LeetCode 1003. Check If Word Is Valid After Substitutions

  Link:- https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/

Summary:

Given a string s, determine if it is valid.

A string s is valid if, starting with an empty string t = "", you can transform t into s after performing the following operation any number of times:

Insert string "abc" into any position in t. More formally, t becomes tleft + "abc" + tright, where t == tleft + tright. Note that tleft and tright may be empty.

Return true if s is a valid string, otherwise, return false. 

For Example:

Input:

s = "aabcbc"

Output:

true

Logic:

This can be solved with the stack. Traverse string and if s[i]!= 'c' then insert into stack.

Else, check if stack element at the top and next to the top is 'b' and 'a' respectively or not.

if no: return false, else remove the top two elements of the stack.

Repeat the above steps.

At last, check if the stack is empty or not.

Time Complexity:

O(n)

To view my solution:

https://github.com/shivam04/codes/blob/master/leetcode/check-if-word-is-valid-after-substitutions.cpp

LeetCode 1530. Number of Good Leaf Nodes Pairs

 Link:- https://leetcode.com/problems/number-of-good-leaf-nodes-pairs/

Summary:

Given a root of the binary tree and an integer distance. Find the number of pairs of leaves such that the shortest path between them is less than or equal to distance.  

For Example:

Input:

root = [1,2,3,null,4], distance = 3

Output:

1

Logic:

Convert the binary tree into a graph using DFS and find all the leaves in that DFS traversal.

Run other DFS from each leaf and check if the shortest distance between current leave and other leaf is less than or equal to distance or not. 

If yes:

ans++

Finally, return ans/2.

Why ans/2?

As we run DFS from each leaf, we end up computing the same pair of leaves twice.

Time Complexity:

O(n^2)

To view my solution:

https://github.com/shivam04/codes/blob/master/leetcode/number-of-good-leaf-nodes-pairs.cpp

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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