Our task is to mirror characters from the N-th position up to the end of a given string, where ‘a’ will be converted into ‘z’, ‘b’ into ‘y’, and so on. This JavaScript problem requires mirroring the characters in a string starting from a specified position. There are various approaches available to accomplish this, such as utilizing loops and maps. In this article, we will explore various methods to find the mirror characters of a string.

Examples:

Input : N = 3
        paradox
Output : paizwlc
We mirror characters from position 3 to end.
Input : N = 6
        pneumonia
Output : pnefnlmrz

Method 1: Creating a String:

In this method, we maintain a string (or a character array) containing the English lowercase alphabet. Starting from the pivot point up to the length of the string, we retrieve the reversed alphabetical counterpart of a character by using its ASCII value as an index. By applying this approach, we convert the given string into the desired mirrored form.

Syntax:

let modifiedString = "";
  
for (let i = 0; i < startPosition; i++)
    modifiedString = modifiedString + inputString[i];
  
for (let i = startPosition; i < inputStringLength; i++)
    modifiedString = modifiedString + reverseAlphabet[inputString[i].charCodeAt() - 'a'.charCodeAt()];

Example: In the code we will implement above approach by creating a string.

function reverseAlphabetFromPosition(
        inputString, startPosition) {
    let reverseAlphabet = 
        "zyxwvutsrqponmlkjihgfedcba";
    let inputStringLength = 
        inputString.length;
    let newString = "";

    for (let i = 0; i < startPosition; i++)
        newString += inputString[i];

    for (let i = startPosition; i < inputStringLength; i++)
        newString += reverseAlphabet[inputString[i].
                    charCodeAt() - 'a'.charCodeAt()];

    return newString;
}

let givenString = "horje";
let startingPosition = 5;
console.log(
    reverseAlphabetFromPosition(
        givenString, startingPosition - 1));

geekhulitvvph

Time Complexity: O(n)

Space Complexity: O(n)

Method 2: Using a for loop

• Initialize an empty string to store the mirrored result.
• Iterate through the characters of the given string using a loop.
• When the loop index is equal to or greater than the starting position, replace the current character with its mirror (e.g., ‘a’ becomes ‘z’ and ‘b’ becomes ‘y’).
• Append the mirrored character to the result string.

Syntax:

for (let i = 0; i < givenString.length; i++) {
// condition
}

Example: In the code we will implement above approach by using single for loop.

let givenString = "horje";
let startingPosition = 5;
let mirroredString = '';
for (let i = 0; i < givenString.length; i++) {
    if (i >= startingPosition - 1) {
        mirroredString += 
        String.fromCharCode(219 - givenString.charCodeAt(i));
    } else {
        mirroredString += givenString[i];
    }
}
console.log(mirroredString);

geekhulitvvph

Time Complexity: O(n)

Space Complexity: O(n)

Method 3: Using Custom Mapping

• Create a character mapping to represent the mirror transformations.
• Initialize an empty string for the mirrored result.
• Iterate through the characters of the given string.
• If the character is in the mapping, replace it with the mapped character; otherwise, keep it as is.

Syntax:

const mirrorMap = {
    'a': 'z', 'b': 'y', 'c': 'x', 'd': 'w', 'e': 'v',
    'f': 'u', 'g': 't', 'h': 's', 'i': 'r', 'j': 'q',
    'k': 'p', 'l': 'o', 'm': 'n', 'n': 'm', 'o': 'l',
    'p': 'k', 'q': 'j', 'r': 'i', 's': 'h', 't': 'g',
    'u': 'f', 'v': 'e', 'w': 'd', 'x': 'c', 'y': 'b',
    'z': 'a'
};

Example: In the code we will implement above approach by using map.

let givenString = "horje";
let startingPosition = 5;
const mirrorMap = {
    'a': 'z', 'b': 'y', 'c': 'x', 'd': 'w', 'e': 'v',
    'f': 'u', 'g': 't', 'h': 's', 'i': 'r', 'j': 'q',
    'k': 'p', 'l': 'o', 'm': 'n', 'n': 'm', 'o': 'l',
    'p': 'k', 'q': 'j', 'r': 'i', 's': 'h', 't': 'g',
    'u': 'f', 'v': 'e', 'w': 'd', 'x': 'c', 'y': 'b',
    'z': 'a'
};

let mirroredString = '';
for (let char of givenString) {
    startingPosition--;
    if (startingPosition > 0) {
        mirroredString += char;
    }
    else mirroredString += mirrorMap[char] || char;
}
console.log(mirroredString);

geekhulitvvph

Time Complexity: O(n)

Space Complexity: O(n)

Method 4: Using Array.reduce() Method

Using the Array.reduce() method to mirror a string involves converting the string into an array of characters, then applying reduce to accumulate the characters in reverse order, effectively reversing the string.

Example: In this example The function mirrorString uses split(”) and reduce() to reverse a string by accumulating characters in reverse order.

function mirrorString(str) {
    return str.split('').reduce((reversed, char) => char + reversed, '');
}

console.log(mirrorString("Hello, World!"));

!dlroW ,olleH

Method 5: Using Array.map() Method

In this method, we convert the string into an array of characters using split(”). Then, we use the map() method to iterate through each character. For characters after the specified starting position, we calculate their mirrored counterparts based on their ASCII values. Finally, we join the characters back into a string.

Example:

function mirrorStringFromPosition(inputString, startPosition) {
    const startIdx = startPosition - 1;
    const mirroredString = inputString
        .split('')
        .map((char, index) => {
            if (index >= startIdx) {
                // Calculate mirrored character based on ASCII values
                const mirroredCharCode = 219 - char.charCodeAt();
                return String.fromCharCode(mirroredCharCode);
            }
            return char;
        })
        .join('');

    return mirroredString;
}

const givenString = "horje";
const startingPosition = 5;
console.log(mirrorStringFromPosition(givenString, startingPosition));

geekhulitvvph

Time Complexity: O(n)

Space Complexity: O(n)

Method 6: Using Alphabet Calculation

In this approach we calculates the mirrored character by subtracting the character’s position in the alphabet from the position of the last letter (‘z’ or ‘Z’).

Example:

function mirrorCharacters(str) {
    const mirror = char => {
        if (char >= 'a' && char <= 'z') {
            return String.fromCharCode('a'.charCodeAt(0) + 
            ('z'.charCodeAt(0) - char.charCodeAt(0)));
        }
        if (char >= 'A' && char <= 'Z') {
            return String.fromCharCode('A'.charCodeAt(0) + 
            ('Z'.charCodeAt(0) - char.charCodeAt(0)));
        }
        return char;
    };

    return str.split('').map(mirror).join('');
}


const input = "horje";
const result = mirrorCharacters(input);
console.log(result);

tvvphulitvvph

Method 7: Using ASCII Value Manipulation

In this approach, we directly manipulate the ASCII values of the characters to find their mirrored counterparts. This method involves calculating the mirrored character by subtracting the character’s ASCII value from the ASCII value of ‘z’ and adding the ASCII value of ‘a’. This approach ensures constant time complexity for each character transformation.

Example: This example demonstrates the use of the ASCII value manipulation approach to mirror characters in a string from a specified position.

function mirrorStringFromN(inputString, N) {
    let result = '';

    for (let i = 0; i < inputString.length; i++) {
        if (i < N) {
            result += inputString[i];
        } else {
            let mirroredChar = String.fromCharCode('a'.charCodeAt(0) + ('z'.charCodeAt(0) - inputString[i].charCodeAt(0)));
            result += mirroredChar;
        }
    }

    return result;
}

// Test cases
console.log(mirrorStringFromN('paradox', 3)); // Output: paizwlc
console.log(mirrorStringFromN('pneumonia', 6)); // Output: pnefnlmrz

parzwlc
pneumomrz