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First_Come_First_Serve CPU Scheduling

#include<stdio.h> #include<conio.h> struct proc {     int id;     int burst;     int at; }; int main() {     int n,i,j,total_waiting_time,waiting_time;     printf("Enter the number of processess: ");     scanf("%d",&n);     struct proc p[n],temp;     f or(i=0;i<n;i++)     {         p[i].id=i;         printf("Enter the arrival time of the process: ");         scanf("%d",&p[i].at);         printf("Enter the burst time of the process: ");         scanf("%d",&p[i].burst);     }     for(i=0;i<n;i++)     {         for(j=i+1;j<n;j++)         {             if(p[j].at<p[i].at)             {                 ...
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Reversing stack Method 2 !! (One Helper Stack only)

#include   <bits/stdc++.h> using   namespace   std ; stack < int >  rev ( stack < int >  s ) {      int   size  =  s . size (),  x ,  y ,  t ;      stack < int >  s2 ;      size  -=  1 ;      while  ( size --)     {          t  =  size  +  1 ;          x  =  s . top ();          s . pop ();          while  ( t --)         {              s2 . push ( s . top ());              s . pop ();         }  ...
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Populating Next Right Pointers in Each Node in O(1) space (without queue and level order)

Image
Example 1: Input: root = [1,2,3,4,5,6,7] Output: [1,#,2,3,#,4,5,6,7,#] Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level. Example 2: Input: root = [] Output: [] class Solution { public:     Node * connect ( Node * root )     {         Node * p1 = root , * p2 ;         while ( p1 != NULL )         {             p2 = p1 ;             while ( p2 != NULL )             {                 if ( p2 -> left )                     p2 -> left -> next = p2 -> right ;       ...
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Calculate factorial of large numbers !! (Using Arrays)

#include   <bits/stdc++.h> using   namespace   std ; #define   max   10000 int   main () {      int   n ,  c ,  j ,  k ,  x ,  i ;      int   a [ max ];      cout   <<   "Enter the number: " ;      cin   >>   n ;      j  =  0 ;  // j will be pointing to new index created to stored carry      a [ 0 ] =  1 ;      for  ( i  =  2 ;  i  <=  n ;  i ++)     {          c  =  0 ;  //carry when each number of array is multiplied by i          //(i is a...
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Multiplication of large numbers (Given in string format)

  class   Solution { public:      string   multiply ( string   num1 ,  string   num2 )     {          if  ( num1  ==  "0"  ||  num2  ==  "0" )              return   "0" ;          if  ( num1  ==  "1" )              return   num2 ;          if  ( num2  ==  "1" )              return   num1 ;          if  ( num1 . size () ==  1  &&  num2 . size () ==  1 )         {           ...
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Left View of Binary Tree (Method 1 using recursion)

#include   <iostream> #include   <vector> using   namespace   std ; struct   TreeNode {      int   val ;      TreeNode  * left ;      TreeNode  * right ;      TreeNode () :  val ( 0 ),  left ( nullptr ),  right ( nullptr ) {}      TreeNode ( int   x ) :  val ( x ),  left ( nullptr ),  right ( nullptr ) {}      TreeNode ( int   x ,  TreeNode   * left ,  TreeNode   * right ) :                              val ( x ),  left ( left ),  right ( right ) {} }; TreeNode   * BuildTree () {      int   d ;      cin   >>   d ;      if  ( d  == -...
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Check Bracket Sequence

#include<bits/stdc++.h> using namespace std; int main() {     int i,f,top=-1,size;     string s;     cin>>s;     size=s.size();     char a[size];     f=1;     for(i=0;i<s.size() && f==1;i++)     {         if(s[i]=='(')         a[++top]='(';         else if(s[i]==')')         {             if(top==-1)             f=0;             else             top-=1;         }     }     if(top!=-1 || f==0)     cout<<"invalid\n";     else     cout<<"Valid"; }
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Image Multiplication

https://practice.geeksforgeeks.org/problems/image-multiplication0627/1 Input: 1 / \ 3 2 / \ / \ 7 6 5 4 / \ \ / \ \ 11 10 15 9 8 12 Output: 332 Explanation: Sum = (1*1) + (3*2) + (7*4) + (6*5) + (11*12) + (15*9) = 332 class Solution {     public:     void treeHelper ( Node * a , Node * b , long long int & ans )     {         if (a== NULL ||b== NULL )         return ;         long long int temp=( a -> data * b -> data )%( 1000000007 );         ans=(ans+temp)%( 1000000007 );         treeHelper ( a -> left , b -> right , ans);         treeHelper ( a -> right , b -> left , ans);     }   ...
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Boundary Traversal of binary tree

struct Node {     int data;     Node * left, * right; }; class Solution { public:     void call_left ( vector < int > & v , Node * root )     {         if ( root == NULL )             return ;         if ( root ->left == NULL && root ->right == NULL )             return ; // don't push the leaf nodes.         v . push_back ( root ->data);         if ( root ->left)             call_left ( v , root ->left);         else             call_left ( v , root ->right);     }     void call_leaves ( vector < int > & v , Node * root )     {         if ( root == NULL )             return ;   ...
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BST to greater sum tree

  // Input: //            2 //          /    \ //         1      6 //               /  \ //              3    7 // Output: 18 16 13 7 0 (InOrder) // Explanation: // Every node is replaced with the sum of nodes greater than itself. // The resultant tree is: //                16 //              /    \ //            18       7 //                   /   \ //                  13    0 //Bhaskar Varshney #include <iostream> #include <queue> using namespace std; struct TreeNode {     int val;     TreeNode *left;     TreeNode *rig...
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