Skip to main content

Number of paths Recursion Basic Problem

Number of paths Recursion Basic Problem-

Problem Statement-

The problem is to count all the possible paths from top left to bottom right of a MxN matrix with the constraints that from each cell you can either move to right or down.

The first line of input contains an integer T, denoting the number of test cases. The first line of each test case is M and N, M is number of rows and N is number of columns.

For each test case, print the number of paths.

1 ≤ T ≤ 30
1 ≤ M,N ≤ 10

3 3
2 8


Testcase 1:
 Let the given input 3*3 matrix is filled as such:

The possible paths which exists to reach 'I' from 'A' following above conditions are as follows:

This is Strongly recommended that You should try This question to your own before jumping to direct solution

Try this question Yourself Practice here-gfg

Number of paths Rat in a maze Recursion Basic Problem
Rat in A mace

Now My Solution-

We can solve this question  recusively 

There can be 2 type of problems here one we have to print path(needed 2d array) and one we have to tell number of total path.

In this question I will focus only number of total path If you want print the Number of paths also 

i will write solution for that too comment below this post for that printing path .

now Lets jump to Solution 

My Approach -

let's think about base condition 

if we have only one element then we have only one path (return 1) and if we go out of m,n then we don't have any solution so we will return 0 in that case.


        return 0;



        return 1;


See recustion is  a topic where you have to think about sub problem what i am thinking now  if i am standing at a generic position lets say at index i,j  i have 2 choices form my poison either i can go to right j+1th poison where j  denoting column index or i can go down at i+1th row.

So here i got my induction step -

return solution(i+1,j,m,n)+solution(i,j+1,m,n);

according to these concept i can write my code . Here is my solution -


Popular posts from this blog

codeforces rating system | Codeforces rating Newbie to Legendary Grandmaster

 Codeforces rating system | Codeforces rating Newbie to Legendary Grandmaster- Codeforces is one of the most popular platforms for competitive programmers and  codeforces rating matters a lot  .  Competitive Programming  teaches you to find the easiest solution in the quickest possible way. CP enhances your problem-solving and debugging skills giving you real-time fun. It's brain-sport. As you start solving harder and harder problems in live-contests your analytical and rational thinking intensifies. To have a good codeforces profile makes a good impression on the interviewer. If you have a good  codeforces profile so it is very easy to get a referral for product base company like amazon, google , facebook etc.So in this blog I have explained everything about codeforces rating system. What are different titles on codeforces- based on rating codeforces divide rating into 10 part. Newbie Pupil Specialist Expert Candidate Codemaster Master International Master Grandmaster Internat

Apple Division CSES Problem Set Solution | CSES Problem Set Solution Apple division with code

 Apple Division CSES Problem Set Solution | CSES Problem Set Solution Apple division with code - Apple Division CSES Problem Solution Easy Explanation. Apple division is problem is taken form cses introductory problem set.Let's Read Problem statement first. Problem Statement- Time limit:  1.00 s   Memory limit:  512 MB There are  n n  apples with known weights. Your task is to divide the apples into two groups so that the difference between the weights of the groups is minimal. Input The first input line has an integer  n n : the number of apples. The next line has  n n  integers  p 1 , p 2 , … , p n p 1 , p 2 , … , p n : the weight of each apple. Output Print one integer: the minimum difference between the weights of the groups. Constraints 1 ≤ n ≤ 20 1 ≤ n ≤ 20 1 ≤ p i ≤ 10 9 1 ≤ p i ≤ 10 9 Example Input: 5 3 2 7 4 1 Output: 1 Explanation: Group 1 has weights 2, 3 and 4 (total weight 9), and group 2 has weights 1 and 7 (total weight 8). Join Telegram channel for code discussi

Concert Tickets Cses Problem set solution | Concert Tickets Cses Problem set solution Using multiset

 Concert Tickets Cses Problem set solution- Porblem statement- Time limit:  1.00 s   Memory limit:  512 MB There are  n n  concert tickets available, each with a certain price. Then,  m m  customers arrive, one after another. Each customer announces the maximum price he or she is willing to pay for a ticket, and after this, they will get a ticket with the nearest possible price such that it does not exceed the maximum price. Input The first input line contains integers  n n  and  m m : the number of tickets and the number of customers. The next line contains  n n  integers  h 1 , h 2 , … , h n h 1 , h 2 , … , h n : the price of each ticket. The last line contains  m m  integers  t 1 , t 2 , … , t m t 1 , t 2 , … , t m : the maximum price for each customer. Output Print, for each customer, the price that they will pay for their ticket. After this, the ticket cannot be purchased again. If a customer cannot get any ticket, print  − 1 − 1 . Constraints 1 ≤ n , m ≤ 2 ⋅ 10 5 1 ≤ n , m ≤ 2 ⋅