Given a string S containing letters and digits, and an integer array Shift where, 
and for each element of Shift array ![0 \leq Shift[i] \leq 10^{9}](https://geeksforgeeks.org/wp-content/uploads/2023/10/wardslaus_quicklatex.com-eb9e5d7d710aec172109c7c7771aad61_l3.png "Rendered by QuickLaTeX.com")
. The task is, for each Shift[i] = X, you have to shift the first i+1 letters of S, X times. Return the final string after all applying all such shift to S.
Note: Shift means cyclically increment ASCII value.
Examples:
Input: S = “abc789”, Shift = [2, 5, 9]
Output: “qpl706”
Explanation: Starting with “abc”.
After shifting the first 1 letters of S by 2, we have “cbc”.
After shifting the first 2 letters of S by 5, we have “hgc”.
After shifting the first 3 letters of S by 9, we have “qpl”.
Input : S = “neveropen”, Shift[] = [ 11, 10000, 9999999 ]
Output : qdnyulaufkuug
Approach: The i-th character of S is shifted Shift[i] + Shift[i+1] + … + Shift[Shift.length – 1] times.
So we update the Shift array backward to know the exact number of shifts to be applied to each element of string S.
Now,
- Traverse the given text (S) one character at a time.
- For each character, transform the given character as per the rule, i:e apply shift, Shift[i] times.
- Return the new string generated.
Below is the implementation of the above approach:
C++
// CPP implementation of above approach#include <bits/stdc++.h>using namespace std;// Function to find S after shifting each letterstring shift_S(string S, int Shift[], int n){ // update shift array for each element for (int i = n - 2; i >= 0; --i) Shift[i] += Shift[i + 1]; // finding the new shifted string string result = ""; // traverse S and shift letters according to shift array for (int i = 0; i < S.length(); i++) { // For upper letters if (isupper(S[i])) { result += char((int(S[i]) + Shift[i] - 'A') % 26 + 'A'); } // For lower letters else if (islower(S[i])) { result += char((int(S[i]) + Shift[i] - 'a') % 26 + 'a'); } // For digits else { result += char((int(S[i]) + Shift[i] - '0') % 10 + '0'); } } // Return the shifted string return result;}// Driver programint main(){ string S = "abc"; int Shift[] = { 2, 5, 9 }; int n = sizeof(Shift) / sizeof(Shift[0]); // function call to print required answer cout << shift_S(S, Shift, n); return 0;}// This code is written by Sanjit_Prasad |
Java
// Java implementation of the above approach public class GfG{ // Function to find S after shifting each letter public static String shift_S(String S, int Shift[], int n) { // update shift array for each element for (int i = n - 2; i >= 0; --i) Shift[i] += Shift[i + 1]; // finding the new shifted string String result = ""; // traverse S and shift letters according to shift array for (int i = 0; i < S.length(); i++) { // For upper letters if (Character.isUpperCase(S.charAt(i))) { result += (char)(((int)(S.charAt(i)) + Shift[i] - 'A') % 26 + 'A'); } // For lower letters else if (Character.isLowerCase(S.charAt(i))) { result += (char)(((int)(S.charAt(i)) + Shift[i] - 'a') % 26 + 'a'); } // For digits else { result += (char)(((int)(S.charAt(i)) + Shift[i] - '0') % 10 + '0'); } } // Return the shifted string return result; } public static void main(String []args){ String S = "abc"; int Shift[] = { 2, 5, 9 }; int n = Shift.length; // Function call to print the required answer System.out.println(shift_S(S, Shift, n)); }} // This code is contributed by Rituraj Jain |
Python3
# Python3 implementation of above approach# Function to find S after shifting# each letterdef shift_S(S, Shift, n): # update shift array for # each element for i in range(n - 2, -1, -1): Shift[i] = Shift[i] + Shift[i + 1] # finding the new shifted string result = "" # traverse S and shift letters # according to shift array for i in range(len(S)): # For upper letters if(S[i].isupper()): result = result + chr((ord(S[i]) + Shift[i] - ord('A')) % 26 + ord('A')) # For lower letters elif (S[i].islower()): result = result + chr((ord(S[i]) + Shift[i] - ord('a')) % 26 + ord('a')) # For digits else: result = result + chr((ord(S[i]) + Shift[i] - ord('0')) % 10 + ord('0')) # Return the shifted string return result # Driver CodeS = "abc"Shift = [2, 5, 9]n = len(Shift)# Function call to print the required answerprint(shift_S(S, Shift, n))# This code is contributed # by Shashank_Sharma |
C#
// C# implementation of the above approach using System;class GfG{ // Function to find S after // shifting each letter public static String shift_S(string S, int[] Shift, int n) { // Update shift array for each element for(int i = n - 2; i >= 0; --i) Shift[i] += Shift[i + 1]; // Finding the new shifted string string result = ""; // Traverse S and shift letters // according to shift array for(int i = 0; i < S.Length; i++) { // For upper letters if (Char.IsUpper(S[i])) { result += (char)(((int)(S[i]) + Shift[i] - 'A') % 26 + 'A'); } // For lower letters else if (Char.IsLower(S[i])) { result += (char)(((int)(S[i]) + Shift[i] - 'a') % 26 + 'a'); } // For digits else { result += (char)(((int)(S[i]) + Shift[i] - '0') % 10 + '0'); } } // Return the shifted string return result; } // Driver codepublic static void Main(){ string S = "abc"; int[] Shift = { 2, 5, 9 }; int n = Shift.Length; // Function call to print // the required answer Console.WriteLine(shift_S(S, Shift, n)); } }// This code is contributed by sanjoy_62 |
Javascript
<script>// JavaScript implementation of above approach// Function to find S after shifting// each letterfunction shift_S(S, Shift, n){ // update shift array for // each element for(let i = n - 2;i>=0;i--){ Shift[i] = Shift[i] + Shift[i + 1] } // finding the new shifted string let result = "" // traverse S and shift letters // according to shift array for(let i=0;i<S.length;i++){ // For upper letters if(S.charCodeAt(i) >= 65 && S.charCodeAt(i) <= 90) result = result + String.fromCharCode((S.charCodeAt(i) + Shift[i] -'A'.charCodeAt(0)) % 26 + 'A'.charCodeAt(0)) // For lower letters else if (S.charCodeAt(i) >= 97 && S.charCodeAt(i) <= 122) result = result + String.fromCharCode((S.charCodeAt(i) + Shift[i] -'a'.charCodeAt(0)) % 26 + 'a'.charCodeAt(0)) // For digits else result = result + String.fromCharCode((S.charCodeAt(i) + Shift[i] -'0'.charCodeAt(0)) % 26 + '0'.charCodeAt(0)) } // Return the shifted string return result}// Driver Codelet S = "abc"let Shift = [2, 5, 9]let n = Shift.length// Function call to print the required answerdocument.write(shift_S(S, Shift, n))// This code is contributed// by Shinjanpatra</script> |
Output:
qpl
Time Complexity: O(N), where N is the length of string S.
Auxiliary Space: O(N)
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