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)); Similar Reads
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