Find missing number in an array

If the array doesn't contain any missing than

mast vide thi, sare methods samajh aa gye. Thanq.

sir very good explain thankyou sir ji

SOLUTION : Here is the solution of all these method in javascript

1) Find Missing Number By Hashing :
const findMissingNumberByHashing = (array: number[]) => {
const hash: Record<string, number> = {};
for (let i = 0; i < array.length; i++) {
hash[array[i]] = 1;
}
for (let i = 1; i <= array.length + 1; i++) {
if (!hash[i]) return i;
}
};

findMissingNumberByHashing([1, 3, 4, 5, 6]);

2) Find Missing Number by Gauss's Method for Summing :
const findMissingNumber = (array: number[]) => {
// (Gauss's Method for Summing)
const n = array.length + 1; // 7 (adding 1 because one number is missing in the array)
const sum = (n * (n + 1)) / 2; // 28
const arraySum = array.reduce((acc, curr) => acc + curr, 0); // 21
return sum - arraySum; // 28 - 21 = 7
};


findMissingNumber([1, 2, 3, 4, 5, 6])

3) Find Missing Number By XOR :
const findMissingNumberByXOR = (array: number[]) => {
let x1 = array[0]; // 1
let x2 = 1; // 1
for (let i = 1; i < array.length; i++) { // in this loop we are calculating the XOR of all the elements of the array
x1 = x1 ^ array[i]; // 1 ^ 3 ^ 4 ^ 5 ^ 6 = 3
}
for (let i = 2; i <= array.length + 1; i++) { // in this loop we are calculating the XOR of all the numbers from 1 to n + 1
x2 = x2 ^ i; // 1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 6 = 1
}
return x1 ^ x2; // 3 ^ 1 = 2
}

findMissingNumberByXOR([1, 3, 4, 5, 6]);

i found out better approach T.C->O(LOGN) (IF ARRAY IS SORTED) WE CAN USE BINARY SEARCH
int s = 0;
int e = N - 1;

while (s <= e) {
int mid = s + (e - s) / 2;

// If the element at index mid is not in its correct position
// i.e., arr[mid] != mid + 1
if (arr[mid] != mid + 1) {
// If arr[mid] is greater than mid + 1 or arr[mid] is negative,
// the missing number is on the left side
if (arr[mid] > mid + 1 || arr[mid] <= 0) {
e = mid - 1;
}
// If arr[mid] is less than mid + 1, the missing number is on the right side
else {
s = mid + 1;
}
}
// If the element is in its correct position, search on the right side
else {
s = mid + 1;
}
}

return s + 1; // The missing number is s + 1

New 2 Method
1.Match the array position no OR
2.find Even or Odd no

can you pls give its source code.. it would be really helpful

code tera baap sikayege kya

great

Thank you sir , Just one question that how can I build my mind to think for the solution in such various approaches

We can Also Use Cyle Sort

XORing all the elements & then XORing the numbers till n.
And then XORing both tof the obtained results.
This brings us the left out element as XOR of a no with itself is zero(like in binary).

Good video

that was just amazing, before this video i was doing like sorting all the element in the array and finding all the element whether if it is at correct index for not

Yrr 7 8 9 11 12 ke liye work nahi karega :)

As well as please explain code for the problem.

Lovely

Methode 2 won't work if 0 present in array

Code for XOR:
//N value=5:
int n;
cin>>n;
//arr=[1,2,3,4]
int arr[n-1],int ans=0;
for(int i=0;i<n-1;i++){
cin>>arr[i];
ans^=arr[i]^(i+1);
}
cout<<ans^n;

I think this one is more efficient, in very few cases you have to iterate over whole array. If array is in order.

for(int i=0;i<arr.length;i++){
if(arr[i] != i+1){
S.o.p.l(i+1);
break;
}

brute better optimal -------> awesome.

Sir, kindly explain what is overflow for this case.

There are lot of numbers in a list
All of them are whole numbers below 100
They are in not any order
Just random whole numbers below 100
We have to find list of numbers which are missing in the series?
Anyone solve it ?

From where n=5 comes ??

note that xor of any number repeats every four digits. so you ca use mod to predetermine the xor of Len(nums). then to find the missing number find the xor of the nums values ONLY. finally return the xor of Len(nums).^ xor of the nums values

nice explanation sir

use binary search, time complexity will be logN

GOOD BLOG

Sir can you explain the code for XOR method ??

Your voice is same as apna college channel...

will help if the array is in order

function missingNum(arr) {
for(i=arr.length-1;i>=0;i--) {
if(arr[i] - arr[i-1] == 2) {
return arr[i]
}
}
return -1;
}

Can you please share the code of the third approach?

XOR one is really cool . I tried with -negative value and array is not sorted still working perfect . Thank a lot sir ji

Great video ! 🙌 I understand the XOR operation by this video.

Sir, what about the time it takes to make the second array(the complete one)?

Sort the array then

For(int i=0;i<n;i++)
if(arr[i]!=i+1)
printf("%d",i+1);

Your explaining level ....😍😍😍😍

in second method the n=4 not 5

nice solution!

Method 2 sum formula not work for all
As if 1, 2,5 now answer must be 3,4 but using tis formula we get 7 wrong

How to find more then one missing element???

find missing number in arere

Write a Swift program to check if one of the first 4 elements in a given array of integers is a 7. The length of the array may be less than 4. PLS MAKE VIDEO ON IT

Thanks! for the video

For i=0 to n-1
if(arr[i]!=i+1)
return i+1;

Can’t we just: 1. initialise counter to array’s zero-th element. 2. Iterate through array increment this counter, wherever the counter and array element does not match - is the answer.

Number of terms toh aapne 4 le rkhi hai likha 5 hai

If I subtract numbers instead of xor-ing , then will it be as efficient as the xor solution?
i.e for the array 1 2 3 5
(1-1) + (2-2) + (3-3) + (4-5) + (5) = 4

Thanks