Given two arrays of numbers, find if each of the two arrays have the same set of ntegers ? Suggest an algo which can run faster than NlogN without extra space?

Design a data structure that offers the following operations in O(1) time: e.g. in contant time * insert * remove * contains * get random element (one of Only those elemnt currently present in data structure

we have to think of all possible DS that can do some of the operation in desired complexity then we can come up with some combinations of Data structure so that we can get what question is asking for :)

So its clear O(1) lookup implies a hashed data structure.

By comparison:

* O(1) insert/delete with O(N) lookup implies a linked list.
* O(1) insert, O(N) delete, and O(N) lookup implies an array-backed list
* O(logN) insert/delete/lookup implies a tree(BST) or heap.

we basically need to find out some kind of combinations from above DS .

First I thought About Doubly Linked List instead of Array but soon realized that we won't get desired complexity isn't it ? even we can achieve some of operation in O(1) but how u will make sure all will be done in O(1) ??

so lets Consider a data structure composed of a hashtable H and an array A. The hashtable keys are the elements in the data structure, and the values are their positions in the array.

here can be possible way to achieve the desired time complexity

1. insert(value): append the value to array and let i be it's index in A. Set H[value]=i.

2. remove(value): We are going to replace the cell that contains value in A with the last element in A. let d be the last element in the array A at index m. let i be H[value], the index in the array of the value to be removed. Set A[i]=d, H[d]=i, decrease the size of the array by one, and remove value from H.

3. contains(value): return H.contains(value) no doubt in this :)

4. getRandomElement(): let r=random(current size of A). return A[r].
e.g. index=random()%size of current array will generate index in range & then we can return element at given index . this will make sure that this function will generate the random number from  present elements of data structure only.

Time Complexity Will be O(1)
Space Complexity O(N) , N is size of hashtable

Design a data structure that offers the following operations in O(1) time: * insert * remove * contains * get random element (one of Only those elemnt currently present in data structures)

design the algorithm which computs the placements of 21 triomino that cover 8*8 chess board

Here is Another One :)

A triomino is formed by joining three unit-sized squares in m L-shape.
Amutiled chessboard(henceforth 8 x 8M board) is made up of 64 unitsized
squares arranged in an 8 × 8 square minus the topleft square. design the algorithm which computs the placements of 21 triomino that cover 8*8 chess board.

Design a program that takes an Image and a collection of m x m-sized tiles and produce a mosaic from the tiles that resembles the image.

Friend of Mine Who Appeared For Google sometimes back Told Me about this intersteting Problem:) I did some lazyness to post the question here . Are you guys are able to think about approach ? I Not Going to do no More Spoon Feeding Here :) so Try It Out

Hint :) There is Interesteing Math Behind It. Are You able to Think out the cloest point k-dimensional ? if you are able to come up with approach & complexity analysis thats enough to Crack Google

Joseph Perumtation Problem

“n” people are seated around in a circle. At each step, the “m”th person is eliminated from the circle. The next iteration continues from the person next to the eliminated person. In the end, there will be only one person remaining. Given “n” and “m”, write a program to find out the person who will be remaining at the end

Given a list of Tagalog words as a String[], return the same list in Tagalog alphabetical order

A Reader send me this problem . Thanks to Him for this intersteing Problem. so i am going to posting here.

In the first half of the 20th century, around the time that Tagalog became the national language of the Philippines, a standardized alphabet was introduced to be used in Tagalog school books (since then, the national language has changed to a hybrid "Pilipino" language, as Tagalog is only one of several major languages spoken in the Philippines).

Tagalog's 20-letter alphabet is as follows:

a b k d e g h i l m n ng o p r s t u w y

Note that not all letters used in English are present, 'k' is the third letter, and 'ng' is a single letter that comes between 'n' and 'o'.

You are compiling a Tagalog dictionary, and for people to be able to find words in it, the words need to be sorted in alphabetical order with reference to the Tagalog alphabet. Given a list of Tagalog words as a String[], return the same list in Tagalog alphabetical order

1)


{"ang","ano","anim","alak","alam","alab"}

Returns: {"alab", "alak", "alam", "anim", "ano", "ang" }

A few "A" words that are alphabetically close together.
2)


{"siya","niya","kaniya","ikaw","ito","iyon"}

Returns: {"kaniya", "ikaw", "ito", "iyon", "niya", "siya" }

Common Tagalog pronouns.
3)


{"kaba","baka","naba","ngipin","nipin"}

Returns: {"baka", "kaba", "naba", "nipin", "ngipin" }

Find three numbers in an array which forms a maximum product (for signed integers)

Naive Solution Requires sorting the input data & then finding product of 3 largest from the last isn't it can't we do it linear time :0

Data Structure Used: Array

Algorithm:
The solution involves in finding three maximum and two minimum numbers. If the minimum numbers are negatives and if their product is greater than the two maximum number's product, then they have to considered for maximum product.

so let l1,l2,l3 are the three maximum number & s1,s2 are the minimum numbers
Our Solution will be maxproduct=max( l1*s1*s2,l1*l2*l2)

#include <iostream>
using namespace std;
 
#define MAX 10
 
int* MaxProduct(const int input[], const int size)
{
 int* output = new int[3];
 int negative = 0;
 for(int i = 0; i < 3; i++)
 {
  output[i] = -999999;
 }
 int min[2] = {0,0};
 
 for(int i=0;i<size;i++)
 {
  // find two smallest negative numbers
  if(input[i] <= 0)
  {
   if(input[i] < min[0])
   {
    min[1] = min[0];
    min[0] = input[i];
   }
   else if(input[i] < min[1])
   {
    min[1] = input[i];
   }
   negative++;
  }
 
  // find three largest positive numbers
  if(input[i] > output[0])
  {
   output[2] = output[1];
   output[1] = output[0];
   output[0] = input[i];
  }
  else if(input[i] > output[1])
  {
   output[2] = output[1];
   output[1] = input[i];
  }
  else if(input[i] > output[2])
  {
   output[2] = input[i];
  }
 }
 
 if(size != negative)
 {
  if((min[0] * min[1]) > (output[0] * output[1]) ||  

(min[0] * min[1]) > (output[1] * output[2]))
  {
   output[1] = min[0];
   output[2] = min[1];
  }
 }
 
 return output;
}
 
int main()
{ 

const int input[MAX]={-6,-1,-2,-33,-4,-15,-7,-28,-9,-10};
 int* result = 0;
 result = MaxProduct(input,MAX);
 
 for(int i = 0; i < 3; i++)
 {
  cout << (i+1) << "# element : " << result[i] << endl;
 }
 

const int input1[MAX] = {0,-1,-2,-33,4,15,-7,-28,-9,-10};
 int* result1 = 0;
 result1 = MaxProduct(input1,MAX);
 
 for(int i = 0; i < 3; i++)
 {
  cout << (i+1) << "# element : " << result1[i] << endl;
 }
 
 const int input2[MAX] = {0,-1,-2,-33,-4,-15,-7,-28,-9 

,-10};
 int* result2 = 0;
 result2 = MaxProduct(input2,MAX);
 
 for(int i = 0; i < 3; i++)
 {
  cout << (i+1) << "# element : " << result2[i] << endl;
 }
 
 const int input3[MAX] = {6,1,2,33,4,15,7,28,9,10};
 int* result3 = 0;
 result3 = MaxProduct(input3,MAX);
 
 for(int i = 0; i < 3; i++)
 {
  cout << (i+1) << "# element : " << result3[i] << endl;
 }
 
 return 0;
}

Time Complexity O(N) As we Done only Single Pass Over Array
Space Complexity O(1)
Run Here https://ideone.com/4zhiE

Painter Partition Problem

You have to paint N boards of length {A0, A1, A2 … AN-1}. There are K painters available and you are also given how much time a painter takes to paint 1 unit of board. You have to get this job done as soon as possible under the constraints that any painter will only paint continuous sections of board, say board {2, 3, 4} or only board {1} or nothing but not board {2, 4, 5}.

Find The Number of Unique Path in Maze Where Robot is Sitting at Top-Left Corner & Can Move According to Given Constraints

A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid . marked ‘end' in the diagram below.How many possible unique paths are there?

I posted the solution with post because i was aware of such problem earlier :)

Basic Solution,Approach,Algoriothms Using BackTracing

Start form top left corner (say we are at position 1,1 in starting deonted by row=r,column=c (e.g. r=1, c=1 initially) & then will will keep moving untill reach when r=m and c=n e.g. bottom-left corner) writing code for this is simple.

int backtrack(int r, int c, int m, int n) {
if (r == m && c == n)
return 1;
if (r > m || c > n)
return 0;

return backtrack(r+1, c, m, n) + backtrack(r, c+1, m, n);
}

but as normal backtracking problems it inovolves repeated calculations
so is very inefficient in the sense that it recalculates the same solutio n over patah again & again. so we need to use a dynamic programming (DP) technique called memoization akso called top down approach.

const int M_MAX = 10;
const int N_MAX = 10;

int backtrack(int r, int c, int m, int n, int mat[][N_MAX+2]) {
if (r == m && c == n)
return 1;
if (r > m || c > n)
return 0;

if (mat[r+1][c] == -1)
mat[r+1][c] = backtrack(r+1, c, m, n, mat);
if (mat[r][c+1] == -1)
mat[r][c+1] = backtrack(r, c+1, m, n, mat);

return mat[r+1][c] + mat[r][c+1];
}

int bt(int m, int n) {
int mat[M_MAX][N_MAX];
memset(mat, -1, sizeof(mat));
return backtrack(1, 1, m, n, mat);
}

Time Complexity O(M+N)
Space Complexity O(M*N)



Bottom-Up Dynamic Programming More Efficient

As we know The total unique paths at above matrix (r,c) is equal to the sum of total unique paths from matrix to the right (r,c+1) and the matrix below (r+1,c).

(For clarity, we will solve this part assuming an X*Y Matrix)
Each path has (X-1)+(Y-1) steps. Imagine the following paths:

X X Y Y X (move right -> right -> down -> down -> right)
X Y X Y X (move right -> down -> right -> down -> right)
...& so on

Each path can be fully represented by the moves at which we move
right. That is, if I were to ask you which path you took, you could
simply say “I moved right on step 3 and 4.”
Since you must always move right X-1 times, and you have X-1 + Y-1
total steps, you have to pick X-1 times to move right out of X-1+Y-1
choices. Thus, there are C(X-1, X-1+Y-1) paths (e.g., X-1+Y-1 choose
X-1):

(X-1 + Y-1)! / ((X-1)! * (Y-1)!)..

const int M_MAX = 100;
const int N_MAX = 100;

int dp(int m, int n) {
int mat[M_MAX][N_MAX] = {0};
mat[m][n+1] = 1;

for (int r = m; r >= 1; r--)
for (int c = n; c >= 1; c--)
mat[r][c] = mat[r+1][c] + mat[r][c+1];

return mat[1][1];
}

Lets Assume you have M×N sample matrix or grid. so it doen't matter how you traverse the grids, you always traverse a total of M steps. To be more exact, you always have to choose M-N steps to the right (R) and N steps to the bottom (B). Therefore, the problem can be transformed to a question of how many ways can you choose M-N R‘s and N B‘s in these M+N-2 steps. The answer is C(M,N) (or C(M,M-N)). Therefore, the general solution for a m x n grid is C(m+n-2, m-1) & this is our answer.

Time Complexity O(M*N)
Space Complexity O(M*N)


Follow Up: Using Above We Have Only Calculated the number of paths can we print those path as well if yes what will be time complexity.