Print Concatenation of Zig-Zag String in 'n' Rows - GeeksforGeeks
Given a string and number of rows 'n'. Print the string formed by concatenating n rows when input string is written in row-wise Zig-Zag fashion.
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Given a string and number of rows 'n'. Print the string formed by concatenating n rows when input string is written in row-wise Zig-Zag fashion.
Input: str = "ABCDEFGH" n = 2 Output: "ACEGBDFH" Explanation: Let us write input string in Zig-Zag fashion in 2 rows. A C E G B D F H Now concatenate the two rows and ignore spaces in every row. We get "ACEGBDFH" Input: str = "GEEKSFORGEEKS" n = 3 Output: GSGSEKFREKEOE Explanation: Let us write input string in Zig-Zag fashion in 3 rows. G S G S E K F R E K E O E Now concatenate the two rows and ignore spaces in every row. We get "GSGSEKFREKEOE"
1) Create an array strings, arr[n] 2) Initialize direction as "down" and row as 0. The direction indicates whether we need to move up or down in rows. 3) Traverse the input string, do following for every character. a) Append current character to string of current row. b) If row number is n-1, then change direction to 'up' c) If row number is 0, then change direction to 'down' d) If direction is 'down', do row++. Else do row--. 4) One by one print all strings of arr[].
void
printZigZagConcat(string str,
int
n)
{
// Corner Case (Only one row)
if
(n == 1)
{
cout << str;
return
;
}
// Find length of string
int
len = str.length();
// Create an array of strings for all n rows
string arr[n];
// Initialize index for array of strings arr[]
int
row = 0;
bool
down;
// True if we are moving down in rows,
// else false
// Travers through given string
for
(
int
i = 0; i < len; ++i)
{
// append current character to current row
arr[row].push_back(str[i]);
// If last row is reached, change direction to 'up'
if
(row == n-1)
down =
false
;
// If 1st row is reached, change direction to 'down'
else
if
(row == 0)
down =
true
;
// If direction is down, increment, else decrement
(down)? (row++): (row--);
}
// Print concatenation of all rows
for
(
int
i = 0; i < n; ++i)
cout << arr[i];
}