Shifting Letters LeetCode Solution - The Coding Shala

Home >> LeetCode >> Shifting Letters

 In this post, we will learn how to solve LeetCode's Shifting Letters Problem and will implement its solution in Java.

Shifting Letters Problem

We have a string S of lowercase letters, and an integer array shifts. Call the shift of a letter, the next letter in the alphabet, (wrapping around so that 'z' becomes 'a'). For example, shift('a') = 'b', shift('t') = 'u', and shift('z') = 'a'. Now for each shifts[i] = x, we want to shift the first i+1 letters of S, x times. Return the final string after all such shifts to S are applied.

Example 1:
Input: S = "abc", shifts = [3,5,9]
Output: "rpl"
Explanation: 
We start with "abc".
After shifting the first 1 letters of S by 3, we have "dbc".
After shifting the first 2 letters of S by 5, we have "igc".
After shifting the first 3 letters of S by 9, we have "rpl", the answer.

Practice this problem on LeetCode(Click Here).

Shifting Letters Java Solution

Approach 1

The last character of the string only shifts number shifts[length-1] times. The second last character of string shifts number shifts[length-1] + shifts[length-2] times, and like this first character shifts the sum of all the numbers in shifts array times.

Java Program: 

class Solution {
    public String shiftingLetters(String S, int[] shifts) {
        long currShift = 0;
        for(int i=0; i<shifts.length; i++) {
            currShift += shifts[i];
        }
        StringBuilder sb = new StringBuilder();
        long ch = S.charAt(0) - 97;
        if(currShift%26 != 0)
        ch = (ch + currShift) % 26;
        sb.append((char)( ch+97));
        for(int i=1; i<shifts.length; i++){
            currShift -= shifts[i-1];
            ch = S.charAt(i) - 97;
            if(currShift%26 != 0)
            ch = (ch + currShift)%26;
            sb.append((char) (ch+97));
        }
        return sb.toString();
    }
}


Other Posts You May Like
Please leave a comment below if you like this post or found some errors, it will help me to improve my content.
By

Comments

Post a Comment

Popular Posts from this Blog

Shell Script to Create a Simple Calculator - The Coding Shala

Home >> Scripting >> simple calculator Shell Script to Create a Simple Calculator Write a shell script to create a simple calculator using a switch-case statement? Code:  #! / bin / bash echo "simple calculator" sum = 0 i = "y" echo "enter first number" read n1 echo "enter second number" read n2 while [ $ i = "y" ] do echo "1.Addition" echo "2.Subtraction" echo "3.Multiplication" echo "4.Division" echo "Enter choice" read ch case $ch in 1 ) sum =$(echo " $n1 + $n2" | bc - l) echo "Addition is =" $ sum ;; 2 ) sum =$(echo "$n1 - $n2" | bc - l) echo "Sub is =" $ sum ;; 3 ) sum =$(echo "$n1 * $n2" | bc - l) echo "Mul is =" $ sum ;; 4 ) sum =$(echo "$n1 / $n2" | bc - l) echo "div is =" $ sum ;; * )echo "invalid choice" esac echo "Do y...
Read more

N-th Tribonacci Number Solution - The Coding Shala

Home >> Programming >> N-th Tribonacci Number  Hey there, welcome back to another post. In this post, we will learn how to solve the N-th Tribonacci Number problem and will implement its solution in Java. N-th Tribonacci Number Problem Statement The Tribonacci sequence Tn is defined as follows: T0 = 0, T1 = 1, T2 = 1, and Tn+3 = Tn + Tn+1 + Tn+2 for n >= 0. Given n, return the value of Tn.  Example 1:  Input: n = 4 Output: 4 Explanation: T_3 = 0 + 1 + 1 = 2 T_4 = 1 + 1 + 2 = 4 N-th Tribonacci Number Java Solution using Bottom-Up DP This problem is similar to the Fibonacci series. We can solve it using the Bottom-Up approach of dynamic programming. Time Complexity: O(n) Space Complexity: O(n) The space complexity can be reduced to O(1) by using variables to store the previous three values instead of using an array. Java Program:  class Solution { public int tribonacci( int n) { if (n == 0) return 0; ...
Read more

Java Program to Convert Binary to Decimal - The Coding Shala

Home >> Java Programs >> Convert Binary to Decimal  Hey there, welcome back to another post. In this post, we will learn how to convert Binary to Decimal in Java. Java Program to Convert Binary to Decimal You have given a binary number, convert it to a decimal number. Example:  Input: 101 Output: 5 Input: 111 Output: 7 Binary to Decimal Conversion in Java Approach We will the below formula to convert binary to decimal: decimal number = (2^0) * (rightmost digit) + (2^1) * (second rightmost digit) + ...  for example:  binary number = 110 decimal number = 2^0 * 0 + 2^1 * 1 + 2^2 * 1 = 1 * 0 + 2 * 1 + 4 * 1 = 0 + 2 + 4 = 6 Java Program:  import java.util.Scanner; /** * https://www.thecodingshala.com/ */ public class Main { public static void printDecimal( int num) { int decimal = 0; int twos = 1; // 2^0 = 1 initial value while (num > 0) { ...
Read more

LeetCode - Shuffle the Array Solution - The Coding Shala

Home >> LeetCode >> Shuffle the Array  In this post, we will learn how to solve LeetCode's Shuffle the Array problem and will implement its solution in Java. Shuffle the Array Given the array nums consisting of 2n elements in the form [x1,x2,...,xn,y1,y2,...,yn]. Return the array in the form [x1,y1,x2,y2,...,xn,yn]. Example 1: Input: nums = [2,5,1,3,4,7], n = 3 Output: [2,3,5,4,1,7]  Explanation: Since x1=2, x2=5, x3=1, y1=3, y2=4, y3=7 then the answer is [2,3,5,4,1,7]. Example 2: Input: nums = [1,2,3,4,4,3,2,1], n = 4 Output: [1,4,2,3,3,2,4,1] Practice this problem on LeetCode: Click Here. Shuffle the Array Java Solution Approach 1: Using a new Array. Space complexity: O(n). Java Program:  class Solution { public int [] shuffle ( int [] nums , int n ) { int [] ans = new int [ 2 * n ]; int j = 0 ; for ( int i = 0 ; i < 2 * n ; i = i + 2 ){ ans [ i ] = nums [ j ]; ans...
Read more

Java Program to Find GCD or HCF of Two Numbers - The Coding Shala

Home >> Java Programs >> Find GCD of Two Numbers  Hey there, welcome back to another post. In this post, we will learn how to find GCD or HCF of Two Numbers in Java. Java Program to Find GCD or HCF of Two Numbers The GCD (Greatest Common Divisor) or HCF (Highest Common Factor) of two numbers is the largest number that divides both of them. For example:  Number1: 6 (2 * 3) Number2: 9 (3 * 3) GCD or HCF of 6 and 9 is: 3 Find GCD of Two Numbers in Java using for Loop As we already know, GCD is the largest number that divides both numbers. We can start a for loop from 1 (because 1 divides all the numbers) to both numbers and if it divides both numbers then update GCD. Java Program:  import java.util.Scanner; /** * https://www.thecodingshala.com/ */ public class Main { public static void findGCD( int num1, int num2) { int gcd = 1; if (num1 == 0) { gcd = num2; } if (num2 == 0) { ...
Read more