Write a program to find factorial of the given number.
1.
Recursion: A function is called 'recursive' if a statement within the body of a
function calls the same function. It is also called 'circular definition'. Recursion is thus a process of defining something in terms of itself.
Program: To calculate the factorial value using recursion.
#include <stdio.h>
int fact(int n);
int main() {
int x, i;
printf("Enter a value for x: \n");
scanf("%d", &x);
i = fact(x);
printf("\nFactorial of %d is %d", x, i);
return 0;
} int fact(int n) {
/* n=0 indicates a terminating condition */
if (n <= 0) {
return (1);
} else {
/* function calling itself */
return (n * fact(n - 1));
/*n*fact(n-1) is a recursive expression */
}
}
Output:
Enter a value for x:
4
Factorial of 4 is 24
Explanation:
fact(n) = n * fact(n-1)
If n=4
fact(4) = 4 * fact(3) there is a call to fact(3)
fact(3) = 3 * fact(2)
fact(2) = 2 * fact(1)
fact(1) = 1 * fact(0)
fact(0) = 1
fact(1) = 1 * 1 = 1
fact(2) = 2 * 1 = 2
fact(3) = 3 * 2 = 6
Thus fact(4) = 4 * 6 = 24
Terminating condition(n <= 0 here;) is a must for a recursive program. Otherwise
the program enters into an
infinite loop.
2. Write a program to check whether the given number is even or odd.
Program:
#include <stdio.h>
int main() {
int a;
printf("Enter a: \n");
scanf("%d", &a);
/* logic */
if (a % 2 == 0) {
printf("The given number is EVEN\n");
}
else {
printf("The given number is ODD\n");
}
return 0;
}
Output:
Enter a: 2
The given number is EVEN
Explanation with examples:
Example 1: If entered number is an even number
Let value of 'a' entered is 4
if(a%2==0) then a is an even number, else odd.
i.e. if(4%2==0) then 4 is an even number, else odd.
To check whether 4 is even or odd, we need to calculate (4%2).
/* % (modulus) implies remainder value. */
/* Therefore if the remainder obtained when 4 is divided by 2 is 0, then 4 is even.
*/
4%2==0 is true
Thus 4 is an even number.
Example 2: If entered number is an odd number.
Let value of 'a' entered is 7
if(a%2==0) then a is an even number, else odd.
i.e. if(7%2==0) then 4 is an even number, else odd.
To check whether 7 is even or odd, we need to calculate (7%2).
7%2==0 is false /* 7%2==1 condition fails and else part is executed */
Thus 7 is an odd number.
3. Write a program to swap two numbers using a temporary variable.
Swapping interchanges the values of two given variables.
Logic:
step1: temp=x;
step2: x=y;
step3: y=temp;
Example:
if x=5 and y=8, consider a temporary variable temp.
step1: temp=x=5;
step2: x=y=8;
step3: y=temp=5;
Thus the values of the variables x and y are interchanged.
Program:
#include <stdio.h>
int main() {
int a, b, temp;
printf("Enter the value of a and b: \n");
scanf("%d %d", &a, &b);
printf("Before swapping a=%d, b=%d \n", a, b);
/*Swapping logic */
temp = a;
a = b;
b = temp;
printf("After swapping a=%d, b=%d", a, b);
return 0;
}
Output:
Enter the values of a and b: 2 3
Before swapping a=2, b=3
After swapping a=3, b=2
4. Write a program to swap two numbers without using a temporary variable.
Swapping interchanges the values of two given variables.
Logic:
step1: x=x+y;
step2: y=x-y;
step3: x=x-y;
Example:
if x=7 and y=4
step1: x=7+4=11;
step2: y=11-4=7;
step3: x=11-7=4;
Thus the values of the variables x and y are interchanged.
Program:
#include <stdio.h>
int main() {
int a, b;
printf("Enter values of a and b: \n");
scanf("%d %d", &a, &b);
printf("Before swapping a=%d, b=%d\n", a,b);
/*Swapping logic */
a = a + b;
b = a - b;
a = a - b;
printf("After swapping a=%d b=%d\n", a, b);
return 0;
}
Output:
Enter values of a and b: 2 3
Before swapping a=2, b=3
The values after swapping are a=3 b=2
5. Write a program to swap two numbers using bitwise operators.
Program:
#include <stdio.h>
int main() {
int i = 65;
int k = 120;
printf("\n value of i=%d k=%d before swapping", i, k);
i = i ^ k;
k = i ^ k;
i = i ^ k;
printf("\n value of i=%d k=%d after swapping", i, k);
return 0;
}
Explanation:
i = 65; binary equivalent of 65 is 0100 0001
k = 120; binary equivalent of 120 is 0111 1000
i = i^k;
i...0100 0001
k...0111 1000
---------
val of i = 0011 1001
---------
k = i^k
i...0011 1001
k...0111 1000
---------
val of k = 0100 0001 binary equivalent of this is 65
---------(that is the initial value of i)
i = i^k
i...0011 1001
k...0100 0001
---------
val of i = 0111 1000 binary equivalent of this is 120
--------- (that is the initial value of k)
6. Write a program to find the greatest of three numbers.
Program:
#include <stdio.h>
int main(){
int a, b, c;
printf("Enter a,b,c: \n");
scanf("%d %d %d", &a, &b, &c);
if (a > b && a > c) {
printf("a is Greater than b and c");
}
else if (b > a && b > c) {
printf("b is Greater than a and c");
}
else if (c > a && c > b) {
printf("c is Greater than a and b");
}
else {
printf("all are equal or any two values are equal");
}
return 0;
}
Output:
Enter a,b,c: 3 5 8
c is Greater than a and b
Explanation with examples:
Consider three numbers a=5,b=4,c=8
if(a>b && a>c) then a is greater than b and c
now check this condition for the three numbers 5,4,8 i.e.
if(5>4 && 5>8) /* 5>4 is true but 5>8 fails */
so the control shifts to else if condition
else if(b>a && b>c) then b is greater than a and c
now checking this condition for 5,4,8 i.e.
else if(4>5 && 4>8) /* both the conditions fail */
now the control shifts to the next else if condition
else if(c>a && c>b) then c is greater than a and b
now checking this condition for 5,4,8 i.e.
else if(8>5 && 8>4) /* both conditions are satisfied */
Thus c is greater than a and b.
7. Write a program to find the greatest among ten numbers.
Program:
#include <stdio.h>
int main() {
int a[10];
int i;
int greatest;
printf("Enter ten values:");
//Store 10 numbers in an array
for (i = 0; i < 10; i++) {
scanf("%d", &a[i]);
}
//Assume that a[0] is greatest
greatest = a[0];
for (i = 0; i < 10; i++) {
if (a[i] > greatest) {
greatest = a[i];
}
}
printf("\nGreatest of ten numbers is %d", greatest);
return 0;
}
Output:
Enter ten values: 2 53 65 3 88 8 14 5 77 64 Greatest of ten numbers is 88
Explanation with example:
Entered values are 2, 53, 65, 3, 88, 8, 14, 5, 77, 64
They are stored in an array of size 10. let a[] be an array holding these values.
/* how the greatest among ten numbers is found */
Let us consider a variable 'greatest'. At the beginning of the loop, variable 'greatest'
is assinged with the value of
first element in the array greatest=a[0]. Here variable 'greatest' is assigned 2 as a[0]
=2.
Below loop is executed until end of the array 'a[]';.
for(i=0; i<10; i++)
{
if(a[i]>greatest)
{
greatest= a[i];
}
}
For each value of 'i', value of a[i] is compared with value of variable 'greatest'. If
any value greater than the value
of 'greatest' is encountered, it would be replaced by a[i]. After completion of 'for'
loop, the value of variable
'greatest' holds the greatest number in the array. In this case 88 is the greatest of all
the numbers.
8. Write a program to check whether the given number is a prime.
A prime number is a natural number that has only one and itself as factors.
Examples: 2, 3, 13 are prime
numbers.
Program:
#include <stdio.h>
main() {
int n, i, c = 0;
printf("Enter any number n: \n");
scanf("%d", &n);
/*logic*/
for (i = 1; i <= n; i++) {
if (n % i == 0) {
c++;
}
}
if (c == 2) {
printf("n is a Prime number");
}
else {
printf("n is not a Prime number");
}
return 0;
}
Output:
Enter any number n: 7
n is Prime
Explanation with examples:
consider a number n=5
for(i=0;i<=n;i++) /* for loop is executed until the n value equals i */
i.e. for(i=0;i<=5;i++) /* here the for loop is executed until i is equal to n */
1st iteration: i=1;i<=5;i++
here i is incremented i.e. i value for next iteration is 2
now if(n%i==0) then c is incremented
i.e.if(5%1==0)then c is incremented, here 5%1=0 thus c is incremented.
now c=1;
2nd iteration: i=2;i<=5;i++
here i is incremented i.e. i value for next iteration is 3
now if(n%i==0) then c is incremented
i.e.if(5%2==0) then c is incremented, but 5%2!=0 and so c is not incremented, c
remains 1
c=1;
3rd iteration: i=3;i<=5;i++
here i is incremented i.e. i value for next iteration is 4
now if(n%i==0) then c is incremented
i.e.if(5%3==0) then c ic incremented, but 5%3!=0 and so c is not incremented, c
remains 1
c=1;
4th iteration: i=4;i<=5;i++
here i is incremented i.e. i value for next iteration is 5
now if(n%i==0) then c is incremented
i.e. if(5%4==0) then c is incremented, but 5%4!=0 and so c is not incremented, c
remains 1
c=1;
5th iteration: i=5;i<=5;i++
here i is incremented i.e. i value for next iteration is 6
now if(n%i==0) then c is incremented
i.e. if(5%5==0) then c is incremented, 5%5=0 and so c is incremented.
i.e. c=2
6th iteration: i=6;i<=5;i++
here i value is 6 and 6<=5 is false thus the condition fails and control leaves the for
loop.
now if(c==2) then n is a prime number
we have c=2 from the 5th iteration and thus n=5 is a Prime number.
9. Write a program to check whether the given number is a palindromic
number.
If a number, which when read in both forward and backward way is same, then
such a number is called a
palindrome number.
Program:
#include <stdio.h>
int main() {
int n, n1, rev = 0, rem;
printf("Enter any number: \n");
scanf("%d", &n);
n1 = n;
/* logic */
while (n > 0){
rem = n % 10;
rev = rev * 10 + rem;
n = n / 10;
}
if (n1 == rev){
printf("Given number is a palindromic number");
}
else{
printf("Given number is not a palindromic number");
}
return 0;
}
Output:
Enter any number: 121
Given number is a palindrome
Explanation with an example:
Consider a number n=121, reverse=0, remainder;
number=121
now the while loop is executed /* the condition (n>0) is satisfied */
/* calculate remainder */
remainder of 121 divided by 10=(121%10)=1;
now reverse=(reverse*10)+remainder
=(0*10)+1 /* we have initialized reverse=0 */
=1
number=number/10
=121/10
=12
now the number is 12, greater than 0. The above process is repeated for
number=12.
remainder=12%10=2;
reverse=(1*10)+2=12;
number=12/10=1;
now the number is 1, greater than 0. The above process is repeated for number=1.
remainder=1%10=1;
reverse=(12*10)+1=121;
number=1/10 /* the condition n>0 is not satisfied,control leaves the while loop */
Program stops here. The given number=121 equals the reverse of the number. Thus
the given number is a
palindrome number.
10.Write a program to check whether the given string is a palindrome.
Palindrome is a string, which when read in both forward and backward way is
same.
Example: radar, madam, pop, lol, rubber, etc.,
Program:
#include <stdio.h>
#include <string.h>
int main() {
char string1[20];
int i, length;
int flag = 0;
printf("Enter a string: \n");
scanf("%s", string1);
length = strlen(string1);
for(i=0;i < length ;i++){
if(string1[i] != string1[length-i-1]){
flag = 1;
break;
}
}
if (flag) {
printf("%s is not a palindrome\n", string1);
}
else {
printf("%s is a palindrome\n", string1);
}
return 0;
}
Output:
Enter a string: radar
"radar" is a palindrome
Explanation with example:
To check if a string is a palindrome or not, a string needs to be compared with the
reverse of itself.
Consider a palindrome string: "radar",
---------------------------
index: 0 1 2 3 4
value: r a d a r
---------------------------
To compare it with the reverse of itself, the following logic is used:
0th character in the char array, string1 is same as 4th character in the same string.
1st character is same as 3rd character.
2nd character is same as 2nd character.
....
ith character is same as 'length-i-1'th character.
If any one of the above condition fails, flag is set to true(1), which implies that the
string is not a palindrome.
By default, the value of flag is false(0). Hence, if all the conditions are satisfied,
the string is a palindrome.
1.
Recursion: A function is called 'recursive' if a statement within the body of a
function calls the same function. It is also called 'circular definition'. Recursion is thus a process of defining something in terms of itself.
Program: To calculate the factorial value using recursion.
#include <stdio.h>
int fact(int n);
int main() {
int x, i;
printf("Enter a value for x: \n");
scanf("%d", &x);
i = fact(x);
printf("\nFactorial of %d is %d", x, i);
return 0;
} int fact(int n) {
/* n=0 indicates a terminating condition */
if (n <= 0) {
return (1);
} else {
/* function calling itself */
return (n * fact(n - 1));
/*n*fact(n-1) is a recursive expression */
}
}
Output:
Enter a value for x:
4
Factorial of 4 is 24
Explanation:
fact(n) = n * fact(n-1)
If n=4
fact(4) = 4 * fact(3) there is a call to fact(3)
fact(3) = 3 * fact(2)
fact(2) = 2 * fact(1)
fact(1) = 1 * fact(0)
fact(0) = 1
fact(1) = 1 * 1 = 1
fact(2) = 2 * 1 = 2
fact(3) = 3 * 2 = 6
Thus fact(4) = 4 * 6 = 24
Terminating condition(n <= 0 here;) is a must for a recursive program. Otherwise
the program enters into an
infinite loop.
2. Write a program to check whether the given number is even or odd.
Program:
#include <stdio.h>
int main() {
int a;
printf("Enter a: \n");
scanf("%d", &a);
/* logic */
if (a % 2 == 0) {
printf("The given number is EVEN\n");
}
else {
printf("The given number is ODD\n");
}
return 0;
}
Output:
Enter a: 2
The given number is EVEN
Explanation with examples:
Example 1: If entered number is an even number
Let value of 'a' entered is 4
if(a%2==0) then a is an even number, else odd.
i.e. if(4%2==0) then 4 is an even number, else odd.
To check whether 4 is even or odd, we need to calculate (4%2).
/* % (modulus) implies remainder value. */
/* Therefore if the remainder obtained when 4 is divided by 2 is 0, then 4 is even.
*/
4%2==0 is true
Thus 4 is an even number.
Example 2: If entered number is an odd number.
Let value of 'a' entered is 7
if(a%2==0) then a is an even number, else odd.
i.e. if(7%2==0) then 4 is an even number, else odd.
To check whether 7 is even or odd, we need to calculate (7%2).
7%2==0 is false /* 7%2==1 condition fails and else part is executed */
Thus 7 is an odd number.
3. Write a program to swap two numbers using a temporary variable.
Swapping interchanges the values of two given variables.
Logic:
step1: temp=x;
step2: x=y;
step3: y=temp;
Example:
if x=5 and y=8, consider a temporary variable temp.
step1: temp=x=5;
step2: x=y=8;
step3: y=temp=5;
Thus the values of the variables x and y are interchanged.
Program:
#include <stdio.h>
int main() {
int a, b, temp;
printf("Enter the value of a and b: \n");
scanf("%d %d", &a, &b);
printf("Before swapping a=%d, b=%d \n", a, b);
/*Swapping logic */
temp = a;
a = b;
b = temp;
printf("After swapping a=%d, b=%d", a, b);
return 0;
}
Output:
Enter the values of a and b: 2 3
Before swapping a=2, b=3
After swapping a=3, b=2
4. Write a program to swap two numbers without using a temporary variable.
Swapping interchanges the values of two given variables.
Logic:
step1: x=x+y;
step2: y=x-y;
step3: x=x-y;
Example:
if x=7 and y=4
step1: x=7+4=11;
step2: y=11-4=7;
step3: x=11-7=4;
Thus the values of the variables x and y are interchanged.
Program:
#include <stdio.h>
int main() {
int a, b;
printf("Enter values of a and b: \n");
scanf("%d %d", &a, &b);
printf("Before swapping a=%d, b=%d\n", a,b);
/*Swapping logic */
a = a + b;
b = a - b;
a = a - b;
printf("After swapping a=%d b=%d\n", a, b);
return 0;
}
Output:
Enter values of a and b: 2 3
Before swapping a=2, b=3
The values after swapping are a=3 b=2
5. Write a program to swap two numbers using bitwise operators.
Program:
#include <stdio.h>
int main() {
int i = 65;
int k = 120;
printf("\n value of i=%d k=%d before swapping", i, k);
i = i ^ k;
k = i ^ k;
i = i ^ k;
printf("\n value of i=%d k=%d after swapping", i, k);
return 0;
}
Explanation:
i = 65; binary equivalent of 65 is 0100 0001
k = 120; binary equivalent of 120 is 0111 1000
i = i^k;
i...0100 0001
k...0111 1000
---------
val of i = 0011 1001
---------
k = i^k
i...0011 1001
k...0111 1000
---------
val of k = 0100 0001 binary equivalent of this is 65
---------(that is the initial value of i)
i = i^k
i...0011 1001
k...0100 0001
---------
val of i = 0111 1000 binary equivalent of this is 120
--------- (that is the initial value of k)
6. Write a program to find the greatest of three numbers.
Program:
#include <stdio.h>
int main(){
int a, b, c;
printf("Enter a,b,c: \n");
scanf("%d %d %d", &a, &b, &c);
if (a > b && a > c) {
printf("a is Greater than b and c");
}
else if (b > a && b > c) {
printf("b is Greater than a and c");
}
else if (c > a && c > b) {
printf("c is Greater than a and b");
}
else {
printf("all are equal or any two values are equal");
}
return 0;
}
Output:
Enter a,b,c: 3 5 8
c is Greater than a and b
Explanation with examples:
Consider three numbers a=5,b=4,c=8
if(a>b && a>c) then a is greater than b and c
now check this condition for the three numbers 5,4,8 i.e.
if(5>4 && 5>8) /* 5>4 is true but 5>8 fails */
so the control shifts to else if condition
else if(b>a && b>c) then b is greater than a and c
now checking this condition for 5,4,8 i.e.
else if(4>5 && 4>8) /* both the conditions fail */
now the control shifts to the next else if condition
else if(c>a && c>b) then c is greater than a and b
now checking this condition for 5,4,8 i.e.
else if(8>5 && 8>4) /* both conditions are satisfied */
Thus c is greater than a and b.
7. Write a program to find the greatest among ten numbers.
Program:
#include <stdio.h>
int main() {
int a[10];
int i;
int greatest;
printf("Enter ten values:");
//Store 10 numbers in an array
for (i = 0; i < 10; i++) {
scanf("%d", &a[i]);
}
//Assume that a[0] is greatest
greatest = a[0];
for (i = 0; i < 10; i++) {
if (a[i] > greatest) {
greatest = a[i];
}
}
printf("\nGreatest of ten numbers is %d", greatest);
return 0;
}
Output:
Enter ten values: 2 53 65 3 88 8 14 5 77 64 Greatest of ten numbers is 88
Explanation with example:
Entered values are 2, 53, 65, 3, 88, 8, 14, 5, 77, 64
They are stored in an array of size 10. let a[] be an array holding these values.
/* how the greatest among ten numbers is found */
Let us consider a variable 'greatest'. At the beginning of the loop, variable 'greatest'
is assinged with the value of
first element in the array greatest=a[0]. Here variable 'greatest' is assigned 2 as a[0]
=2.
Below loop is executed until end of the array 'a[]';.
for(i=0; i<10; i++)
{
if(a[i]>greatest)
{
greatest= a[i];
}
}
For each value of 'i', value of a[i] is compared with value of variable 'greatest'. If
any value greater than the value
of 'greatest' is encountered, it would be replaced by a[i]. After completion of 'for'
loop, the value of variable
'greatest' holds the greatest number in the array. In this case 88 is the greatest of all
the numbers.
8. Write a program to check whether the given number is a prime.
A prime number is a natural number that has only one and itself as factors.
Examples: 2, 3, 13 are prime
numbers.
Program:
#include <stdio.h>
main() {
int n, i, c = 0;
printf("Enter any number n: \n");
scanf("%d", &n);
/*logic*/
for (i = 1; i <= n; i++) {
if (n % i == 0) {
c++;
}
}
if (c == 2) {
printf("n is a Prime number");
}
else {
printf("n is not a Prime number");
}
return 0;
}
Output:
Enter any number n: 7
n is Prime
Explanation with examples:
consider a number n=5
for(i=0;i<=n;i++) /* for loop is executed until the n value equals i */
i.e. for(i=0;i<=5;i++) /* here the for loop is executed until i is equal to n */
1st iteration: i=1;i<=5;i++
here i is incremented i.e. i value for next iteration is 2
now if(n%i==0) then c is incremented
i.e.if(5%1==0)then c is incremented, here 5%1=0 thus c is incremented.
now c=1;
2nd iteration: i=2;i<=5;i++
here i is incremented i.e. i value for next iteration is 3
now if(n%i==0) then c is incremented
i.e.if(5%2==0) then c is incremented, but 5%2!=0 and so c is not incremented, c
remains 1
c=1;
3rd iteration: i=3;i<=5;i++
here i is incremented i.e. i value for next iteration is 4
now if(n%i==0) then c is incremented
i.e.if(5%3==0) then c ic incremented, but 5%3!=0 and so c is not incremented, c
remains 1
c=1;
4th iteration: i=4;i<=5;i++
here i is incremented i.e. i value for next iteration is 5
now if(n%i==0) then c is incremented
i.e. if(5%4==0) then c is incremented, but 5%4!=0 and so c is not incremented, c
remains 1
c=1;
5th iteration: i=5;i<=5;i++
here i is incremented i.e. i value for next iteration is 6
now if(n%i==0) then c is incremented
i.e. if(5%5==0) then c is incremented, 5%5=0 and so c is incremented.
i.e. c=2
6th iteration: i=6;i<=5;i++
here i value is 6 and 6<=5 is false thus the condition fails and control leaves the for
loop.
now if(c==2) then n is a prime number
we have c=2 from the 5th iteration and thus n=5 is a Prime number.
9. Write a program to check whether the given number is a palindromic
number.
If a number, which when read in both forward and backward way is same, then
such a number is called a
palindrome number.
Program:
#include <stdio.h>
int main() {
int n, n1, rev = 0, rem;
printf("Enter any number: \n");
scanf("%d", &n);
n1 = n;
/* logic */
while (n > 0){
rem = n % 10;
rev = rev * 10 + rem;
n = n / 10;
}
if (n1 == rev){
printf("Given number is a palindromic number");
}
else{
printf("Given number is not a palindromic number");
}
return 0;
}
Output:
Enter any number: 121
Given number is a palindrome
Explanation with an example:
Consider a number n=121, reverse=0, remainder;
number=121
now the while loop is executed /* the condition (n>0) is satisfied */
/* calculate remainder */
remainder of 121 divided by 10=(121%10)=1;
now reverse=(reverse*10)+remainder
=(0*10)+1 /* we have initialized reverse=0 */
=1
number=number/10
=121/10
=12
now the number is 12, greater than 0. The above process is repeated for
number=12.
remainder=12%10=2;
reverse=(1*10)+2=12;
number=12/10=1;
now the number is 1, greater than 0. The above process is repeated for number=1.
remainder=1%10=1;
reverse=(12*10)+1=121;
number=1/10 /* the condition n>0 is not satisfied,control leaves the while loop */
Program stops here. The given number=121 equals the reverse of the number. Thus
the given number is a
palindrome number.
10.Write a program to check whether the given string is a palindrome.
Palindrome is a string, which when read in both forward and backward way is
same.
Example: radar, madam, pop, lol, rubber, etc.,
Program:
#include <stdio.h>
#include <string.h>
int main() {
char string1[20];
int i, length;
int flag = 0;
printf("Enter a string: \n");
scanf("%s", string1);
length = strlen(string1);
for(i=0;i < length ;i++){
if(string1[i] != string1[length-i-1]){
flag = 1;
break;
}
}
if (flag) {
printf("%s is not a palindrome\n", string1);
}
else {
printf("%s is a palindrome\n", string1);
}
return 0;
}
Output:
Enter a string: radar
"radar" is a palindrome
Explanation with example:
To check if a string is a palindrome or not, a string needs to be compared with the
reverse of itself.
Consider a palindrome string: "radar",
---------------------------
index: 0 1 2 3 4
value: r a d a r
---------------------------
To compare it with the reverse of itself, the following logic is used:
0th character in the char array, string1 is same as 4th character in the same string.
1st character is same as 3rd character.
2nd character is same as 2nd character.
....
ith character is same as 'length-i-1'th character.
If any one of the above condition fails, flag is set to true(1), which implies that the
string is not a palindrome.
By default, the value of flag is false(0). Hence, if all the conditions are satisfied,
the string is a palindrome.
