Good one! below is the solution where it's not required to use 2 for loops: Integer[] arr = new Integer[]{2, 3, 4, 6, 7}; Integer missingNum=null; for(int i=0; i
Hi Naveen, What if u want to print two or more missing numbers in an array when there is no declaration and if variables(user defined) are taken at runtime. Please explain this scenario?. And thanks for ur knowledge sharing on the above video.
This is brute force method. As an optimization you can also use (n*(n+1))/2 - sum of first n natural numbers, to get that one missing number. Also the array has to be sorted and contain only natural numbers.
Great work Naveen!! Instead of changing the values in for loop we can use like this package Test; public class Learn { public static void main(String args[]){ int a[]={100,101,103,104,105}; int sum=0; int sum1=0; for(int i=0;i
instead of 2 arrays we can also make use of series sum formula and 1 loop: private static int missingElements(int[] x, int no) { int a1 = 1; int an = no; int expectedSum = (no * (a1 + an)) / 2; int actualSum = 0; for (int i = 0; i < x.length; i++) { actualSum=actualSum+x[i];
Great tutorials. Your logic won't work when 0 is missing in an array starting from -1 to like 10 .. below will work public static void FindMissingNumInArray(int arr1[]){ for (int i=1;i
for negative number u cant use this method try this private static void printMissingNumber(int[] numbers, int count) { int missingCount = count - numbers.length; BitSet bitSet = new BitSet(count);
for (int number : numbers) { bitSet.set(number - 1); }
System.out.printf("Missing numbers in integer array %s, with total number %d is %n", Arrays.toString(numbers), count); int lastMissingIndex = 0; for (int i = 0; i < missingCount; i++) { lastMissingIndex = bitSet.nextClearBit(lastMissingIndex); System.out.println(++lastMissingIndex); }
Hi Naveen, Below code can check for multiple missing numbers public static void main(String[] args) { int [] arr = {-2,0,1,2,4,5,6,8,9,11}; int t=arr[0]+1; // Get the first number in array for(int i=1; i
Hi,But if two or more numbers are missing one after the other,all the missing numbers are not getting printed like eg: int[] nums = {-3,0,1,2,4,5,7,9,10} . here,-1 is not getting printed.
Hi Naveen, Nicely break down and put together. Allow me to contribute to extend the search of missing number from any given sequence. I look forward to hear from you at your earliest convenience. ------------------ public static void main(String[] args) { int a[] = {55,56,57,59,60}; // 58 is missing. int sum = 0; int sum1 = 0; arrLen = a.length; // length of integer array.
for (int i = 0; i < arrLen; i++) { sum += a[i]; // similar to :- sum = sum + a[i]; } System.out.println("sum =" + sum);
I used arrays.binarysearch to reduce the time complexity. Works for multiple missing numbers in sorted order. int [] a = {2,3,4,6,7,8}; int searchnumber = a[0]; for (int i = 0; i < a.length; i++) { int found = Arrays.binarySearch(a, searchnumber); if (found
Hi Naveen, Thanks for sharing. what if we asked to write a program for more than one missing numbers? do they ask such questions? if so what would be the logic. thanks in advance
Well explanation Naveen thank you very much , I have question in second for loop you gave j=1, not sure why i am thinking 0 because index start from 0 right like first loop?
Set the starting index as x = 1. Check whether the previous number array[x-1] is less than of the current number array[x] by 1. If it's not lesser by 1.. then subtract the value at array[x] by array[x-1]. You will get the a number. If the number is 2, then the missing number must be array[x]-1 or array[x-1]+1. In other words.. it's 1 step lesser than the number on the right or 1 step greater than the number on the left. Suppose if there are multiple numbers i.e the difference gives more than 2. 1,2,6,7 6-2 = 4 Consider 4 as n. Which means there are 3 (n-1 -> 4-1) elements missing between 2 and 6. Those three elements are 3,4,5.
Assumes values are assumed sorted and hard to read: public static void main(String [] args) { int a[] = {1, 2, 3, 4, 6, 7}; Collection values = new HashSet(); for (int i : a) { values.add(i); } int max = Arrays.stream(a).max().getAsInt(); int min = Arrays.stream(a).min().getAsInt(); for (int i = min; i < max; i++) { if (!values.contains(i)) { System.out.println(i + " is missing."); } } }
Easiest way & this will print more than one missing numbers as well. package arrays; import java.util.HashSet; import java.util.Set; public class findMissingNumber { public static void main(String[] args) { int [] abc = {1,2,4,7}; int min = Integer.MAX_VALUE; int max = Integer.MIN_VALUE; Set seen = new HashSet(); for(int num:abc) { seen.add(num); min = Math.min(min,num); max = Math.max(max,num); } for(int i=min;i
This code works with your example: In this case the missing number is 23 public static void main(String[] args) { final int[] n = {20, 21, 22, 24, 25}; int sum1 = Arrays.stream(n).sum(); int sum2 = IntStream.rangeClosed(n[0], n[n.length - 1]).sum(); System.out.println(sum2 - sum1); }
I think this Solution is not generic, Let say if user forget to add 2 no instead of single no, then this solution is failed, Plz fix this video a{1,2,3,5,7} ex 4 and 6 were missing
use this method private static void printMissingNumber(int[] numbers, int count) { int missingCount = count - numbers.length; BitSet bitSet = new BitSet(count);
for (int number : numbers) { bitSet.set(number - 1); }
System.out.printf("Missing numbers in integer array %s, with total number %d is %n", Arrays.toString(numbers), count); int lastMissingIndex = 0; for (int i = 0; i < missingCount; i++) { lastMissingIndex = bitSet.nextClearBit(lastMissingIndex); System.out.println(++lastMissingIndex); }
public static void main(String[] args) { // TODO Auto-generated method stub int[] numbers = new int[] { 211, 212, 213, 214, 215, 217, 218, 219}; findMissingNumber(numbers); } //Write a java program to find missing number in an array? public static void findMissingNumber(int[] numbers) { int arrayTotal = 0; int missingTotal = 0; int min = 0; int max = 0; for(int i=0;i numbers[i]) { min = numbers[i]; } if(max < numbers[i]) { max = numbers[i]; } arrayTotal = arrayTotal + numbers[i]; } for(int i=min;i
Good one!
below is the solution where it's not required to use 2 for loops:
Integer[] arr = new Integer[]{2, 3, 4, 6, 7};
Integer missingNum=null;
for(int i=0; i
Hi Naveen,
What if u want to print two or more missing numbers in an array when there is no declaration and if variables(user defined) are taken at runtime. Please explain this scenario?. And thanks for ur knowledge sharing on the above video.
This is brute force method. As an optimization you can also use (n*(n+1))/2 - sum of first n natural numbers, to get that one missing number. Also the array has to be sorted and contain only natural numbers.
i ,agree with you @ssandeep79 your method is correct ! :)
Read the comments only to see if someone already gave the formula for sum of n natural numbers. Was not disappointed 😁
Great work Naveen!!
Instead of changing the values in for loop we can use like this
package Test;
public class Learn {
public static void main(String args[]){
int a[]={100,101,103,104,105};
int sum=0;
int sum1=0;
for(int i=0;i
yes u can do that. I have taken this example just to explain the concept. U can remove the hard coded values in the second loop.
What if the list is not sorted?
But does this work if boundary values are missing?
This code doesn't work for the boundary values!
instead of 2 arrays we can also make use of series sum formula and 1 loop:
private static int missingElements(int[] x, int no) {
int a1 = 1;
int an = no;
int expectedSum = (no * (a1 + an)) / 2;
int actualSum = 0;
for (int i = 0; i < x.length; i++) {
actualSum=actualSum+x[i];
}
return expectedSum - actualSum;
}
public class missingnumber
{
public static void main(String[] args)
{
int arr[]= {1,2,3,4,5,6,8,9,10};
for(int i=0;i
Great tutorials. Your logic won't work when 0 is missing in an array starting from -1 to like 10 .. below will work
public static void FindMissingNumInArray(int arr1[]){
for (int i=1;i
Hi, I have tried it is working
for negative number u cant use this method
try this
private static void printMissingNumber(int[] numbers, int count) {
int missingCount = count - numbers.length;
BitSet bitSet = new BitSet(count);
for (int number : numbers) {
bitSet.set(number - 1);
}
System.out.printf("Missing numbers in integer array %s, with total number %d is %n",
Arrays.toString(numbers), count);
int lastMissingIndex = 0;
for (int i = 0; i < missingCount; i++) {
lastMissingIndex = bitSet.nextClearBit(lastMissingIndex);
System.out.println(++lastMissingIndex);
}
}
Hello, your code works but can you please give some explanation of your logic, i can see it in the code but please explain.Thanks.
Your code fails when the numbers are not sorted
@@abhifast555
Arrays.sort(arr1);
will solve the problem
but if there negative numbers
wont work
only with positive numbers
Hi Naveen, thanks for the code. If we missed more than one number, we will not get output right?
Hi Naveen,
Below code can check for multiple missing numbers
public static void main(String[] args) {
int [] arr = {-2,0,1,2,4,5,6,8,9,11};
int t=arr[0]+1; // Get the first number in array
for(int i=1; i
Hi Pradeep, simple we can write down program like below
//Program to identify missing multiple numbers
int[] nums = {1,2,4,5,7,9,10};
for(int i=0;i
Hi,But if two or more numbers are missing one after the other,all the missing numbers are not getting printed like eg: int[] nums = {-3,0,1,2,4,5,7,9,10} . here,-1 is not getting printed.
Hi Naveen, what if there is two missing number in array?. How we can calculate, can you explain.
Naveen, thanks for the videos. What if there are 2 or more missing numbers in the array how would you do it?
Use XOR Operation. Here is the code:
int[] a = {100,101,103,104,105,102,107};
int res = a.length ;
for(int i = 0 ;i
Hi Naveen, Nicely break down and put together. Allow me to contribute to extend the search of missing number from any given sequence. I look forward to hear from you at your earliest convenience.
------------------
public static void main(String[] args) {
int a[] = {55,56,57,59,60}; // 58 is missing.
int sum = 0;
int sum1 = 0;
arrLen = a.length; // length of integer array.
for (int i = 0; i < arrLen; i++) {
sum += a[i];
// similar to :- sum = sum + a[i];
}
System.out.println("sum =" + sum);
for (int j = a[0]; j
this doesn't work for boundary values
I used arrays.binarysearch to reduce the time complexity. Works for multiple missing numbers in sorted order. int [] a = {2,3,4,6,7,8};
int searchnumber = a[0];
for (int i = 0; i < a.length; i++) {
int found = Arrays.binarySearch(a, searchnumber);
if (found
Thanks for sharing good questions, what if there are more than one missing numbers in the array?
Doesn't work
Why don't you use arithmetic progression formula instead of running for loop.good explanation by the way
Hi Naveen,
Thanks for sharing. what if we asked to write a program for more than one missing numbers? do they ask such questions? if so what would be the logic.
thanks in advance
Well explanation Naveen thank you very much , I have question in second for loop you gave j=1, not sure why i am thinking 0 because index start from 0 right like first loop?
@@vinaykumarsharma9493 thanks lot Vinaykumar got it it's just number count
@@vinaykumarsharma9493 hmm right thanks Vinay :)
Hi Naveen, Thanks for wonderful tutorial.I have a question
How to find missing number from unsorted array?
you can use this same method for an unsorted array as well
because here we are only calculating the difference between sums
Hi Naveen,
Wonderful explanation! Also can you start with alogirthms and Data structures in Java sessions, which will be helpful for many of us.
Hi Naveen,
In this video i got to know how to find one missing number. If there are multiple numbers are missing what logic we need to follow?
Set the starting index as x = 1.
Check whether the previous number array[x-1] is less than of the current number array[x] by 1.
If it's not lesser by 1.. then subtract the value at array[x] by array[x-1]. You will get the a number.
If the number is 2, then the missing number must be array[x]-1 or array[x-1]+1. In other words.. it's 1 step lesser than the number on the right or 1 step greater than the number on the left.
Suppose if there are multiple numbers i.e the difference gives more than 2.
1,2,6,7
6-2 = 4
Consider 4 as n.
Which means there are 3 (n-1 -> 4-1) elements missing between 2 and 6.
Those three elements are 3,4,5.
Please include Anagram Program in interview programs
what if the 2 or 3 numbers will be miss at same time means...?
Very helpful
if i want to write code for two or three missing letters then whats was the logic
Assumes values are assumed sorted and hard to read:
public static void main(String [] args) {
int a[] = {1, 2, 3, 4, 6, 7};
Collection values = new HashSet();
for (int i : a) {
values.add(i);
}
int max = Arrays.stream(a).max().getAsInt();
int min = Arrays.stream(a).min().getAsInt();
for (int i = min; i < max; i++) {
if (!values.contains(i)) {
System.out.println(i + " is missing.");
}
}
}
Please add trailing of white space. . .
how to find if more than 1 number is missing ?
Thanks You , I Understand 😍
why did you start second loop j
Thank you very much!!
Hi Naveen ji, if two or more numbers are missing in an array..how can we find out those missing numbers..
Hi Lakshmi, did you get any luck on this? Plz share if so
two number is missing thn what will we do ? like 3 and 5
See this
sc = new Scanner(System.in);
String input []= new String[5];
int inputNum []= new int[5];
int smallestNum=0;
System.out.println("Enter the 5 numbers:");
for(int i=0; i
How to find different missing number in java please write this program sir
Easiest way & this will print more than one missing numbers as well.
package arrays;
import java.util.HashSet;
import java.util.Set;
public class findMissingNumber {
public static void main(String[] args) {
int [] abc = {1,2,4,7};
int min = Integer.MAX_VALUE;
int max = Integer.MIN_VALUE;
Set seen = new HashSet();
for(int num:abc)
{
seen.add(num);
min = Math.min(min,num);
max = Math.max(max,num);
}
for(int i=min;i
What if 0 is missing?
Hi Naveen,
Thank you boss
This will only work if numbers are from 1 to x. Not like 20, 21, 22, 23, 24, 25.
This code works with your example:
In this case the missing number is 23
public static void main(String[] args) {
final int[] n = {20, 21, 22, 24, 25};
int sum1 = Arrays.stream(n).sum();
int sum2 = IntStream.rangeClosed(n[0], n[n.length - 1]).sum();
System.out.println(sum2 - sum1);
}
It works. You have to manipulate the loops:
int a[] = {22,23,25,26,27,28,29,30};
//find sum of array elements
int sum=0;
for(int i=0;i
/* MULTIPLE MISSING NUMBERS AND DUPLICATES IN THE ARRAY */
package javaInterview;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
public class FindMissingNumbers {
static int i = 0;
static ArrayList al = new ArrayList();
public static void main(String[] args) {
Integer[] numArr = { 100, 100, 90, 101, 90, 91, 92, 98, 92 };
al = FindMissingNumbers.findMissingNumbers(numArr);
System.out.println("MISSING NUMBERS ARE");
System.out.println("--------------------");
for (Integer i : al) {
System.out.println(i);
}
}
public static ArrayList findMissingNumbers(Integer arr[]) {
Set sObj = new HashSet(Arrays.asList(arr));
List lObj = new ArrayList(sObj);
Collections.sort(lObj);
Integer iSmallest = lObj.get(0);
Integer iLargest = lObj.get(lObj.size() - 1);
for (int i = iSmallest + 1; i
how can u find two numbers is missing?
I think this Solution is not generic, Let say if user forget to add 2 no instead of single no, then this solution is failed, Plz fix this video a{1,2,3,5,7} ex 4 and 6 were missing
int arr[] = {1,2,3,4,5,6,8,10,11,12,13,15};
int len = arr.length;
int j = 1;
for(int i = 0; i< len; i++)
{
if(arr[i] != j)
{
System.out.println(j);
j++;
}
j++;
}
use this method
private static void printMissingNumber(int[] numbers, int count) {
int missingCount = count - numbers.length;
BitSet bitSet = new BitSet(count);
for (int number : numbers) {
bitSet.set(number - 1);
}
System.out.printf("Missing numbers in integer array %s, with total number %d is %n",
Arrays.toString(numbers), count);
int lastMissingIndex = 0;
for (int i = 0; i < missingCount; i++) {
lastMissingIndex = bitSet.nextClearBit(lastMissingIndex);
System.out.println(++lastMissingIndex);
}
}
What if multiple numbers are missing in the sequence?
Doesnt work with zeros
⭐⭐⭐⭐⭐
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] numbers = new int[] { 211, 212, 213, 214, 215, 217, 218, 219};
findMissingNumber(numbers);
}
//Write a java program to find missing number in an array?
public static void findMissingNumber(int[] numbers) {
int arrayTotal = 0;
int missingTotal = 0;
int min = 0;
int max = 0;
for(int i=0;i numbers[i]) {
min = numbers[i];
}
if(max < numbers[i]) {
max = numbers[i];
}
arrayTotal = arrayTotal + numbers[i];
}
for(int i=min;i