@@unknow2096 one simple approach could be, create a set containing (0-n) elements, loop via that list and remove the element from the set. The elements remaining in the set are your missing elements.
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).
We can also try subtracting method. if (arr[i] - arr[i-1] == 2) then missing number = arr[i] - 1. Time complexity is O(n) and Space Complexity is O(1).
@@surajtopal9940 Here's my python code which isn't working. def func(N,A): new=[] for n in range(max(A)+1): new.append(-1) for i in range(N): if A[i]>=0: new[A[i]]=True for j in range(1,max(A)+1): if new[j]==True: return j+1 break L=[] N=int(input('Enter the number of elements: ')) print('Now, enter the',N, 'elements:') for n in range(N): L.append(int(input())) print(func(N,L))
@@har.preet.singh. brother this is different approach use XOR operation. It is very easy. You just to XOR all the value of the array one by one and put it in the result variable after than you need to do the loop till n and XOR all in the result1 then result ^ result1 and you will get your answer.
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
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
Overflow occurs when we sum large numbers. In java integer has some range. If sum goes beyond range, it will cause overflow and doesn't give proper results.
For input int[] a = {-1, 1, 2, 3, 4, 5, 6, 7} missing Nums : 18 but it should be '0' can you suggest why I am getting .because as per the algo explain by instructor it should be pass for boundary value also : public class MissingNumber2 { private static int missingNumber(int[] input, int n) { int x1 = input[0]; int x2 = 1; for (int i = 1; i < n; i++) { x1 = x1 ^ input[i]; } for (int i = 2; i
Hi, have you ever come across below question? Given a stream of timestamp and CPU utilisations we need to return the time interval where CPU utilisations was max. Ex input 2d array. [[StartTime, EndTime, cpuUtilization]]
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
No. You will get your answer = (4 XOR 5). You won't get 4,5 as individual answers and so you will never know about it. If you have only one number missing and you knew the original array then only you can get your answer as missing element. Here, in your example you have 2 missing numbers and so you won't get the missing elements. I hope you got it :)
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.
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 ?
// Function to get the missing number int getMissingNo(int a[], int n) { // For xor of all the elements in array int x1 = a[0]; // For xor of all the elements from 1 to n+1 int x2 = 1; for (int i = 1; i < n; i++) x1 = x1 ^ a[i]; for (int i = 2; i
@@vinaykumar-sg7xd because second loop is for XOR elements from 1 to n. x2 is initialized with 1 so if we start with 1 in second loop then it will become 0 in XOR process. so to eliminate that issue, x2 is initialized with 1 and loop started from 2
The problem statement is supposed to be "Find the missing number form an array containing n distinct numbers in the range [0, n]" , Good one from Tech Dose.
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; }
Why you are returning the arr[i] ?this will not give us missing number in array for example 1,2,4 and there is 3 missing number then how we get 3 number by returning the arr[i]
