C program to left rotate an array

Write a program to left rotate an array by n position

Example

Input

Input 10 elements in array: 1 2 3 4 5 6 7 8 9 10
Input number of times to rotate: 3

Output

Array after left rotation: 4 5 6 7 8 9 10 1 2 3
Solution

    

      

        

  

    
    


        


  


  

        

          

          
#include <stdio.h>
        
        

          

          
#include <stdlib.h>
        
        

          

          
#define size 5
        
        

          

          


        
        

          

          
void rotate_array_left(int *array,int times)
        
        

          

          
{
        
        

          

          
    int temp=0;
        
        

          

          
    for(int x=0;x<times;x++)
        
        

          

          
    {
        
        

          

          
        temp=array[0];
        
        

          

          
        for(int i=0;i<=size-2;i++)
        
        

          

          
        {
        
        

          

          
            array[i]=array[i+1];
        
        

          

          
        }
        
        

          

          
        array[size-1]=temp;
        
        

          

          
    }   
        
        

          

          
}
        
        

          

          


        
        

          

          
int main(int argc, char** argv) {
        
        

          

          
    int data[size]={1,2,3,4,5};
        
        

          

          
    rotate_array_left(data,2);
        
        

          

          
    for(int j=0;j<=size-1;j++)
        
        

          

          
    {
        
        

          

          
        printf("%d\t",data[j]);
        
        

          

          
    }
        
        

          

          
    return (EXIT_SUCCESS);
        
        

          

          
}
        
  



    

  


      
      

        view raw
        
          rotatearray.c
        
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Count Set bits in an array of integers

assume you have an array of 5 integers and you want to count the numbers of set bits in all array elements , write a function to do that

Solution

	#include <stdio.h>
	#include <stdlib.h>
	#define size 5
	int count_set_bits(int *array)
	{
	int count=0;
	for(int i=0;i<size;i++)
	{
	for(int j=0;j<32;j++)
	{
	if(array[i]&(1<<j))
	{
	count++;
	}
	}
	}
	return count;
	}
	int main(int argc, char** argv) {
	int data[size]={1,2,3,4,5};
	printf("Number of set bits in this array = %d bits \n",count_set_bits(data));
	return (EXIT_SUCCESS);
	}

view raw setbits.c hosted with ❤ by GitHub

Circular Bit Shift

Write two functions RL(data,n) which performs circular left shift on variable data by n bits , RR(data,n) which performs circular right shift on variable data by n bits
note: data is 1 byte variable

refer to this video to know more about circular bit shifts

Solution

	#include <stdio.h>
	#include <stdlib.h>
	char RL(unsigned char byte,unsigned int n)
	{
	unsigned char temp=0;
	for(int x=0;x<n;x++)
	{
	temp=byte&(1<<7);//save last bit
	byte=byte<<1;//shift n left by 1 bit
	byte|=temp;//save last bit in first position from right
	}
	return byte;
	}
	char RR(unsigned char byte,unsigned int n)
	{
	unsigned char temp=0;
	for(int x=0;x<n;x++)
	{
	temp=byte&(1<<0);//save first bit
	byte=byte>>1;//shift n left by 1 bit
	byte|=(temp<<7);//save first bit in last position from right
	}
	return byte;
	}
	int main(int argc, char** argv) {
	printf("%d\n%d\n",RL(16,2),RR(16,2));
	return (EXIT_SUCCESS);
	}

view raw circularshift.c hosted with ❤ by GitHub

Beautiful Matrix

Problem 
codeforces.com/problemset/problem/263/A

Solution

	#include <iostream>
	#include <cstdlib>
	#include <math.h>
	using namespace std;
	int main(int argc, char** argv) {
	int input;
	// read matrix
	for(int r=0;r<=4;r++)
	{
	for(int c=0;c<=4;c++)
	{
	cin>>input;
	if(input==1)
	{
	cout<<abs(r-2)+abs(c-2)<<endl;
	break;
	}
	}
	if(input==1){break;}
	}
	return 0;
	}

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Teams Problem

Problem 
http://codeforces.com/contest/231/problem/A

Solution

	#include <iostream>
	#include <cstdlib>
	using namespace std;
	int main(int argc, char** argv) {
	unsigned int n;// number of problems
	unsigned int o1,o2,o3;// team memebrs
	unsigned int pass=0;
	unsigned int fail=0;
	cin>>n;
	for(int i=0;i<n;i++)
	{
	cin>>o1>>o2>>o3;
	if(o1+o2+o3>=2){pass++;}
	else if(o1+o2+3<2){fail++;}
	}
	cout<<pass;
	return 0;
	}

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What is dangling pointer

The following code snippet illustrate the idea of dangling pointers

	#include <iostream>
	#include <cstdlib>
	using namespace std;
	int main(int argc, char** argv) {
	int *p=(int*)malloc(sizeof(int));
	*p=1;
	/*
	* now p is an integer pointer pointes to space(4 bytes)allocated
	* by malloc function , also the allocated space is initilized by 1
	*/
	cout<<*p<<endl;//result will be 1
	//now we will free or deallocate reserved 4 bytes
	free(p);
	/*
	* now p is a dangling pointer , pointes to deallocated space
	* so data at this pointer is garbage
	*/
	cout<<*p;// output will be garbage
	return 0;
	}

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