CONCEPTS USED:
Basic Mathematics
DIFFICULTY LEVEL:
Medium
PROBLEM STATEMENT(SIMPLIFIED):
Given an array
AwithN-1elements, with elements from1toNpresent into it.
Find a single missing element.Note – Do not use sorting or any in-built function to solve the question.
For Example:
Input : A[] = [1, 2, 5, 4, 6, 7]
Output : 3
Explanation : Since 3 is the only element not present in the array.
SOLVING APPROACH:
Given a number
N, and elements from1-Nare only present in the array, out of which exactly only1element is missing.We can make use of the formula of sum of
Nnatural numbers from the value ofN.We can also traverse the array and find the sum of all
the elements in the array.Finally, the difference of the value produced by the formula and the sum value will give us our desired missing number.
Note: Take special care of type conversions between int and long long int.
ILLUSTRATION:
A[] = [1, 2, 5, 4, 6, 7]
Sum of n natural numbers = n*(n+1)/2 = 7*(7+1)/2 = 7*8/2 = 28
Sum of elements of Array = 25
Missing number = 28 - 25 = 3
SOLUTIONS:
#includeint main() { int t; scanf("%d",&t); while(t--) { long long n; scanf("%lld",&n); int arr[n-1]; long long sum=0; long long sum_of_n = (n*(n+1))/2; for(int i=0;i
#includeusing namespace std; #define ll long long int main() { int t;cin>>t; while(t--) { long long n;cin>>n; int arr[n-1]; long long sum=0; long long sum_of_n = (n*(n+1))/2; for(int i=0;i >arr[i]; sum += arr[i]; } long long res = sum_of_n - sum; cout<
import java.util.*;
import java.io.*;
public class Main {
public static void main(String args[]) throws IOException {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t!=0)
{
int n = sc.nextInt();
int arr[] = new int[n-1];
long sum = 0;
long sum_of_n = ((long)n*(n+1))/2;
for(int i=0;i
Space Complexity:
O(1)
[forminator_quiz id="635"]
This article tried to discuss Basic Mathematics. Hope this blog helps you understand and solve the problem. To practice more problems on Basic Mathematics you can check out .
