Given a string str, our task is to print its reversed pattern. The Pattern will be incremental in reversed order, and when the string is completed fill ‘*’ in the remaining places.
Like we enter a string “abcd”, now in first line we have to print “a” then in next line we have to print “cb” and then in third line we will print “**d”.
Example
Input: str[] = { “abcd” } Output: a c b * * d
Explanation −
- In first line print 1 character
- In second line print 2 characters in reverse order
- In third line Print 3 characters in reverse order, if the string is less than 3 then print the characters and fill the blank spaces with *.
Input: str[] = {“tutorialspoint”} Output:
The approach used below is as follows −
- We will traverse the string from i=0 and will check i<n && k<(n-i)*2 is true
- Then we will take a variable k and Set k as ((i*(i+1))/2)-1
- We will check If k >= n-1 then, will Print "* " else will print the string’s value in reverse order
Algorithm
Start In function int reverse_it(char str[], int n) Step 1-> Declare and Initialize i, j=0 , k=0 Step 2-> Loop For i=0 and i<n && k<(n-i)*2 and i++ Set k as ((i*(i+1))/2)-1 Loop For j=0 and j<i && k<(n-i)*2 and j++ If k >= n-1 then, Print "* " Else Print "%c ",str[k] Decrement k by 1 End loop Print new line End loop In Function int main(int argc, char const *argv[]) Step 1-> Declare and initialize string str[] Step 2-> Declare and Initialize size as sizeof(str)/sizeof(str[0]) Step 3-> Call function reverse_it(str, size); Stop
Example
#include <stdio.h> int reverse_it(char str[], int n) { int i, j=0 , k=0; for(i=0; i<n && k<(n-i)*2; i++) { //Assigning k k = ((i*(i+1))/2)-1; for(j=0; j<i && k<(n-i)*2; j++) { //will check if k is greater than the total number of characters //then we will print * for filling the extra characters if(k >= n-1) printf("* "); //print the string in reverse order else printf("%c ",str[k]); k--; } //for line break after reverse sequence printf("\n"); } return 0; } //Main Function int main(int argc, char const *argv[]) { char str[] = {"tutorialspoint"}; int size = sizeof(str)/sizeof(str[0]); reverse_it(str, size); return 0; }
Output
If run the above code it will generate the following output −