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>// Function to compute the answerchar* ReFormatString(char* S, int K){ // Create a temporary string to store // the alphanumeric characters only 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]); } } // String ans is created to store // the final string. 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;}// Driver codeint main(){ char s[] = "5F3Z-2e-9-w"; int K = 4; // Function Call printf("%s", ReFormatString(s, K)); return 0;} |
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;// Function to compute the answerstring ReFormatString(string S, int K){ // Create a temporary string to store // the alphanumeric characters only 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 is created to store // the final string. 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;}// Driver codeint main(){ string s = "5F3Z-2e-9-w"; int K = 4; // Function Call cout << ReFormatString(s, K); return 0;} |
Java
class GFG { // Function to compute the answer public static String ReFormatString(String S, int K) { // Create a temporary String to store // the alphanumeric characters only 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 is created to store // the final String. 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; // Function Call System.out.println(ReFormatString(s, K)); }}// This code is contributed by Saurabh Jaiswal |
Python3
# Python code for the above approach# Function to compute the answerdef ReFormatStrings(s,k): # Create a temporary string to store the alphanumeric characters only temp = "" n = len(s) for i in range(0,n): if(s[i] != '-'): temp += s[i].upper() length = len(temp) # String ans is created to store the final string. 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 codeif __name__ == "__main__": s = "5F3Z-2e-9-w" k = 4 # Function call print(ReFormatStrings(s,k)) # This code is contributed by ajaymakvana |
C#
using System;public class GFG{ // Function to compute the answer public static string ReFormatString(string S, int K) { // Create a temporary string to store // the alphanumeric characters only 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 is created to store // the final string. 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; // Function Call Console.WriteLine(ReFormatString(s, K)); }}// This code is contributed by akashish__ |
Javascript
// JS code for the above approachfunction reverse(s){ let splitString = s.split(""); let reverseArray = splitString.reverse(); let joinArray = reverseArray.join(""); return joinArray;}// Function to compute the answerfunction ReFormatString(S,K){ // Create a temporary string let to store // the alphanumeric characters only 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 is created to store // the final string. 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;}// Driver codelet s = "5F3Z-2e-9-w";let K = 4;// Function Callconsole.log(ReFormatString(s, K));// This code is contributed by akashish__. |
Output
5F3Z-2E9W
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++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;// Function to compute the answerstring 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;}// Driver codeint main(){ string s = "5F3Z-2e-9-w"; int K = 4; // Function Call cout << ReFormatString(s, K); return 0;} |
Java
/*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); // extracts each character nstr = ch + nstr; // adds each character in // front of the existing string } return nstr; } // Function to compute the answer 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; // Function Call System.out.println(ReFormatString(s, K)); }}// This code is contributed by akashish__ |
Python3
# Python3 code for the above approach# Function to reverse a stringdef reverse(string): string = string[::-1] return string# Function to compute the answerdef ReFormatString( S, K): length = len(S) cnt = 0 x = 0 # Move the characters to the # back of the string. 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# Driver codes = "5F3Z-2e-9-w"K = 4# Function Callprint(ReFormatString(s, K))# This code is contributed by akashish__ |
C#
using System;using System.Collections.Generic;// C# code for the above approachpublic class GFG { public static String reverseS(String str) { String nstr = ""; for (int i = 0; i < str.Length; i++) { char ch = str[i]; // extracts each character nstr = ch + nstr; // adds each character in // front of the existing string } return nstr; } // Function to compute the answer public static String ReFormatString(String S, int K) { int len = S.Length; int cnt = 0; 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; } // Driver code static public void Main() { string s = "5F3Z-2e-9-w"; int K = 4; // Function Call Console.WriteLine(ReFormatString(s, K)); }}// This code is contributed by akashish__ |
Javascript
// JS code for the above approach// Function to compute the answerfunction 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;}// Driver codelet s = "5F3Z-2e-9-w";let K = 4;// Function Callconsole.log(ReFormatString(s, K));// This code is contributed by akashish__ |
Output
5F3Z-2E9W
Time Complexity: O(N)
Auxiliary Space: O(1)
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