How do you find the missing number in an unsorted array?

Algorithm:

1. Calculate the sum of first n natural numbers as sumtotal= n*(n+1)/2.
2. Create a variable sum to store the sum of array elements.
3. Traverse the array from start to end.
4. Update the value of sum as sum = sum + array[i]
5. Print the missing number as sumtotal – sum.

Which algorithm is the most appropriate to use when searching for a number in an unsorted array?

Sequential search
Sequential search is the best that we can do when trying to find a value in an unsorted array.

Which method is fast for insertion and removal of unsorted data?

SortedDictionaryTValue> has faster insertion and removal operations for unsorted data, O(log n) as opposed to O(n) for SortedList.

How do you find the missing number in a given array from number 1 to N in Java?

How to find the missing number in a given Array from number 1 to n in Java?

1. The sum of n sequential numbers will be [n*(n+1)]/2. Using this get the sum of the numbers the n numbers.
2. Add all the elements in the array.
3. Subtract the sum of the numbers in the array from the sum of the n numbers.

How are duplicates removed from an array without using any library?

“how are duplicates removed from an array without using any library in java” Code Answer

1. // Must sort arrays first –> Arrays.sort(arrayName)
2. public class RemoveDuplicateInArrayExample{
3. public static int removeDuplicateElements(int arr[], int n){
4. if (n==0 || n==1){
5. return n;
6. }
7. int[] temp = new int[n];
8. int j = 0;

Which searching algorithm you will prefer when the data is not sorted and less in number?

Binary search is an efficient algorithm for finding an item from a sorted list of items. It works by repeatedly dividing in half the portion of the list that could contain the item, until you’ve narrowed down the possible locations to just one.

How are duplicates removed from an array?

Approach: Take a Set. Insert all array elements in the Set. Set does not allow duplicates and sets like LinkedHashSet maintains the order of insertion so it will remove duplicates and elements will be printed in the same order in which it is inserted.

How to find the first and second Missing numbers from an array?

We can find the first missing number as a sum of natural numbers from 1 to avg, i.e., avg* (avg+1)/2 minus the sum of array elements smaller than avg We can find the second missing number by subtracting the first missing number from the sum of missing numbers

What is the average of missing integers in an array?

Average of missing integers = 6/2 = 3. Sum of array elements less than or equal to average = 1 + 3 = 4 Sum of natural numbers from 1 to avg = avg* (avg + 1)/2 = 3*4/2 = 6 First missing number = 6 – 4 = 2 Second missing number = Sum of missing integers-First missing number Second missing number = 6-2= 4

How to count all distinct elements in an unsorted array?

Given an unsorted array, count all distinct elements in it. Examples: Input : arr[] = {10, 20, 20, 10, 30, 10} Output : 3 There are three distinct elements 10, 20 and 30.

How to print the missing number from an array in C++?

Print the missing number as a sum. 1 Create a variable sum = 1 to which will store the missing number and a counter c = 2. 2 Traverse the array from start to end. 3 Update the value of sum as sum = sum – array [i] + c and update c as c++. 4 Print the missing number as a sum.