1;
// for(int i=0; i
I think using binary search algorithm
By using a foreach loop we can compare all the numbers with another list
Then we can find n missing numbers
let arr = [1, 3, 5];
arr.sort((a,b)=>a-b);
let result = [];
for(let i=0;i
Sum of numbers you can do this way .... n*n+1/2
appreciation for ur efforts that u put in this series
what if multiple numbers are missing ? in line like (1,4,5,6,7)
This is very hard-coded. A program should be as flexible as possible.
Its for the interviews of freshers.
Very very simpl explanation ❤
Thanks
int a[]= {2,3,5,7};
for(int i=0;i
n = a.length + 1
sum2 = n(n+1)/2;
return sum2 % sum1;
Sir why so complicated, use int c=arr[0] and compare using for loop for arr[i] with c++ values, if mismatch then missing.. please Let me know on this
Sir this program is for 1 missing number..if 2 numbers are missing does it show 2 missing numbers??
int arr[] = { 15 ,6 , 7, 10, 11, 13 };
System.out.println("array without sort: "+Arrays.toString(arr));
//sort array ascending
Arrays.sort(arr);
System.out.println("array with ascending:"+Arrays.toString(arr));
//sort array ascending without Arrays.sort(arr)
// int t1;
// for(int i=0; i
What to do if multiplie numbers are missing
this need to be sorted or use smallest to find the starting number. you made it to start from 1 always.
what if multiple numbers are missing ?
package Array;
import java.util.Arrays;
public class FindMissingElementInArray {
public static void main(String[] args) {
int a[]= {1,4,5,6,3};
Arrays.sort(a);
int num=a[0];
for(int i=0; i
super explanation sir
Thanks
This is only for one missing number in an array.
Am I right?
Yup
How to find missing num if numbers are not in sequence (we dont knw range values)and array length is very large?
To find missing number the array should be in the sorted order and the difference between each element should be same
The above solution is only for any one value is missing in an array what if multiple values are missing in an array, If any one has the solution for my quiry please respond
Really it's veryful...but I have qsn..if suppose if the number in odd or even sequence will it work ?
Or suppose prime no .
Make i++ in sum2 as i+x where x is x=a[1]-a[0];
according to your code input: [8,9,11,12,13] output:10 it will come means i will give 5000.
Obviously it will come bro but while iteration 2nd for loop we have to give I value as 8 and then in the condition section we have to pass stopping range called 13. So final result we will get 10
Give me 5000 thousand bro😂😂😂😂😂 please
Thanks Sir ..for this simple approach ..🕺
Keep watching
hello sir if it is depends on user if they type some other numbers. then how can we find the missing numbers? Your logic is only when input is given in the program. please clarify
int arr[] = { 15 ,6 , 7, 10, 11, 13 };
System.out.println("array without sort: "+Arrays.toString(arr));
//sort array ascending
Arrays.sort(arr);
System.out.println("array with ascending:"+Arrays.toString(arr));
//sort array ascending without Arrays.sort(arr)
// int t1;
// for(int i=0; i
I. Have the same doubt?
package javaapplication23;
import java.util.ArrayList;
import java.util.Scanner;
public class JavaApplication23 {
public static void main(String[] args) {
int sum = 0, sum2 = 0;
int[] a = {1,2,3};
int[] b = {1,2,3,4};
for (int i = 0; i < a.length; i++){
sum += a[i];
}
for (int i = 0; i < b.length; i++){
sum2 += b[i];
}
System.out.println(Math.abs(sum-sum2));
}
}
Commenting for better reach
if there are more than one missing number.how to find them?
int arr[] = { 15 ,6 , 7, 10, 11, 13 };
System.out.println("array without sort: "+Arrays.toString(arr));
//sort array ascending
Arrays.sort(arr);
System.out.println("array with ascending:"+Arrays.toString(arr));
//sort array ascending without Arrays.sort(arr)
// int t1;
// for(int i=0; i
Sir.If Array series contains multiple missing numbers.this logic works or not
❤
ty
We can have below solution as well in more simpler
public class FindMissingNumberInArray {
public static void main(String[] args) {
int arr[] = { 1, 2, 3, 4, 5, 6, 8 };
for (int i = 0; i < arr.length; i++) {
for (int j = i + 1; j
//Secound for loop modified
public class MissingNumberFromArray {
public static void main(String[] args) {
//Array should not have duplicates
//Array no need to be sorted order
//value should be in range
int a[]= {11,12,14,15};
int sum=0;
for(int value:a)
{
sum=sum+value;
}
System.out.println("Sum of element in array: "+sum);
int sum2=0;
for(int i=a[0];i
it is fine but what will happen if user gives array as a = {15 , 11 , 14 , 12 } as sir said that array can of any order ?
package FA_Java_Programs;
public class ArrayMissingNumber {
public static void main(String[] args) {
int[] a= {6,3,4,7};
int sum=0;
for(int x:a)
sum+=x;
int start=a[0];
for(int i=0;i
for missing numbers:
int arr[] = { 15 ,6 , 7, 10, 11, 13 };
System.out.println("array without sort: "+Arrays.toString(arr));
//sort array ascending
Arrays.sort(arr);
System.out.println("array with ascending:"+Arrays.toString(arr));
//sort array ascending without Arrays.sort(arr)
// int t1;
// for(int i=0; i
Thnks