Skip to main content

Decreasing Srrnmieeda October Cook off challenge Solution | Easy Explain Decreasing Srrmieeda

 Decreasing Srrnmieeda October Cook off challenge Solution | Easy Explain Decreasing Srrmieeda-

Easy Explanation of Decreasing Srrnmieeda October Cook-off challenge Solution  With Code.Here Is Problem Statement.

Problem Statement-

You are given two integers L and R. Find the smallest non-negative integer N such that


Here, % is the modulo operator, so A%B is the remainder of A after division by B. For example, 11%3=2.


  • The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
  • The first and only line of each test case contains two space-separated integers L and R.


For each test case, print a single line containing one integer ― the smallest possible N, or 1 if no solution exists.

It is guaranteed that when a solution exists, the smallest solution does not exceed 1018.


  • 1T105
  • 1L<R106

Example Input

4 6
1 2

Example Output



Example case 1: N=6 satisfies the given condition, since 6%4(=2)>6%5(=1)>6%6(=0). Notice that N=7 also satisfies the condition, but it is larger.

Example case 2: It is impossible to find a valid solution because for any non-negative integer N, we have N%1(=0)N%2.

Problem Link-

Join Telegram For Competetive Programming Guide and Update-

Decreasing Srrnmieeda  Solution -

In this Problem  given that N%L>N%(L+1)>>N%(R1)>N%R.  and L< R

1-So for satisfying first relationship N should be between >=R otherwise condition one cannot be satisfied.
2-But we know that if N is divisible by L so we  if N> R it is not possible to obtain the relation 1 that is 

According to those two statements, What can be the max value of N%L? Maxmimum value of N%L cannot
 exceed 2*L so N<2*L.
and if if(R >= 2*L) we cannot obtain sequence .

so Final Logic is -
if(R >= 2*L) cout << -1 << '\n'; else cout << R << '\n';

Here is m code-


Popular posts from this blog

Vaccine Distribution Codechef December Long challenge solution 2020

 Vaccine Distribution  Codechef December Long challenge solution 2020- This Problem vaccine distribution is taken from December long challenge 2020. lets read problem statement. Problem statement- Finally, a COVID vaccine is out on the market and the Chefland government has asked you to form a plan to distribute it to the public as soon as possible. There are a total of  N N  people with ages  a 1 , a 2 , … , a N a 1 , a 2 , … , a N . There is only one hospital where vaccination is done and it is only possible to vaccinate up to  D D  people per day. Anyone whose age is  ≥ 80 ≥ 80  or  ≤ 9 ≤ 9  is considered to be  at risk . On each day, you may not vaccinate both a person who is at risk and a person who is not at risk. Find the smallest number of days needed to vaccinate everyone. Input The first line of the input contains a single integer  T T  denoting the number of test cases. The description of  T T  test cases follows. The first line of each test case contains two space-separ

Codechef February long challenge solution 2021

   Codechef February  long challenge solution 2021- Hey guys Welcome Again, A small update CodeChef February  long challenge solution 2021 starting from 1st week of beb  2021 . Codechef long challenge is one of the popular contests of CodeChef you should participate in codechef Feb  long challenge.This is first contest of codechef in 2021 so good luck. And wish you happy new year. Contest link- I will post the solution after Contest.Stay tuned. About CodeChef  February  long challenge solution 2021: CodeChef Long Challenge is a 10-day monthly coding contest where you can show off your computer programming skills. The significance being - it gives you enough time to think about a problem, try different ways of attacking the problem, read the concepts etc. If you’re usually slow at solving problems and have ample time at hand, this is ideal for you. We also put in a lot of efforts in getting quality problems, w

Merge Two sorted array Without Extra Space

Problem statement-  Given two sorted arrays arr1[] and arr2[] of sizes N and M in non-decreasing order. Merge them in sorted order without using any extra space. Modify arr1 so that it contains the first N elements and modify arr2 so that it contains the last M elements.    Example 1: Input:  N = 4, arr1[] = [1 3 5 7]  M = 5, arr2[] = [0 2 6 8 9] Output:  arr1[] = [0 1 2 3] arr2[] = [5 6 7 8 9] Explanation: After merging the two  non-decreasing arrays, we get,  0 1 2 3 5 6 7 8 9.   Example 2: Input:  N = 2, arr1[] = [10, 12]  M = 3, arr2[] = [5 18 20] Output:  arr1[] = [5 10] arr2[] = [12 18 20] Explanation: After merging two sorted arrays  we get 5 10 12 18 20. Your Task: You don't need to read input or print anything. You only need to complete the function merge() that takes arr1, arr2, N and M as input parameters and modifies them in-place so that they look like the sorted merged array when concatenated. Expected Time Complexity:  O((n+m) log(n+m)) Expected Auxilliary Space: O(1