Given a string S that consists of only alphanumeric characters and dashes. The string is separated into N + 1 groups by N dashes. Also given an integer K.
We want to reformat the string S, such that each group contains exactly K characters, except for the first group, which could be shorter than K but still must contain at least one character. Furthermore, a dash must be inserted between two groups, and you should convert all lowercase letters to uppercase.
Return the reformatted string.
Examples:
Input: S = “5F3Z-2e-9-w”, K = 4
Output: “5F3Z-2E9W”
Explanation: The string S has been split into two parts,
each part has 4 characters.
Note that two extra dashes are not needed and can be removed.
Input: S = “2-5g-3-J”, K = 2
Output: “2-5G-3J”
Explanation: The string s has been split into three parts,
each part has 2 characters except the first part
as it could be shorter as mentioned above
Naive Approach: To solve the problem follow the below idea:
- We will have a Greedy approach in which we will create a temporary string with only the alphanumeric characters(but in reverse) and then add the dashes after every K step.
- The reversal is necessary at the beginning because each group contains exactly K characters, except for the first group as mentioned in the problem.
Follow the steps to solve the problem:
- Create an empty string temp and push only the characters (in upper-case) that are different than ‘-‘.
- Now reverse the string obtained. Also, create a string ‘ans’ to store the final string.
- Iterate over the string and whenever ‘K’ characters are pushed in ‘ans’ push a dash “-” into the string.
- Return ‘ans’ as the result.
Below is the implementation to solve the problem:
C
#include <ctype.h>
#include <stdio.h>
#include <string.h>
char * ReFormatString( char * S, int K)
{
char temp[100];
int n = strlen (S);
int len = 0;
for ( int i = 0; i < n; i++) {
if (S[i] != '-' ) {
temp[len++] = toupper (S[i]);
}
}
char * ans
= ( char *) malloc ( sizeof ( char ) * (len + len / K + 1));
int val = K;
int j = 0;
for ( int i = len - 1; i >= 0; i--) {
if (val == 0) {
val = K;
ans[j++] = '-' ;
}
ans[j++] = temp[i];
val--;
}
ans[j] = '\0' ;
int i = 0, k = j - 1;
while (i < k) {
char t = ans[i];
ans[i++] = ans[k];
ans[k--] = t;
}
return ans;
}
int main()
{
char s[] = "5F3Z-2e-9-w" ;
int K = 4;
printf ( "%s" , ReFormatString(s, K));
return 0;
}
|
C++
#include <bits/stdc++.h>
using namespace std;
string ReFormatString(string S, int K)
{
string temp;
int n = S.length();
for ( int i = 0; i < n; i++) {
if (S[i] != '-' ) {
temp.push_back( toupper (S[i]));
}
}
int len = temp.length();
string ans;
int val = K;
for ( int i = len - 1; i >= 0; i--) {
if (val == 0) {
val = K;
ans.push_back( '-' );
}
ans.push_back(temp[i]);
val--;
}
reverse(ans.begin(), ans.end());
return ans;
}
int main()
{
string s = "5F3Z-2e-9-w" ;
int K = 4;
cout << ReFormatString(s, K);
return 0;
}
|
Java
class GFG {
public static String ReFormatString(String S, int K)
{
String temp = "" ;
int n = S.length();
for ( int i = 0 ; i < n; i++) {
if (S.charAt(i) != '-' ) {
temp += (Character.toUpperCase(S.charAt(i)));
}
}
int len = temp.length();
String ans = "" ;
int val = K;
for ( int i = len - 1 ; i >= 0 ; i--) {
if (val == 0 ) {
val = K;
ans += '-' ;
}
ans += temp.charAt(i);
val--;
}
char [] charArray = ans.toCharArray();
reverse(charArray, charArray.length);
String res = new String(charArray);
return res;
}
static void reverse( char a[], int n)
{
char t;
for ( int i = 0 ; i < n / 2 ; i++) {
t = a[i];
a[i] = a[n - i - 1 ];
a[n - i - 1 ] = t;
}
}
public static void main(String args[]) {
String s = "5F3Z-2e-9-w" ;
int K = 4 ;
System.out.println(ReFormatString(s, K));
}
}
|
Python3
def ReFormatStrings(s,k):
temp = ""
n = len (s)
for i in range ( 0 ,n):
if (s[i] ! = '-' ):
temp + = s[i].upper()
length = len (temp)
ans = ""
val = k
for i in range (length - 1 , - 1 , - 1 ):
if (val = = 0 ):
val = k
ans + = '-'
ans + = temp[i]
val - = 1
ans = ans[:: - 1 ]
return ans
if __name__ = = "__main__" :
s = "5F3Z-2e-9-w"
k = 4
print (ReFormatStrings(s,k))
|
C#
using System;
public class GFG{
public static string ReFormatString( string S, int K)
{
string temp= "" ;
int n = S.Length;
for ( int i = 0; i < n; i++) {
if (S[i] != '-' ) {
temp+=( char .ToUpper(S[i]));
}
}
int len = temp.Length;
string ans= "" ;
int val = K;
for ( int i = len - 1; i >= 0; i--) {
if (val == 0) {
val = K;
ans+= '-' ;
}
ans+=temp[i];
val--;
}
char [] charArray = ans.ToCharArray();
Array.Reverse( charArray );
string res = new string (charArray);
return res;
}
static public void Main (){
string s = "5F3Z-2e-9-w" ;
int K = 4;
Console.WriteLine(ReFormatString(s, K));
}
}
|
Javascript
function reverse(s)
{
let splitString = s.split( "" );
let reverseArray = splitString.reverse();
let joinArray = reverseArray.join( "" );
return joinArray;
}
function ReFormatString(S,K)
{
let temp = "" ;
let n = S.length;
for (let i = 0; i < n; i++) {
if (S[i] != '-' ) {
temp+=S[i].toUpperCase();
}
}
let len = temp.length;
let ans = "" ;
let val = K;
for (let i = len - 1; i >= 0; i--) {
if (val == 0) {
val = K;
ans += '-' ;
}
ans += temp[i];
val--;
}
ans = reverse(ans);
return ans;
}
let s = "5F3Z-2e-9-w" ;
let K = 4;
console.log(ReFormatString(s, K));
|
Time Complexity: O(n).
Space Complexity: O(n) as extra space has been used by creating temp string.
Efficient approach: To solve the problem follow the below idea:
- Without creating any other string we will move all the dashes to the front and remove them then we will make use of the mathematical formula to calculate the number of dashes at the right of all the alphanumeric characters.
- Number of Dashes=(Total alphanumeric elements)/(number of elements in every group)
Formula:
Number of Dashes at any step = (Total alphanumeric elements to the right of the current index) / (number of elements in every group)
Follow the steps to solve the problem:
- Iterate from the back of the string and move all the alphanumeric characters to the back of the string.
- Delete all the dashes from the beginning.
- Calculate the number of dashes(rounded-up) that would be present in the final string and append free it to the original string.
- Iterate from the front and depending on the number of dashes that would be present up to that character, move the character by that amount in the left direction.
- Delete all the extra dashes that would have accumulated in the front of the string
- Return the string after all the modifications as the answer.
Below is the implementation for the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
string ReFormatString(string S, int K)
{
int len = S.length();
int cnt = 0;
int x = 0;
for ( int i = len - 1; i >= 0; i--) {
if (S[i] == '-' ) {
x++;
}
else {
S[i + x] = toupper (S[i]);
}
}
int slen = len - x;
int step = slen / K;
reverse(S.begin(), S.end());
int val = x;
while (val--) {
S.pop_back();
}
int temp = step;
while (temp--)
S.push_back( ' ' );
reverse(S.begin(), S.end());
len = S.length();
int i = slen, j = step, f = 0;
while (j < len) {
step = i / K;
if (f == 1)
step--;
int rem = i % K;
if (rem == 0 and f == 0) {
S[j - step] = '-' ;
f = 1;
continue ;
}
S[j - step] = S[j];
i--;
j++;
f = 0;
}
len = S.length();
reverse(S.begin(), S.end());
for ( int i = len - 1; i >= 0; i--) {
if (S[i] != '-' ) {
break ;
}
if (S[i] == '-' )
S.pop_back();
}
reverse(S.begin(), S.end());
return S;
}
int main()
{
string s = "5F3Z-2e-9-w" ;
int K = 4;
cout << ReFormatString(s, K);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
public static String reverseS(String str)
{
String nstr = "" ;
for ( int i = 0 ; i < str.length(); i++) {
char ch
= str.charAt(i);
nstr
= ch + nstr;
}
return nstr;
}
public static String ReFormatString(String S, int K)
{
int len = S.length();
int cnt = 0 ;
int x = 0 ;
for ( int i = len - 1 ; i >= 0 ; i--) {
if (S.charAt(i) == '-' ) {
x++;
}
else {
S = S.substring( 0 , i + x)
+ Character.toUpperCase(S.charAt(i))
+ S.substring(i + x + 1 );
}
}
int slen = len - x;
int step = ( int )(slen / K);
S = reverseS(S);
int val = x;
while (val > 0 ) {
S = S.substring( 0 , S.length() - 1 );
val--;
}
int temp = step;
while (temp > 0 ) {
S += " " ;
temp--;
}
S = reverseS(S);
len = S.length();
int i = slen, j = step, f = 0 ;
while (j < len) {
step = ( int )(i / K);
if (f == 1 )
step--;
int rem = i % K;
if (rem == 0 && f == 0 ) {
S = S.substring( 0 , j - step) + "-"
+ S.substring(j - step + 1 );
f = 1 ;
continue ;
}
S = S.substring( 0 , j - step) + S.charAt(j)
+ S.substring(j - step + 1 );
i--;
j++;
f = 0 ;
}
len = S.length();
S = reverseS(S);
for ( int m = len - 1 ; m >= 0 ; m--) {
if (S.charAt(m) != '-' ) {
break ;
}
if (S.charAt(m) == '-' )
S = S.substring( 0 , S.length() - 1 );
}
S = reverseS(S);
return S;
}
public static void main(String[] args)
{
String s = "5F3Z-2e-9-w" ;
int K = 4 ;
System.out.println(ReFormatString(s, K));
}
}
|
Python3
def reverse(string):
string = string[:: - 1 ]
return string
def ReFormatString( S, K):
length = len (S)
cnt = 0
x = 0
for i in range (length - 1 , - 1 , - 1 ):
if (S[i] = = '-' ):
x + = 1
else :
S = S[:i + x] + S[i].upper() + S[i + x + 1 :]
slen = length - x
step = slen / K
S = reverse(S)
val = x
while (val> 0 ):
S = S[: len (S) - 1 ]
val - = 1
temp = step
while (temp> 0 ):
S + = ' '
temp - = 1
S = reverse(S)
length = len (S)
i = slen
j = step
f = 0
while (j < length):
step = int (i / K)
if (f = = 1 ):
step - = 1
rem = i % K
if (rem = = 0 and f = = 0 ):
step = int (step)
j = int (j)
S = S[: int (j - step)] + '-' + S[ int (j - step) + 1 :]
f = 1
continue
S = S[: int (j - step)] + S[ int (j)] + S[ int (j - step) + 1 :]
i - = 1
j + = 1
f = 0
length = len (S)
S = reverse(S)
for char in reversed (S):
if (char ! = '-' ):
break
if (char = = '-' ):
S = S[: len (S) - 1 ]
S = reverse(S)
return S
s = "5F3Z-2e-9-w"
K = 4
print (ReFormatString(s, K))
|
C#
using System;
using System.Collections.Generic;
public class GFG {
public static String reverseS(String str)
{
String nstr = "" ;
for ( int i = 0; i < str.Length; i++) {
char ch = str[i];
nstr = ch + nstr;
}
return nstr;
}
public static String ReFormatString(String S, int K)
{
int len = S.Length;
int cnt = 0;
int x = 0;
int i;
for (i = len - 1; i >= 0; i--) {
if (S[i] == '-' ) {
x++;
}
else {
S = S.Substring(0, i + x)
+ Char.ToUpper(S[i])
+ S.Substring(i + x + 1);
}
}
int slen = len - x;
int step = ( int )(slen / K);
S = reverseS(S);
int val = x;
while (val > 0) {
S = S.Substring(0, S.Length - 1);
val--;
}
int temp = step;
while (temp > 0) {
S += " " ;
temp--;
}
S = reverseS(S);
len = S.Length;
i = slen;
int j = step, f = 0;
while (j < len) {
step = ( int )(i / K);
if (f == 1)
step--;
int rem = i % K;
if (rem == 0 && f == 0) {
S = S.Substring(0, j - step) + "-"
+ S.Substring(j - step + 1);
f = 1;
continue ;
}
S = S.Substring(0, j - step) + S[j]
+ S.Substring(j - step + 1);
i--;
j++;
f = 0;
}
len = S.Length;
S = reverseS(S);
for ( int m = len - 1; m >= 0; m--) {
if (S[m] != '-' ) {
break ;
}
if (S[m] == '-' )
S = S.Substring(0, S.Length - 1);
}
S = reverseS(S);
return S;
}
static public void Main()
{
string s = "5F3Z-2e-9-w" ;
int K = 4;
Console.WriteLine(ReFormatString(s, K));
}
}
|
Javascript
function ReFormatString(S, K) {
let len = S.length;
let cnt = 0;
let x = 0;
for (let i = len - 1; i >= 0; i--) {
if (S[i] == '-' ) {
x++;
}
else {
let c = (S[i].toUpperCase());
let arr1 = S.split( '' );
arr1[i + x] = c;
S = arr1.join( "" );
}
}
let slen = len - x;
let step = slen / K;
S = S.split( '' ).reverse().join( '' );
let val = x;
while (val--) {
S = S.substring(0, S.length - 1);
}
let temp = step;
while (temp--)
S += ' ' ;
S = S.split( '' ).reverse().join( '' );
len = S.length;
let i = slen, j = step, f = 0;
while (j < len) {
step = Math.floor(i / K);
if (f == 1)
step--;
let rem = i % K;
if (rem == 0 && f == 0) {
let arr2 = S.split( '' );
arr2[j - step] = '-' ;
S = arr2.join( "" );
f = 1;
continue ;
}
let arr3 = S.split( '' );
arr3[j - step] = S[j];
S = arr3.join( "" );
i--;
j++;
f = 0;
}
len = S.length;
S = S.split( '' ).reverse().join( '' );
for (let i = len - 1; i >= 0; i--) {
if (S[i] != '-' ) {
break ;
}
if (S[i] == '-' )
S = S.substring(0, S.length - 1);
}
S = S.split( '' ).reverse().join( '' );
return S;
}
let s = "5F3Z-2e-9-w" ;
let K = 4;
console.log(ReFormatString(s, K));
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Last Updated :
26 Apr, 2023
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