/* Another Methods of Traversing in a Linear Array  */

#include<stdio.h>         /* Header File that are called standard input output*/
#include<conio.h>        /* Header File that are called consol input output*/

void main(){       /*start the main function and void that are not return type*/

clrscr();            /* Clear the screen at every time when input new element*/

int LA[] = {20, 65, 73, 88, 33, 95, 25};   /*array that will be stored 5 element*/ 

int k,count =0, LB=0,UB=7;    /*only declare that are integer variable*/
k=LB;
printf("\nThe array contain.....\n");/*Only for print that mean get the output  */

for(k=LB;k<UB;k++){                       /*that are loop k=0,UB=5*/
printf("%d\t", LA[k]);
k++;                           /*increase the loop element */
count++;                  /*to be increase for counting in the array*/
} 
printf("\n\n The number of elements=%d", count);  /* Count the element of array */
getch();   /*keep to stay the screen*/   
}


            index=0,1,2,3,4 there are 5 index element since, there have 5 element that are 5,6,7,8,3
                        0 1 2 3 4

Note:    0=5,1=6,2=7,3=8,4=3

So, when 0<5 that are condition is true and print 0 index that are mean 5 print stil condition 4<5 will be print that are 3 element and next condition 5<5 that are false condition and exit loop.
Here, when print o index and increase the value of loop the 1 and continue 4 and will be exit loop when value is 5.

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Traversing in a Linear Array

/* Traversing in a Linear Array  */

#include<stdio.h>         /* Header File that are called standard input output*/
#include<conio.h>        /* Header File that are called consol input output*/

void main(){       /*start the main function and void that are not return type*/

clrscr();            /* Clear the screen at every time when input new element*/

int LA[] = {5,6,7,8,3};   /*array that will be stored 5 element*/ 

int k,count =0, LB=0,UB=5;    /*only declare that are integer variable*/
k=LB;
printf("\nThe array contain.....\n");/*Only for print that mean get the output  */

while(k<UB){                       /*that are loop k=0,UB=5*/
printf("%d\t", LA[k]);
k++;                           /*increase the loop element */
count++;                  /*to be increase for counting in the array*/
} 
printf("\n\n The number of elements=%d", count);  /* Count the element of array */
getch();   /*keep to stay the screen*/   
}


            index=0,1,2,3,4 there are 5 index element since, there have 5 element that are 5,6,7,8,3
                        0 1 2 3 4

Note:    0=5,1=6,2=7,3=8,4=3

So, when 0<5 that are condition is true and print 0 index that are mean 5 print stil condition 4<5 will be print that are 3 element and next condition 5<5 that are false condition and exit loop.
Here, when print o index and increase the value of loop the 1 and continue 4 and will be exit loop when value is 5.

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Procedure P2.9A : CROSSOUT(A, N, K)

1.  If A[K] = 1, then : Return.

2.  Repeat for L = 2K to N by K:

Set A[L] : 1.

[End of loop]

3. Return.

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Complexity;Space-Time Tradeoffs

Briefly describe the notations of (a) the complexity of an algorithm and (b) the space-time tradeoff of algorithms  :

(a)  The complexity of an algorithm is a function f(n) which measures the time and/or space used by an algorithm in terms of the input size n.

(b) The space-time tradeoff refers to a choice between algorithmic solutions of a data processing problem that allows one to decrease the running time of an algorithmic solution by increasing the space to store the data and vice versa.

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Program to convert an Expression from Infix to Postfix

#include<stdio.h>
#include<stdlib.h>

#define MAXCOLS 80
#define TRUE 1
#define FALSE 0

void postfix(char*, char *);
int isoperand(char);
void popandtest(struct stack*, char*, int*);
int prcd(char, char);
void push(struct stack*, char);
char pop(struct stack*);


void main(){
char infix[MAXCOLS];
char postr[MAXCOLS];
int pos = 0;

while ((infix[pos++] = getchar()) != '\n' );
infix[--pos]='\0';
printf("%s%s", "the original infix expression is", infix);
postfix(infix, postr);
printf("%s\n", postr);
}   /* end of main */

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Using One-Dimensional Arrays Of C Programming Language

#define NUMELTS 100
void main(){
  
     int num[NUMELTS ];                        /*   array of numbers     */
    int i;
    int total;                                            /*   sum of the number     */
    float avg;                                         /* average of the number  */
    float diff;                                        /*  number and the average */
   
total = 0;
for(i=0; i<NUMELTS; i++){
/*   read the numbers into the array and add them    */ 
scanf("%d", num[i]);
total +=num[i];   
} /* end for  */ 

avg  = (float)total/NUMELTS;                /* computer the average */
printf("\nnumber difference");      /*    print heading   */
/   * print each number and its difference      */

for(i = 0; i<NUMELTS; i++){
 diff = num[i]-avg;
printf("\n%d%f", num[i], diff);

}   /*  end for  */

printf("\naverage is : %f", avg );
}   /* end main */

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Evaluation of Postfix Expression

Evaluation of Postfix Expression : 

Algorithm : 

1. Add a right parenthesis ")" at the end of P.   [This acts as a sentinel]

2. Scan P from left to right and repeat Steps 3 and 4 for each element of P until the sentinel ")" is encountered.

3. If an operand is encountered, put it on STACK.

4. If an operator  is encountered, then :

(a) Remove the two top elements of STACK, where A is the top element and B is the next-to-top element..

(b) Evaluate B A

(c) Place the result of (b) back on STACK.

[End of If structure.]

[End of Step 2 loop.]

5. Set VALUE equal to the top element on STACK.

6.Exit.

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Arithmetic Expressions;Polish Notation

Arithmetic Expressions;Polish Notation : 

Highest : Exponentiation

Next Highest : Multiplication (*) and Division(/)

Lowest : Addition(+) and Subtruction(-)

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