String – 1


helloName
Given a string name, e.g. “Bob”, return a greeting of the form “Hello Bob!”.
helloName(“Bob”) → “Hello Bob!”
helloName(“Alice”) → “Hello Alice!”
helloName(“X”) → “Hello X!”

public String helloName(String name) {
        return "Hello " + name + "!";
    }

makeAbba
Given two strings, a and b, return the result of putting them together in the order abba, e.g. “Hi” and “Bye” returns “HiByeByeHi”.
makeAbba(“Hi”, “Bye”) → “HiByeByeHi”
makeAbba(“Yo”, “Alice”) → “YoAliceAliceYo”
makeAbba(“What”, “Up”) → “WhatUpUpWhat”

public String makeAbba(String a, String b) {
        return a + b + b + a;
    }

makeTags
The web is built with HTML strings like “<i>Yay</i>” which draws Yay as italic text. In this example, the “i” tag makes <i> and </i> which surround the word “Yay”. Given tag and word strings, create the HTML string with tags around the word, e.g. “<i>Yay</i>”.
makeTags(“i”, “Yay”) → “<i>Yay</i>”
makeTags(“i”, “Hello”) → “<i>Hello</i>”
makeTags(“cite”, “Yay”) → “<cite>Yay</cite>”

public String makeTags(String tag, String word) {
        return "<" + tag + ">" + word + "</" + tag + ">";
    }

makeOutWord
Given an “out” string length 4, such as “<<>>”, and a word, return a new string where the word is in the middle of the out string, e.g. “<<word>>”. Note: use str.substring(i, j) to extract the String starting at index i and going up to but not including index j.
makeOutWord(“<<>>”, “Yay”) → “<<Yay>>”
makeOutWord(“<<>>”, “WooHoo”) → “<<WooHoo>>”
makeOutWord(“[[]]”, “word”) → “[[word]]”

public String makeOutWord(String out, String word) {
        return out.substring(0, 2) + word + out.substring(2, 4);
    }

extraEnd
Given a string, return a new string made of 3 copies of the last 2 chars of the original string. The string length will be at least 2.
extraEnd(“Hello”) → “lololo”
extraEnd(“ab”) → “ababab”
extraEnd(“Hi”) → “HiHiHi”

public String extraEnd(String str) {
    String end = str.substring(str.length()-2);
    return  end+end+end;
}

firstTwo
Given a string, return the string made of its first two chars, so the String “Hello” yields “He”. If the string is shorter than length 2, return whatever there is, so “X” yields “X”, and the empty string “” yields the empty string “”. Note that str.length() returns the length of a string.
firstTwo(“Hello”) → “He”
firstTwo(“abcdefg”) → “ab”
firstTwo(“ab”) → “ab”

public String firstTwo(String str) {
        if (str.length() >= 2) {
            String front = str.substring(0, 2);
            return front;
        }
        return str;
    }

firstHalf
Given a string of even length, return the first half. So the string “WooHoo” yields “Woo”.
firstHalf(“WooHoo”) → “Woo”
firstHalf(“HelloThere”) → “Hello”
firstHalf(“abcdef”) → “abc”

public String firstHalf(String str) {
        return str.substring(0, str.length() / 2);
    }

withoutEnd
Given a string, return a version without the first and last char, so “Hello” yields “ell”. The string length will be at least 2.
withoutEnd(“Hello”) → “ell”
withoutEnd(“java”) → “av”
withoutEnd(“coding”) → “odin”

public String withoutEnd(String str) {
       return str.substring(1, str.length() - 1);
   }

comboString
Given 2 strings, a and b, return a string of the form short+long+short, with the shorter string on the outside and the longer string on the inside. The strings will not be the same length, but they may be empty (length 0).
comboString(“Hello”, “hi”) → “hiHellohi”
comboString(“hi”, “Hello”) → “hiHellohi”
comboString(“aaa”, “b”) → “baaab”

public String comboString(String a, String b) {
    if (a.length() < b.length()) {
        return a + b + a;
    }
    return b + a + b;
}

nonStart
Given 2 strings, return their concatenation, except omit the first char of each. The strings will be at least length 1.
nonStart(“Hello”, “There”) → “ellohere”
nonStart(“java”, “code”) → “avaode”
nonStart(“shotl”, “java”) → “hotlava”

public String nonStart(String a, String b) {
       return a.substring(1) + b.substring(1);
   }

left2
Given a string, return a “rotated left 2” version where the first 2 chars are moved to the end. The string length will be at least 2.
left2(“Hello”) → “lloHe”
left2(“java”) → “vaja”
left2(“Hi”) → “Hi”

public String left2(String str) {
        return str.substring(2, str.length()) + str.substring(0, 2);
    }

right2
Given a string, return a “rotated right 2” version where the last 2 chars are moved to the start. The string length will be at least 2.
right2(“Hello”) → “loHel”
right2(“java”) → “vaja”
right2(“Hi”) → “Hi”

public String right2(String str) {
        return str.substring(str.length() - 2) + str.substring(0, str.length() - 2);
    }

theEnd
Given a string, return a string length 1 from its front, unless front is false, in which case return a string length 1 from its back. The string will be non-empty.
theEnd(“Hello”, true) → “H”
theEnd(“Hello”, false) → “o”
theEnd(“oh”, true) → “o”

public String theEnd(String str, boolean front) {
       if (front == true) {
           return str.substring(0, 1);
       }
       return str.substring(str.length() - 1);
   }

withouEnd2
Given a string, return a version without both the first and last char of the string. The string may be any length, including 0.
withouEnd2(“Hello”) → “ell”
withouEnd2(“abc”) → “b”
withouEnd2(“ab”) → “”

public String withouEnd2(String str) {
        if (str.length() > 2) {
            return str.substring(1, str.length() - 1);
        }
        return "";
    }

middleTwo
Given a string of even length, return a string made of the middle two chars, so the string “string” yields “ri”. The string length will be at least 2.
middleTwo(“string”) → “ri”
middleTwo(“code”) → “od”
middleTwo(“Practice”) → “ct”

public String middleTwo(String str) {
       int len = str.length();
       return str.substring(len / 2 - 1, len / 2 + 1);
   }

endsLy
Given a string, return true if it ends in “ly”.
endsLy(“oddly”) → true
endsLy(“y”) → false
endsLy(“oddy”) → false

public boolean endsLy(String str) {
       if (str.length() >= 2 && str.substring(str.length() - 2).equals("ly")) {
           return true;
       }
       return false;
   }

nTwice
Given a string and an int n, return a string made of the first and last n chars from the string. The string length will be at least n.
nTwice(“Hello”, 2) → “Helo”
nTwice(“Chocolate”, 3) → “Choate”
nTwice(“Chocolate”, 1) → “Ce”

public String nTwice(String str, int n) {
       return str.substring(0, n) + str.substring(str.length() - n);
   }

twoChar
Given a string and an index, return a string length 2 starting at the given index. If the index is too big or too small to define a string length 2, use the first 2 chars. The string length will be at least 2.
twoChar(“java”, 0) → “ja”
twoChar(“java”, 2) → “va”
twoChar(“java”, 3) → “ja”

public String twoChar(String str, int index) {
     if (str.length() >= index + 2 && index > 0) {
         return str.substring(index, index + 2);
     }
     return str.substring(0, 2);
 }

middleThree
Given a string of odd length, return the string length 3 from its middle, so “Candy” yields “and”. The string length will be at least 3.
middleThree(“Candy”) → “and”
middleThree(“and”) → “and”
middleThree(“solving”) → “lvi”

public String middleThree(String str) {
    int len = str.length();
    return str.substring(len/2-1, len/2+2);
}

hasBad
Given a string, return true if “bad” appears starting at index 0 or 1 in the string, such as with “badxxx” or “xbadxx” but not “xxbadxx”. The string may be any length, including 0. Note: use .equals() to compare 2 strings.
hasBad(“badxx”) → true
hasBad(“xbadxx”) → true
hasBad(“xxbadxx”) → false

public boolean hasBad(String str) {
    if( str.indexOf("bad") !=-1 && str.indexOf("bad") <=1 ){
        return true;
    }
    return false;
}

atFirst
Given a string, return a string length 2 made of its first 2 chars. If the string length is less than 2, use ‘@’ for the missing chars.
atFirst(“hello”) → “he”
atFirst(“hi”) → “hi”
atFirst(“h”) → “h@”

public String atFirst(String str) {
    if(str.length()>=2){
        return str.substring(0, 2);
    }else if(str.length() == 1){
        return str+"@";
    }
    return "@"+"@";
}

lastChars
Given 2 strings, a and b, return a new string made of the first char of a and the last char of b, so “yo” and “java” yields “ya”. If either string is length 0, use ‘@’ for its missing char.
lastChars(“last”, “chars”) → “ls”
lastChars(“yo”, “java”) → “ya”
lastChars(“hi”, “”) → “h@”

public String lastChars(String a, String b) {
       if (a.length() >= 1 && b.length() >= 1) {
           return a.substring(0, 1) + b.substring(b.length() - 1);
       } else if (a.length() == 0 && b.length() >= 1) {
           return "@" + b.substring(b.length() - 1);
       } else if (b.length() == 0 && a.length() >= 1) {
           return a.substring(0, 1) + "@";
       } else if (a.length() == 0 && b.length() == 0) {
           return "@" + "@";
       }
       return "";
   }

conCat
Given two strings, append them together (known as “concatenation”) and return the result. However, if the concatenation creates a double-char, then omit one of the chars, so “abc” and “cat” yields “abcat”.
conCat(“abc”, “cat”) → “abcat”
conCat(“dog”, “cat”) → “dogcat”
conCat(“abc”, “”) → “abc”

public String conCat(String a, String b) {
        int lenA = a.length();
        int lenB = b.length();
        if (lenA > 0 && lenB > 0 && a.charAt(lenA - 1) == b.charAt(0)) {
            return a + b.substring(1, b.length());
        }
        return a + b;
    }

lastTwo
Given a string of any length, return a new string where the last 2 chars, if present, are swapped, so “coding” yields “codign”.
lastTwo(“coding”) → “codign”
lastTwo(“cat”) → “cta”
lastTwo(“ab”) → “ba”

public String lastTwo(String str) {
       int len = str.length();
       if (len >= 2) {
           return str.substring(0, len - 2) + str.substring(len - 1) + str.substring(len - 2, len - 1);
       }
       return str;
   }

seeColor
Given a string, if the string begins with “red” or “blue” return that color string, otherwise return the empty string.
seeColor(“redxx”) → “red”
seeColor(“xxred”) → “”
seeColor(“blueTimes”) → “blue”

public String seeColor(String str) {
       if (str.length() >= 3 && str.substring(0, 3).equals("red")) {
           return str.substring(0, 3);
       } else if (str.length() >= 4 && str.substring(0, 4).equals("blue")) {
           return str.substring(0, 4);
       }
       return "";
   }

frontAgain
Given a string, return true if the first 2 chars in the string also appear at the end of the string, such as with “edited”.
frontAgain(“edited”) → true
frontAgain(“edit”) → false
frontAgain(“ed”) → true

public boolean frontAgain(String str) {
      int len = str.length();
      if(len>=2 && str.substring(0,2).equals(str.substring(len-2)) ){
          return true;
      }
      return false;
  }

minCat
Given two strings, append them together (known as “concatenation”) and return the result. However, if the strings are different lengths, omit chars from the longer string so it is the same length as the shorter string. So “Hello” and “Hi” yield “loHi”. The strings may be any length.
minCat(“Hello”, “Hi”) → “loHi”
minCat(“Hello”, “java”) → “ellojava”
minCat(“java”, “Hello”) → “javaello”

public String minCat(String a, String b) {
       int lenA = a.length();
       int lenB = b.length();
       if (lenA > lenB) {
           return a.substring(lenA - lenB, lenA) + b;
       }
       return a + b.substring(lenB - lenA, lenB);
   }

extraFront
Given a string, return a new string made of 3 copies of the first 2 chars of the original string. The string may be any length. If there are fewer than 2 chars, use whatever is there.
extraFront(“Hello”) → “HeHeHe”
extraFront(“ab”) → “ababab”
extraFront(“H”) → “HHH”

public String extraFront(String str) {
       if (str.length() >= 2) {
           String front = str.substring(0, 2);
           return front + front + front;
       }
       return str + str + str;
   }

without2
Given a string, if a length 2 substring appears at both its beginning and end, return a string without the substring at the beginning, so “HelloHe” yields “lloHe”. The substring may overlap with itself, so “Hi” yields “”. Otherwise, return the original string unchanged.
without2(“HelloHe”) → “lloHe”
without2(“HelloHi”) → “HelloHi”
without2(“Hi”) → “”

public String without2(String str) {
       if (str.length() >= 2 && str.substring(0, 2).equals(str.substring(str.length() - 2))) {
           return str.substring(2, str.length());
       }
       return str;
   }

deFront
Given a string, return a version without the first 2 chars. Except keep the first char if it is ‘a’ and keep the second char if it is ‘b’. The string may be any length. Harder than it looks.
deFront(“Hello”) → “llo”
deFront(“java”) → “va”
deFront(“away”) → “aay”

public String deFront(String str) {
       if ((str.length() >= 2 && str.charAt(0) != 'a' && str.charAt(1) != 'b')) {
           return str.substring(2);
       } else if (str.length() >= 2 && str.charAt(1) != 'b' && str.charAt(0) == 'a') {
           return str.substring(0, 1) + str.substring(2);
       } else if (str.length() >= 2 && str.charAt(0) != 'a' && str.charAt(1) == 'b') {
           return str.substring(1);
       }
       return str;
   }

startWord
Given a string and a second “word” string, we’ll say that the word matches the string if it appears at the front of the string, except its first char does not need to match exactly. On a match, return the front of the string, or otherwise return the empty string. So, so with the string “hippo” the word “hi” returns “hi” and “xip” returns “hip”. The word will be at least length 1.
startWord(“hippo”, “hi”) → “hi”
startWord(“hippo”, “xip”) → “hip”
startWord(“hippo”, “i”) → “h”

public String startWord(String str, String word) {
        if (str.length() >= word.length() && str.substring(1, word.length()).equals(word.substring(1, word.length()))) {
            return str.substring(0, word.length());
        }
        return "";
    }

withoutX
Given a string, if the first or last chars are ‘x’, return the string without those ‘x’ chars, and otherwise return the string unchanged.
withoutX(“xHix”) → “Hi”
withoutX(“xHi”) → “Hi”
withoutX(“Hxix”) → “Hxi”

public String withoutX(String str) {
        if (str.length() > 0 && str.charAt(0) == 'x') {
            str = str.substring(1);
        }
        if (str.length() > 0 && str.charAt(str.length() - 1) == 'x') {
            str = str.substring(0, str.length() - 1);
        }
        return str;
    }

withoutX2
Given a string, if one or both of the first 2 chars is ‘x’, return the string without those ‘x’ chars, and otherwise return the string unchanged. This is a little harder than it looks.
withoutX2(“xHi”) → “Hi”
withoutX2(“Hxi”) → “Hi”
withoutX2(“Hi”) → “Hi”

public String withoutX2(String str) {
       if (str.length() > 0 && str.charAt(0) == 'x') {
           str = str.substring(1);
           if (str.length() > 0 && str.charAt(0) == 'x') {
               str = str.substring(1);
           }
       } else if (str.length() > 0 && str.charAt(1) == 'x') {
           str = str.substring(0, 1) + str.substring(2);
       }
       return str;
   }

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