Comma Operator inside for loop Condition in C++

You may have used Comma operator in initilialization and increment part of for loop like below

                 for(init;condition;increment) --> for(int i=0,j=5;i<j;i++,j--)

But can we used Comma operator inside for loop condition ?

The answer is Yes but it behave bit differently.

Predict the output of below program

void commaOperator()
{
    int i=0,j=0;
    cout<<"Using && inside for loop Condition"<<endl;
    for(;i<4&&j<6;)
    {
            cout<<i<<" "<<j<<endl;
            i++;
            j++;
    }
    cout<<"Using comma inside for loop Condition "<<endl;
    i=0,j=0;
    for(;i<4,j<6;)
    {
            cout<<i<<" "<<j<<endl;
            i++;
            j++;
    }

}

Output:

Using && inside for loop Condition
0 0
1 1
2 2
3 3
Using comma inside for loop Condition
0 0
1 1
2 2
3 3
4 4
5 5

As Comma operator is a binary operator it evaluates first operand but discards its result and then evaluates the second operand and returns its result.
In above example , for loop will run till second condition is true (j<6 ) and discards the first condition.

But what if we use comma operator in assignment like below
int a=5,b=7,c;
c=a,b;

Here c will be equal to a (5) as assignment has higher precedence than comma operator.


Please comment if you find anything incorrect.

Find Consecutive Elements in Unsorted Array in O(n) - C++ Code

#include<iostream>
#include<set>

int findConsecutiveElements(int arr[], int n)
{
        std::set<int> s;
        int ans = 0;

        for (int i = 0; i < n; i++)
                s.insert(arr[i]);

        for (int i=0; i<n; i++)
        {
                if (s.find(arr[i]-1) == s.end())
                {
                        int j = arr[i];
                        while (s.find(j) != s.end())
                                j++;

                        ans = std::max(ans, j - arr[i]);
                }
        }
        return ans;
}

int main()
{
        int arr[] = {1,21, 24, 25, 9, 3, 11, 4, 20, 22,23};
        int n = sizeof arr/ sizeof arr[0];
        std::cout << "Length of the Longest consecutive elements is "
                << findConsecutiveElements(arr, n);
        return 0;
}

Kth smallest element in a BST - C++ Code

Do inorder Traversal of BST and count number of elements traversed till now . As inorder traversal gives us the element in sorted Order, when count reaches k , we break out and Print the element.

/**
 * Definition for Binary Search Tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 * };
 */
int kthsmallest(TreeNode* root, int k) {
    if(root==NULL)
      return 0;
    stack<TreeNode*> s;
    bool complete=false;
    TreeNode* cur=root;
    int count=0;
    int val=0;
    while(!complete)
    {
        if(cur!=NULL)
        {
            s.push(cur);
            cur=cur->left;
        }
        else
        {
            if(!s.empty())
            {
                count++;
                cur=s.top();
                s.pop();
                if(count==k)
                {
                    val=cur->val;       
      break;
                }
                  cur=cur->right;
            }
            else
              complete=true;
        }  
    }
    return val;
}

Please comment if you find anything incorrect.

First Non Repeating Character in String

#include<iostream>
#include<string>
#include<memory>
#include<climits>
#define NUMBER_OF_CHARS 256
using namespace std;
struct CountIndex
{

 int count;
 int index;
};
char getFirstNonRepeatngCharacter(const string& str)
{
 char ch=' ';
 unique_ptr<CountIndex[]> ci(new CountIndex[256]());
 //Can also use shared_ptr
 //shared_ptr<CountIndex> ci(new CountIndex[256](),std::default_delete<CountIndex[]>());
 for(auto& it:str)
 {
  ci.get()[it].count++;
  if(ci.get()[it].count==1)
   ci.get()[it].index= &it - &str[0];
  else

   ci.get()[it].index= ci.get()[it].index;

 }
 int maxIndex=INT_MAX;

 for(int it=0;it<NUMBER_OF_CHARS;it++)
 {
  if(ci.get()[it].count==1 && ci.get()[it].index <maxIndex)
  {
   maxIndex= ci.get()[it].index;

  }
 }
 return maxIndex!=INT_MAX?str[maxIndex]:' ';
}
int main()
{    
 string str ;
 cin>>str;
 char ch=getFirstNonRepeatngCharacter(str);
 if(ch !=' ')
  cout<<" First Non repeating Character is "<< ch;
 else
  cout<<" No Repeating Character Exist ";


}

Folder Organizer

import os,errno
import sys,getopt
def main(argv):
    inputDir=''
    try:
        opts,args= getopt.getopt(argv,"hi:",["iDir="])
    except getopt.GetoptError:
        print 'DirectoryOrganizer.py -i <inputDirectory>'
        sys.exit(2)
    for opt,arg in opts:
        if opt == '-h':
            print 'DirectoryOrganizer.py -i <inputDirectory>'
            sys.exit()
        elif opt in ("-i","--iDir"):
            inputDir=arg
    files= os.listdir(inputDir)
    for file in files:
        extension= os.path.splitext(file)[1][1:]
        if extension:
            fileInputDir =inputDir+extension
            try:
                if not os.path.isdir(fileInputDir):
                    os.makedirs(fileInputDir)
                if not os.path.isfile(fileInputDir+"\\"+file):
                    os.rename(inputDir+file,fileInputDir+"\\"+file)
                else:
                    print "file with same name already exist in folder "+ fileInputDir+"\\"+file
            except OSError as e:
                if e.errno != errno.EEXIST:
                    raise

if __name__ == "__main__":
main(sys.argv[1:])

Allocating Void pointer memory using new operator

I was Implementing Generic List in C++ and was able to successfully allocate memory of void*  data using malloc ( malloc return void *) . But how about using new. Can we do below

void Test()
{
           void *f= new void[25];  //void *f=malloc(25); --> Works
           delete f;
}

The Answer is NO.
Because void has no size . How much space has to be allocated is not known.

So if you need to allocate memory using new use operator like

void Test()
{
          void *f=operator new(25);
          delete f;
}

Floyd Washall Algorithm Code in Java

We are given a directed graph and if we need to find out minimum distance between all the sources and destinations, we will use Floyd Warshall All Pair Shortest Path Algorithm.
Suppose we are given below graph

import java.util.*;
import java.lang.*;

public class Floyd
{
 final static int INF=Integer.MAX_VALUE;
 void Path(int graph[][],int n)
 {
   int minDistance[][]=new int[n][n];
   for(int i=0;i<n;i++)
   {
    for(int j=0;j<n;j++)
    {
     minDistance[i][j]=graph[i][j];
     
      
    }
   }
   for(int k=0;k<n;k++)
   {
    minDistance[k][k]=0;
   }
   
   for(int k=0;k<n;k++)
   {
    for(int i=0;i<n;i++)
    {
     for(int j=0;j<n;j++)
     {
          if(minDistance[i][j]>minDistance[i][k]+minDistance[k][j] && minDistance[i][k] !=INF && minDistance[k][j]!=INF)   
          minDistance[i][j]=minDistance[i][k]+minDistance[k][j];
     }
    }
   }
  
  printMatrix(minDistance,n);
  
 }
  void printMatrix(int minDistance[][],int n)
  {
   for(int i=0;i<n;i++)
   {
    for(int j=0;j<n;j++)
    {
         System.out.print(minDistance[i][j]+ "  ");
    }
    System.out.println();
   }
   
  }
 public static void main(String[] args)
 {
  int a[][]={{0,5,INF,10},
             {INF,0,3,INF},
             {INF,INF,0,1},
             {INF,INF,INF,0}};
  Floyd f=new Floyd();
  f.Path(a,4);
 }
}