Given a string S that consists of only alphanumeric characters and dashes. The string is separated into n+1 groups by n dashes. We are also given a number K, the task is 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] Using Extra String - O(n) Time and O(n) Space
The idea is to use Greedy approach where we fill the result (from end) with k characters and the end fill the remaining characters at the beginning.
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.
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;
// Iterate over the string from right
// to left and start pushing
// characters at an interval of K
for (int i = len - 1; i >= 0; i--) {
if (val == 0) {
val = K;
ans.push_back('-');
}
ans.push_back(temp[i]);
val--;
}
// Reverse the final string and
// return it.
reverse(ans.begin(), ans.end());
return ans;
}
int main(){
string s = "5F3Z-2e-9-w";
int K = 4;
cout << ReFormatString(s, K);
return 0;
}
#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;
// Iterate over the string from right
// to left and start pushing
// characters at an interval of K
for (int i = len - 1; i >= 0; i--) {
if (val == 0) {
val = K;
ans[j++] = '-';
}
ans[j++] = temp[i];
val--;
}
// Reverse the final string and
// return it.
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;
}
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;
// Iterate over the String from right
// to left and start pushing
// characters at an interval of K
for (int i = len - 1; i >= 0; i--) {
if (val == 0) {
val = K;
ans += '-';
}
ans += temp.charAt(i);
val--;
}
// Reverse the final String and return it
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));
}
}
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
# Iterate over the string from right to left
# and start pushing characters at an interval of K
for i in range(length - 1,-1,-1):
if(val == 0):
val = k
ans += '-'
ans += temp[i]
val -= 1
# Reverse the final string and return it.
ans = ans[::-1]
return ans
# Driver code
if __name__ == "__main__":
s = "5F3Z-2e-9-w"
k = 4
print(ReFormatStrings(s,k))
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;
// Iterate over the string from right
// to left and start pushing
// characters at an interval of K
for (int i = len - 1; i >= 0; i--) {
if (val == 0) {
val = K;
ans+='-';
}
ans+=temp[i];
val--;
}
// Reverse the final string and
// return it.
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));
}
}
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;
// Iterate over the let from right
// to left and start pushing
// characters at an interval of K
for (let i = len - 1; i >= 0; i--) {
if (val == 0) {
val = K;
ans += '-';
}
ans += temp[i];
val--;
}
// Reverse the final let and
// return it.
ans = reverse(ans);
return ans;
}
let s = "5F3Z-2e-9-w";
let K = 4;
console.log(ReFormatString(s, K));
[Expected Approach] Using One variable - O(n) Time and O(1) Space
This is mainly an optimization over the above Greedy Approach.
- Without creating any other string, we move all the dashes to the front and remove them then.
- We make use of the below mathematical formula to calculate the number of dashes.
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.
C++
#include <bits/stdc++.h>
using namespace std;
string ReFormatString(string S, int K){
int len = S.length();
int cnt = 0;
int x = 0;
// Move the characters to the
// back of the string.
for (int i = len - 1; i >= 0; i--) {
if (S[i] == '-') {
x++;
}
else {
S[i + x] = toupper(S[i]);
}
}
// Calculate total number of
// alphanumeric characters so
// as to get the number of dashes
// in the final string.
int slen = len - x;
int step = slen / K;
// Remove x characters from the
// start of the string
reverse(S.begin(), S.end());
int val = x;
while (val--) {
S.pop_back();
}
// Push the empty spaces in
// the string (slen+step) to get
// the final string length
int temp = step;
while (temp--)
S.push_back(' ');
reverse(S.begin(), S.end());
len = S.length();
// Using simple mathematics
// to push the elements
// in the string at the correct place.
int i = slen, j = step, f = 0;
while (j < len) {
// At every step calculate the
// number of dashes that would be
// present before the character
step = i / K;
if (f == 1)
step--;
int rem = i % K;
// If the remainder is zero it
// implies that the character is a dash.
if (rem == 0 and f == 0) {
S[j - step] = '-';
f = 1;
continue;
}
S[j - step] = S[j];
i--;
j++;
f = 0;
}
// Remove all the dashes that would have
// accumulated in the beginning of the string.
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;
}
/*package whatever //do not write package name here */
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;
// Move the characters to the
// back of the string.
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);
}
}
// Calculate total number of
// alphanumeric characters so
// as to get the number of dashes
// in the final string.
int slen = len - x;
int step = (int)(slen / K);
// Remove x characters from the
// start of the string
S = reverseS(S);
int val = x;
while (val > 0) {
S = S.substring(0, S.length() - 1);
val--;
}
// Push the empty spaces in
// the string (slen+step) to get
// the final string length
int temp = step;
while (temp > 0) {
S += " ";
temp--;
}
S = reverseS(S);
len = S.length();
// Using simple mathematics
// to push the elements
// in the string at the correct place.
int i = slen, j = step, f = 0;
while (j < len) {
// At every step calculate the
// number of dashes that would be
// present before the character
step = (int)(i / K);
if (f == 1)
step--;
int rem = i % K;
// If the remainder is zero it
// implies that the character is a dash.
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;
}
// Remove all the dashes that would have
// accumulated in the beginning of the string.
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));
}
}
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:]
# Calculate total number of
# alphanumeric characters so
# as to get the number of dashes
# in the final string.
slen = length - x
step = slen / K
# Remove x characterclss from the
# start of the string
S = reverse(S)
val = x
while (val>0):
S = S[:len(S)-1]
val-=1
# Push the empty spaces in
# the string (slen+step) to get
# the final string length
temp = step
while (temp>0):
S+=' '
temp-=1
S = reverse(S)
length = len(S)
# Using simple mathematics
# to push the elements
# in the string at the correct place.
i = slen
j = step
f = 0
while (j < length):
# At every step calculate the
# number of dashes that would be
# present before the character
step = int(i / K)
if (f == 1):
step-=1
rem = i % K
# If the remainder is zero it
# implies that the character is a dash.
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
# Remove all the dashes that would have
# accumulated in the beginning of the string.
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))
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 x = 0;
int i;
// Move the characters to the
// back of the string.
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);
}
}
// Calculate total number of
// alphanumeric characters so
// as to get the number of dashes
// in the final string.
int slen = len - x;
int step = (int)(slen / K);
// Remove x characters from the
// start of the string
S = reverseS(S);
int val = x;
while (val > 0) {
S = S.Substring(0, S.Length - 1);
val--;
}
// Push the empty spaces in
// the string (slen+step) to get
// the final string length
int temp = step;
while (temp > 0) {
S += " ";
temp--;
}
S = reverseS(S);
len = S.Length;
// Using simple mathematics
// to push the elements
// in the string at the correct place.
i = slen;
int j = step, f = 0;
while (j < len) {
// At every step calculate the
// number of dashes that would be
// present before the character
step = (int)(i / K);
if (f == 1)
step--;
int rem = i % K;
// If the remainder is zero it
// implies that the character is a dash.
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;
}
// Remove all the dashes that would have
// accumulated in the beginning of the string.
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));
}
}
function ReFormatString(S, K) {
let len = S.length;
let cnt = 0;
let x = 0;
// Move the characters to the
// back of the string.
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("");
}
}
// Calculate total number of
// alphanumeric characters so
// as to get the number of dashes
// in the final string.
let slen = len - x;
let step = slen / K;
// Remove x characters from the
// start of the string
S = S.split('').reverse().join('');
let val = x;
while (val--) {
S = S.substring(0, S.length - 1);
}
// Push the empty spaces in
// the string (slen+step) to get
// the final string length
let temp = step;
while (temp--)
S += ' ';
S = S.split('').reverse().join('');
len = S.length;
// Using simple mathematics
// to push the elements
// in the string at the correct place.
let i = slen, j = step, f = 0;
while (j < len) {
// At every step calculate the
// number of dashes that would be
// present before the character
step = Math.floor(i / K);
if (f == 1)
step--;
let rem = i % K;
// If the remainder is zero it
// implies that the character is a dash.
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;
}
// Remove all the dashes that would have
// accumulated in the beginning of the string.
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));
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