Problem-1 Given a string and a non-negative int n, return a larger string that is n copies of the original string. Example: stringTimes(“Hi”, 2) → “HiHi” stringTimes(“Hi”, 3) → “HiHiHi” stringTimes(“Hi”, 1) → “Hi”
Solution public String stringTimes(String str, int n) { String y=””; // empty string to store the result for (int i=0;i<n; i++){ y=y+str; } return y; }
Problem-2 Given a string and a non-negative int n, we’ll say that the front of the string is the first 3 chars, or whatever is there if the string is less than length 3. Return n copies of the front; Example: frontTimes(“Chocolate”, 2) → “ChoCho” frontTimes(“Chocolate”, 3) → “ChoChoCho” frontTimes(“Abc”, 3) → “AbcAbcAbc”
Solution public String frontTimes(String str, int n) { String result=””; for(int i=0;i<n;i++){ if(str.length()<=3) result+=str; if(str.length()>3) result+=str.substring(0,3); } return result; }
Problem-3 Count the number of “xx” in the given string. We’ll say that overlapping is allowed, so “xxx” contains 2 “xx”. Example: countXX(“abcxx”) → 1 countXX(“xxx”) → 2 countXX(“xxxx”) → 3
Solution int countXX(String str) { int count=0; for(int i=0; i<str.length()-1;i++){ if(str.charAt(i)==’x’ && str.charAt(i+1)==’x’) count++; } return count; }
Problem-4 Given a string, return true if the first instance of “x” in the string is immediately followed by another “x”. Example doubleX(“axxbb”) → true doubleX(“axaxax”) → false doubleX(“xxxxx”) → true
Solution boolean doubleX(String str) { int e=str.indexOf(‘x’); // check the position of the first occurrence of ‘x’ if (e==-1) return false; // if there is no x return false if(e+1>=str.length()) return false; //if the first ‘x’ is at the end return false return (str.substring(e,e+2).equals(“xx”)); } Another Solution boolean doubleX(String str) { int e=str.indexOf(‘x’); if (e==-1) return false; if(e+1>=str.length()) return false; String f=str.substring(e,e+2); // create a substring starting from the first ‘x’ return (f.startsWith(“xx”)); // compare the substring f with “xx” }
Problem -5 Given a non-empty string like “Code” return a string like “CCoCodCode”. Example: stringSplosion(“Code”) → “CCoCodCode” stringSplosion(“abc”) → “aababc” stringSplosion(“ab”) → “aab”
Solution public String stringSplosion(String str) { String new1=””; for (int i=0;i<str.length();i++){ new1+=str.substring(0,i+1); } return new1; }
Problem -6 Given a string, return the count of the number of times that a substring length 2 appears in the string and also as the last 2 chars of the string, so “hixxxhi” yields 1 (we won’t count the end substring). Example: last2(“hixxhi”) → 1 last2(“xaxxaxaxx”) → 1 last2(“axxxaaxx”) → 2
Solution public int last2(String str) { if(str.length()<2) return 0; int count=0; for (int i=0;i<str.length()-1;i++){ if(str.charAt(i)==str.charAt(str.length()-2) && str.charAt(i+1)==str.charAt(str.length()-1)) count++; } return count-1; } Solution Two (better solution use .equals()) public int last2(String str) { if(str.length()<2) return 0; int count=0; String fin=str.substring(str.length()-2);//create a substring of the final two chars for (int i=0;i<str.length()-2;i++){ String new1=str.substring(i,i+2); if(new1.equals(fin)) count++;// compare using .equals() } return count; }
Problem-7 Given an array of ints, return the number of 9’s in the array. Example: arrayCount9({1, 2, 9}) → 1 arrayCount9({1, 9, 9}) → 2 arrayCount9({1, 9, 9, 3, 9}) → 3
Solution public int arrayCount9(int[] nums) { int count=0; for(int i=0;i<nums.length;i++){ if(nums[i]==9) count++; } return count; }
Problem-8 Given an array of ints, return true if one of the first 4 elements in the array is a 9. The array length may be less than 4. Example: arrayFront9({1, 2, 9, 3, 4}) → true arrayFront9({1, 2, 3, 4, 9}) → false arrayFront9({1, 2, 3, 4, 5}) → false
Solution public boolean arrayFront9(int[] nums) { int four=nums.length; if (four>4) four=4; // limit the array length to be four for (int i=0;i<four;i++){ if (nums[i]==9) return true; } return false; }
Problem-9 Given an array of ints, return true if .. 1, 2, 3, .. appears in the array somewhere. Example array123({1, 1, 2, 3, 1}) → true array123({1, 1, 2, 4, 1}) → false array123({1, 1, 2, 1, 2, 3}) → true
Solution public boolean array123(int[] nums) { for (int i=0;i<nums.length-2;i++){ if(nums[i]==1 && nums[i+1]==2 && nums[i+2]==3) return true; } return false; }
Problem-10 Given 2 strings, a and b, return the number of the positions where they contain the same length 2 substring. So “xxcaazz” and “xxbaaz” yields 3, since the “xx”, “aa”, and “az” substrings appear in the same place in both strings. Example stringMatch(“xxcaazz”, “xxbaaz”) → 3 stringMatch(“abc”, “abc”) → 2 stringMatch(“abc”, “axc”) → 0
Solution public int stringMatch(String a, String b) { int count=0; int r =Math.min(a.length(),b.length()); for(int i=0 ;i<r-1;i++){ String a1=a.substring(i,i+2); String b2=b.substring(i,i+2); if(a1.equals(b2)) count++; } return count; }
Problem-11 Given a string, return a version where all the “x” have been removed. Except an “x” at the very start or end should not be removed. Example: stringX(“xxHxix”) → “xHix” stringX(“abxxxcd”) → “abcd” stringX(“xabxxxcdx”) → “xabcdx”
Solution public String stringX(String str) { String result=””; for(int i=0;i<str.length();i++){ if(str.charAt(i)==’x’ && i!=0 && i!=str.length()-1) continue; //if the x isn’t first or last jump result+=str.charAt(i); } return result; } Solution Two public String stringX(String str) { String result=””; for(int i=0;i<str.length();i++){ if(str.substring(i,i+1).equals(“x”) && i!=0 && i!=str.length()-1) continue; result+=str.substring(i,i+1); } return result; }
Problem-12 Given a string, return a string made of the chars at indexes 0,1, 4,5, 8,9 … so “kittens” yields “kien”. Example: altPairs(“kitten”) → “kien” altPairs(“Chocolate”) → “Chole” altPairs(“CodingHorror”) → “Congrr”
Solution public String altPairs(String str) { String result=””; for(int i=0;i<str.length();i+=4){ int lim=i+2; if(lim>str.length()){ lim=str.length(); } result+=str.substring(i,lim); } return result; }
Problem-13 Suppose the string “yak” is unlucky. Given a string, return a version where all the “yak” are removed, but the “a” can be any char. The “yak” strings will not overlap. Example stringYak(“yakpak”) → “pak” stringYak(“pakyak”) → “pak” stringYak(“yak123ya”) → “123ya”
Solution public String stringYak(String str) { String result=””; for(int i=0; i<str.length();i++){ if(i+2< str.length() && str.charAt(i)==’y’ && str.charAt(i+2)==’k’){ i=i+2; // jump by 2 if there is a y-k } else{ result+=str.charAt(i); } } return result; }
Problem-14 Given an array of ints, return the number of times that two 6’s are next to each other in the array. Also count instances where the second “6” is actually a 7. Example: array667({6, 6, 2}) → 1 array667({6, 6, 2, 6}) → 1 array667({6, 7, 2, 6}) → 1
Solution public int array667(int[] nums) { int count=0; for(int i=0;i<nums.length-1;i++){ if (nums[i]==6 && (nums[i+1]==6 || nums[i+1]==7)) count++; } return count; }
Problem-15 Given an array of ints, we’ll say that a triple is a value appearing 3 times in a row in the array. Return true if the array does not contain any triples. Example noTriples({1, 1, 2, 2, 1}) → true noTriples({1, 1, 2, 2, 2, 1}) → false noTriples({1, 1, 1, 2, 2, 2, 1}) → false
Solution public boolean noTriples(int[] nums) { for(int i=0;i<nums.length-2;i++){ if (nums[i]==nums[i+1]&& nums[i+1]==nums[i+2]) return false; } return true; }
Problem-16 Given an array of ints, return true if it contains a 2, 7, 1 pattern — a value, followed by the value plus 5, followed by the value minus 1. Additionally the 271 counts even if the “1” differs by 2 or less from the correct value. Example has271({1, 2, 7, 1}) → true has271({1, 2, 8, 1}) → false has271({2, 7, 1}) → true
Solution public boolean has271(int[] nums) { for (int i=0;i<nums.length-2;i++){ int k=nums[i]; int j=nums[i+2]; if(nums[i+1]==k+5 && (Math.abs(j-k+1)<=2)) return true; } return false; }
**All Problems were taken from the website CodingBat
