Append and Delete of String of hackerrank

Question

/* why the two test cases are not passed by this code*/ /*the link of the problem is https://www.hackerrank.com/challenges/append-and-delete/problem */

static String appendAndDelete(String s, String t, int k) {

    if (s.length() + t.length() < k)
        return "Yes";

    int commonlength = 0;
    for (int i = 0; i < Math.min(s.length(), t.length()); i++) {

        if (t.charAt(i) == s.charAt(i))
            commonlength++;
        else
            break;
    }

    if ((k - s.length() - t.length() + 2 * commonlength) % 2 == 0) {
        return "Yes";
    }

    return "No";

}

Answer 1

solution of the problem in golang

package main

import (
    "bufio"
    "fmt"
    "io"
    "math"
    "os"
    "strconv"
    "strings"
)

func appendAndDelete(s string, t string, k int32) string {
    // Write your code here
    yes := "Yes"
    no := "No"
    if len(s)+len(t) <= int(k) {
        return yes
    }
    min := math.Min(float64(len(s)), float64(len(t)))
    length := 0
    for i := 0; i < int(min); i++ {
        if s[i] == t[i] {
            length++
        } else {
            break
        }
    }

    total := (len(s) - length) + (len(t) - length)

    if total <= int(k) && (total-int(k))%2 == 0 {
        return yes
    }
    return no
}

func main() {
    reader := bufio.NewReaderSize(os.Stdin, 16*1024*1024)

    stdout, err := os.Create(os.Getenv("OUTPUT_PATH"))
    checkError(err)

    defer stdout.Close()

    writer := bufio.NewWriterSize(stdout, 16*1024*1024)

    s := readLine(reader)

    t := readLine(reader)

    kTemp, err := strconv.ParseInt(strings.TrimSpace(readLine(reader)), 10, 64)
    checkError(err)
    k := int32(kTemp)

    result := appendAndDelete(s, t, k)

    fmt.Fprintf(writer, "%s\n", result)

    writer.Flush()
}

func readLine(reader *bufio.Reader) string {
    str, _, err := reader.ReadLine()
    if err == io.EOF {
        return ""
    }

    return strings.TrimRight(string(str), "\r\n")
}

func checkError(err error) {
    if err != nil {
        panic(err)
    }
}

Answer 2

A bit long but passes all the test cases. Runtime complexity O(N)

public static String appendAndDelete(String s, String t, int k) {
    // Write your code here
    int sl = s.length();
    int tl = t.length();
    int min_1 = 0, min_2 = 0, min_3 = 0;
    int counter = 0;
    String res = "No";
    if (sl == tl) {
        for (int i = sl - 1; i >= 0; i--) {
            counter++;
            if (s.charAt(i) != t.charAt(i)) {
                min_1 = counter * 2;
            }
        }
        if (min_1 == 0) {
            min_1 = 2;
        }
        min_2 = sl * 2;
        min_3 = (sl * 2) + 1;
        if (min_1 % 2 == 0) {
            if ((k >= min_1 && k <= min_2) && (k % 2 == 0)) {
                res = "Yes";
            }else if (k >= min_3) {
                res = "Yes";
            }
        }
    }else if (sl > tl) {
        min_1 = sl - tl;
        int dif_1 = 0;
        for (int i = (tl - 1); i >= 0; i--) {
            counter++;
            if (s.charAt(i) != t.charAt(i)) {
                dif_1 = counter * 2;
            }
        }
        min_1 += dif_1;
        min_2 = ((sl - (sl - tl)) * 2) + (sl - tl);
        min_3 = ((sl - (sl - tl)) * 2 + 1) + (sl - tl);
        if (min_1 % 2 == 0) {
            if ((k >= min_1 && k <= min_2) && (k % 2) == 0) {
                res = "Yes";
            }else if (k >= min_3) {
                res = "Yes";
            }
        }else{
            if((k >= min_1 && k <= min_2) && (k % 2) == 1) {
                res = "Yes";
            }else if (k >= min_3) {
                res = "Yes";
            }
        }
    }else if (tl > sl) {
        min_1 = tl - sl;
        int dif_1 = 0;
        for (int i = (sl - 1); i >= 0; i--) {
            counter++;
            if (s.charAt(i) != t.charAt(i)) {
                dif_1 = counter * 2;
            }
        }
        min_1 += dif_1;
        min_2 = ((tl - (tl - sl)) * 2) + (tl - sl);
        min_3 = ((tl - (tl - sl)) * 2 + 1) + (tl - sl);
        if (min_1 % 2 == 0) {
            if ((k >= min_1 && k <= min_2) && (k % 2) == 0) {
                res = "Yes";
            }else if (k >= min_3) {
                res = "Yes";
            }
        }else if ((k >= min_1 && k <= min_2) && (k % 2) == 1) {
            res = "Yes";
        }else if (k >= min_3) {
            res = "Yes";
        }
    }
    return res;
    }

Answer 3

Here is my solution for the problem where all the test cases are passed.

public static string appendAndDelete(string s, string t, int k)
{
            char[] sArray = s.ToCharArray();
            char[] tArray = t.ToCharArray();
            var commonLength = 0;
            var result = "";
             if (s.Length < t.Length)
            {
                 var firstChar = s[0];
                var count = t.Count(x => x == firstChar);
                if (count==t.Length)
                {
                    result = "Yes";
                }
                else
                    result = "No";
            }
            else if (string.Compare(s, t) == 0)
            {
                
                    result="Yes";
                
            }
            else
            {
                for (int i = 0; i < tArray.Length; i++)
                {
                    if (sArray[i] == tArray[i])
                    {
                        commonLength++;
                    }
                    else { break; }
                }
                var totalSubRequired = (s.Length - commonLength) + (t.Length - commonLength);
                if (k >= totalSubRequired)
                {
                    if (string.Compare(s, t) == 0)
                    {
                        result = "Yes";
                    }
                    else
                    {
                        var commonString = s.Substring(0, commonLength);
                        var attachString = t.Substring(commonLength, t.Length - commonLength);
                        var combineString = string.Concat(commonString, attachString);
                        if (string.Compare(t, combineString) == 0)
                        {
                            result = "Yes";
                        }
                        else { result = "No"; ; }
                    }
                }
                else { result = "No"; ; }
            }
    return result;
}

Answer 4

You need to add one more condition in your code, because below condition is not enough:

(k - s.length() - t.length() + 2 * commonlength) % 2 == 0

Try this:

int balance = k - s.length() - t.length() + 2 * commonlength;

if (balance >= 0 && (balance) % 2 == 0) {
    return "Yes";
}

you need to have one more addition condition : balance >= 0 as mentioned above.

Here is a working solution that passes all test cases, added comments in code for clear understanding:

static String appendAndDelete(String s, String t, int k) {
    // Check if k is greater or equal to both the lengths
    if (s.length() + t.length() <= k)
        return "Yes";

    int commonlength = 0;
    // Get the common matching character length
    for (int i = 0; i < Math.min(s.length(), t.length()); i++) {
        if (t.charAt(i) == s.charAt(i))
            commonlength++;
        else {
            break;
        }
    }

    // count how many modifications still needed
    int balance = s.length() - commonlength;
    balance += t.length() - commonlength;

    // Check if k is greater than balance count
    if (balance <= k) {
        // Special case, we need to perform exactly k operations
        // so if balance is odd then k should be odd, if balance is even
        // then k must be even.
        if ((balance - k) % 2 == 0) {
            return "Yes";
        }
    }
    return "No";
}

Answer 5

That's pretty straight forward. Here is the solution that pass all of the mentioned test case:

static String appendAndDelete(String s, String t, int k) {

    if (s.equals(t))
        return (k >= s.length() * 2 || k % 2 == 0) ? "Yes" : "No";

    int commonlength = 0;

    for (int i = 0; i < Math.min(s.length(), t.length()); i++) {
        if (t.charAt(i) != s.charAt(i))
            break;
        commonlength++;
    }

    int cs = s.length() - commonlength;
    int ct = t.length() - commonlength;
    int tot = cs + ct;

    return ((tot == k) || (tot < k && (tot - k) % 2 == 0) || (tot + (2 * commonlength) <= k)) ? "Yes" : "No";

}