Monday, October 31, 2016
Number of common divisors of 2 numbers
Problem link: COMDIV - Number of common divisors
For this we need to know
CommonDivisor(a,b) = NumberOf Divisor(gcd(a,b)) proof
Find the number of divisors: Time Complexity O(sqrt(n))
tutorial for finding the number of divisors : progkriya
Solution:
For this we need to know
CommonDivisor(a,b) = NumberOf Divisor(gcd(a,b)) proof
Find the number of divisors: Time Complexity O(sqrt(n))
tutorial for finding the number of divisors : progkriya
Solution:
Sunday, October 30, 2016
Codeforces problem B Two Buttons analysis
Problem Type : BFS , math , greedy , shortest path
Problem link: Two Buttons
this was a very interesting for me.
This problem asks us what is the minimum steps we need to reach from one given number to another given number by multiplying by 2 or subtracting 1
i have found 2 ways to solve this problem
1) using bfs (hard way)
2) simple mathematics
The hard way:
let us try to analyze 2 given test cases
1) 4 ->6 minimum steps: 2
2) 10 ->1 minimum steps: 9
let us make a graph using the 2 test cases
Problem link: Two Buttons
this was a very interesting for me.
This problem asks us what is the minimum steps we need to reach from one given number to another given number by multiplying by 2 or subtracting 1
i have found 2 ways to solve this problem
1) using bfs (hard way)
2) simple mathematics
The hard way:
let us try to analyze 2 given test cases
1) 4 ->6 minimum steps: 2
2) 10 ->1 minimum steps: 9
let us make a graph using the 2 test cases
Wednesday, October 26, 2016
UVA 383 - Shipping Routes analysis
Problem Type : Single-Source Shortest Paths On Unweighted Graph (BFS/DFS)
Its a basic SSSP problem.
"The cost of a shipment is equal to the size of the shipment times the number of shipping legs required times $100."
we just have to do a bfs or dfs from given source if we have't done bfs or dfs on the source before.We can get the cost of the shipment using the above statement using the cost of the destination :D
Its a basic SSSP problem.
"The cost of a shipment is equal to the size of the shipment times the number of shipping legs required times $100."
we just have to do a bfs or dfs from given source if we have't done bfs or dfs on the source before.We can get the cost of the shipment using the above statement using the cost of the destination :D
Tuesday, October 25, 2016
UVA 260 - Il Gioco dell'X analysis
Problem Type : Finding Connected Components (BFS/DFS)
Let us try to understand the problem
"The game ‘Il Gioco dell’ X’ is played on a N by N board (N ≥ 2). The object of both players, say Black and White, is to join opposite sides of the board by placing in turn their pawns on the board in such a way that a path is made from one side to the other by adjacent (neighboring) pawns of their own color."
Though the statement is not very clear how Black and White will win but we can get the idea from the given picture
Let us try to understand the problem
"The game ‘Il Gioco dell’ X’ is played on a N by N board (N ≥ 2). The object of both players, say Black and White, is to join opposite sides of the board by placing in turn their pawns on the board in such a way that a path is made from one side to the other by adjacent (neighboring) pawns of their own color."
Though the statement is not very clear how Black and White will win but we can get the idea from the given picture
Saturday, October 22, 2016
Sifat Shishir's Blog: UVA 11254 : Consecutive Integers
Sifat Shishir's Blog: UVA 11254 : Consecutive Integers: Prob link : https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=22...
Thursday, October 20, 2016
Friday, October 14, 2016
Depth-first search Implementation
1 thing i always remember about dfs is "Using dfs We go to the depth of a node and then we come back"
I learned dfs from my friend and from this blog shafaetsplanet . In shafaetsplanet the algorithm is explained very easily.
Algorithm :
Basic DFS Traversal of a Graph:
Here starting node is 0.
I learned dfs from my friend and from this blog shafaetsplanet . In shafaetsplanet the algorithm is explained very easily.
Algorithm :
Basic DFS Traversal of a Graph:
Here starting node is 0.
Some good beginner problems
IMPLEMENTATION:
01.Beautiful Year (L1)
Solution
02.Queue at the School (L1)
Solution
03.Lights Out (L1)
Solution
04.Arrival of the General (L1)
Solution
05.Twins (L1)
Solution
06.Effective Approach (L2)
Solution
07.Even Odds (L2)
Solution
08. Airport (L2)
Solution
EASY GREEDY:
01.Dragons (L2)
Solution
Monday, October 10, 2016
Some Graph Problems
Here are some Graph problems i have solved over time
Flood Fill/Finding Connected Components using dfs/bfs:
BFS:
DFS:
Tropological sort:
Strongly connected component:
Graph coloring:
MST:
908 - Re-connecting Computer Sites
Solution
Solution
11631 - Dark roads
Solution
Solution
11747 - Heavy Cycle Edges
Solution
Solution
10034 - Freckles
Solution
Solution
11733 - Airports
Solution
Solution
1174 - IP-TV
Solution
Solution
11710 - Expensive subway
Solution
Solution
11857 - Driving Range
Solution
Solution
10048 - Audiophobia (mst+bfs)
Solution
544 - Heavy Cargo
Solution
10099 - Tourist Guide
Solution
10147 - Highways
Solution
10397 - Connect the Campus
Solution
10600 - ACM Contest and Blackout
Solution
11228 - Transportation system.
Solution
11733 - Airports
Solution
Solution
544 - Heavy Cargo
Solution
10099 - Tourist Guide
Solution
10147 - Highways
Solution
10397 - Connect the Campus
Solution
10600 - ACM Contest and Blackout
Solution
11228 - Transportation system.
Solution
11733 - Airports
Solution
11857 - Driving Range
Solution
Dijkstra:
10986 - Sending email
Solution
Solution
929 - Number Maze
Solution
Solution
bellman ford:
558 - Wormholes
Solution
Solution
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