LeetCode 931. Minimum Falling Path Sum

Link:- https://leetcode.com/problems/minimum-falling-path-sum/

Summary:

Given a square array of integers A, we want the minimum sum of a falling path through A.

A falling path starts at any element in the first row and chooses one element from each row.  The next row's choice must be in a column that is different from the previous row's column by at most one.

For Example:
Input:

[[1,2,3],[4,5,6],[7,8,9]]

Output:
12

Logic:

Given array A with row ->n and column ->n

now create matrix say ans[n][m]

copy first row of A into ans

i.e ans[0] = A[0]

Why?

since we can start from any column in first row

now for rest of the row and column say i and j (1<=i<n, 0<=j<m)

ans[i][j] = A[i][j] + min(ans[i][j-1], ans[i-1][j-1], ans[i-1][j+1]

return min(ans[n-1])

i.e minimum of all values in last row

Why?

because we can end at any column of last row.

To View My Solution:

https://github.com/shivam04/codes/blob/master/leetcode/minimum-falling-path-sum.cpp

LeetCode 1140. Stone Game II

Link:- https://leetcode.com/problems/stone-game-ii/

Summary:

Alex and Lee continue their games with piles of stones.  There are a number of piles arranged in a row, and each pile has a positive integer number of stones piles[i].  The objective of the game is to end with the most stones. 

Alex and Lee take turns, with Alex starting first.  Initially, M = 1.

On each player's turn, that player can take all the stones in the first X remaining piles, where 1 <= X <= 2M.  Then, we set M = max(M, X).

The game continues until all the stones have been taken.

Assuming Alex and Lee play optimally, return the maximum number of stones Alex can get.

For Example:
Input:

[2,7,9,4,4]

Output:
10

Logic:

You can easily solve this question recursion.

So Alex will try to get the maximum point as he is playing optimally

if player == 1 (or Alex)

 alexPoint = max(alexPoint, sum[s to s+x-1]  stoneGameIIUtil(piles, s+x, max(x,m), n ,1-p));

also, Lee is playing optimally and try to reduce Alex points

if player == 0 (or Lee)

 alexPoint = min(alexPoint, stoneGameIIUtil(piles, s+x, max(x,m), n ,1-p));

1<=x<2*M

Will this solution gives TLE?

Yes

as 1<=N<=100 (total number of piles)

We can't use recursion here.

How can we optimize it?

One solution that I think of is using memorization. We can create a matrix say dp

and store current state dp[s][m][p]

s -> starting index

m -> for picking stones from piles i.e (1<=x<=2*m)

p -> 0/1 (Lee/Alex)

Now if we reach this state again we don't need to compute again and return dp[s][m][p]

(Yahi battein to bad me yaad  ayengi 😛)

Can we further optimize it?

Yes.

For calulationg sum[s to s+x-1]

we can use prefixSum array 

so sum[s to s+x-1] will be prefixSum[s+x-1] - prefixSum[s-1]

To View My Solution:

https://github.com/shivam04/codes/blob/master/leetcode/stone-game-ii.cpp

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HACKEREARTH: Permutation Swaps

Link:- https://www.hackerearth.com/practice/algorithms/graphs/breadth-first-search/practice-problems/algorithm/permutation-swaps/

Summary:
Given P and Q be two permutation of numbers 1 to N. Given some pair of indexes (ai , bi).
Your task is to find Q from P by swapping (Pai , Pbi).
Print YES if it possible else print NO.

For Example:
Input:
2
4 1
1 3 2 4
1 4 2 3
3 4
4 1
1 3 2 4
1 4 2 3
2 4

Output:
NO
YES

Logic:
Use Breadth First Search(BFS) to find connected components of indexes.
Then check for indexes in each connected component if P and Q are same or not.

To View My Solution:
https://github.com/shivam04/codes/blob/master/hackerearth/algo/graph/medium/permutation-swaps.cpp

HACKEREARTH: Connected Horses

Link:- https://www.hackerearth.com/practice/algorithms/graphs/breadth-first-search/practice-problems/algorithm/connected-horses-10/description/

Summary:
Given N ,M number of rows and columns in a chess board
Given Q queries each query contain co-ordinates of horses in a chess board.
Find the number of ways horses swap their position

For Example:
Sample Input:
2 
4 4 4
1 1
1 2
3 1
3 2
4 4 4
1 1
1 2
3 1
4 4

Sample Output:
4
2

Topic:
Breadth First Search(BFS) , Dynamic Programming(DP)

Logic:
Use Breadth First Search(BFS) to find the total number of horses in each connected groups, then find the factorial of this number. Apply same in each group and multiplication of all gives the result. Use DP for calculation of factorials.

To View my solution:
https://github.com/shivam04/codes/blob/master/hackerearth/algo/graph/medium/connected-horses.cpp

HACKERRANK: Power Of Large Numbers

Link:- https://www.hackerrank.com/challenges/power-of-large-numbers

Summary:
Given A and B.
Find (A^B)%(10^9+7)

Constraints:
1 <= T <= 10 
1 <= A,B <= 10100000 
A % (109 + 7) != 0

For Example:
Sample Input:
4
3 2
4 5
7 4
34534985349875439875439875349875  93475349759384754395743975349573495

Sample Output:
9
1024
2401
735851262

Logic:
(A^B)%MOD = (A%MOD)^B
Since A is very very large number can't be stored in int or long so using modulo technique bring A in the range of int or long as:
Let 'a' be an array contains digit of A so to construct A%MOD. Let A = 0
for i 0 to len(a):
    A = (A*10 + a[i])%MOD
This gives A%MOD

Now B is also very very large but we can't apply same technique as we done in A as
(A%MOD)^B%MOD != (A%MOD)^(B%MOD)%MOD
So we use Fermet's Little Theorem which says:

where p is prime number.
divide by a both side


Since MOD = 10^9+7 is a prime number

we need to find A^B%MOD
so if  B is divisible by (p-1) or (MOD-1) then B = (MOD-1)*x
so (A^((MOD-1)*x))%MOD = (A^(MOD-1)%MOD)^x%MOD = 1^x%MOD = 1

and if B is not divisible by (MOD-1) then B = (MOD-1)*x + y
so
(A^((MOD-1)*x+y))%MOD = ((A^(MOD-1)%MOD)^x)%MOD*(a^y)%MOD =  1^x*(a^y)%MOD
and y = B%(MOD-1)
so y can be computed as:
y = 0
b is an array contains digit of number B.
for i 0 to len(b)
    y = (y*10 + b[i])%(MOD-1)
if y==0
answer = 1
else
answer = A^y%MOD

To View My Solution:
https://github.com/shivam04/codes/blob/master/hackerrank/math/number%20theory/medium/power-of-large-number.py

CODEFORCES: Sereja and Suffixes(368B)

Problem Link: http://codeforces.com/problemset/problem/368/B

Summary:
Given 'n' number of elements in array.

Given 'm' i.e 'm' different queries each query contains an integer 'l'.

Find the number of distinct elements from 'l' to 'n'  in array.

For Example:

input

10 10
1 2 3 4 1 2 3 4 100000 99999
1
2
3
4
5
6
7
8
9
10

output

6
6
6
6
6
5
4
3
2
1

Topic:
Dynamic Programming
Logic:

Let dp[i] be the answer i.e number of distinct elements from i to n.

and dp[i] = dp[i+1] +1 when element at a[i] has not been visited yet or visit[a[i]] = 0 otherwise dp[i] = dp[i+1] i.e visit[a[i]] = 1

For any 'l' print dp[l]  is the answer.

To View my solution:

https://github.com/shivam04/codes/blob/master/codeforces/seerja_and_suffixes.py