Array – 2


countEvens
Return the number of even ints in the given array. Note: the % “mod” operator computes the remainder, e.g. 5 % 2 is 1.

countEvens([2, 1, 2, 3, 4]) → 3
countEvens([2, 2, 0]) → 3
countEvens([1, 3, 5]) → 0

public int countEvens(int[] nums) {
       int count = 0;
       for (int i = 0; i < nums.length; i++) {
           if (nums[i] % 2 == 0) {
               count++;
           }
       }
       return count;
   }

bigDiff
Given an array length 1 or more of ints, return the difference between the largest and smallest values in the array. Note: the built-in Math.min(v1, v2) and Math.max(v1, v2) methods return the smaller or larger of two values.
bigDiff([10, 3, 5, 6]) → 7
bigDiff([7, 2, 10, 9]) → 8
bigDiff([2, 10, 7, 2]) → 8

public int bigDiff(int[] nums) {
       int max = 0;
       int min = nums[0];
       for (int i = 0; i < nums.length; i++) {
           if (max < nums[i]) {
               max = nums[i];
           }
           if (min > nums[i]) {
               min = nums[i];
           }
       }
       return max - min;
   }

centeredAverage
Return the “centered” average of an array of ints, which we’ll say is the mean average of the values, except ignoring the largest and smallest values in the array. If there are multiple copies of the smallest value, ignore just one copy, and likewise for the largest value. Use int division to produce the final average. You may assume that the array is length 3 or more.
centeredAverage([1, 2, 3, 4, 100]) → 3
centeredAverage([1, 1, 5, 5, 10, 8, 7]) → 5
centeredAverage([-10, -4, -2, -4, -2, 0]) → -3

public int centeredAverage(int[] nums) {
    int sum = 0;
    for(int i = 0; i < nums.length; i++){
        sum = sum + nums[i];
    }
    int minMax = min(nums) + max(nums);
    sum -= minMax;
    return sum / (nums.length -2);
}

public int min (int[] nums){
    int min = nums[0];
    for(int i = 0; i < nums.length; i++){
        if(min > nums[i]){
            min = nums[i];
        }
    }
    return min;
}

public int max (int[] nums){
    int max = nums[0];
    for(int i = 0; i < nums.length; i++){
        if(max < nums[i]){
            max = nums[i];
        }
    }
    return max;
}

sum13
Return the sum of the numbers in the array, returning 0 for an empty array. Except the number 13 is very unlucky, so it does not count and numbers that come immediately after a 13 also do not count.
sum13([1, 2, 2, 1]) → 6
sum13([1, 1]) → 2
sum13([1, 2, 2, 1, 13]) → 6

public int sum13(int[] nums) {
       int sum = 0;
       for (int i = 0; i < nums.length; i++) {
           if (nums[i] == 13) {
               nums[i] = 0;
               if (nums.length > i + 1) {
                   nums[i + 1] = 0;
               }
           }
           sum += nums[i];
       }
       return sum;
   }

sum67
Return the sum of the numbers in the array, except ignore sections of numbers starting with a 6 and extending to the next 7 (every 6 will be followed by at least one 7). Return 0 for no numbers.
sum67([1, 2, 2]) → 5
sum67([1, 2, 2, 6, 99, 99, 7]) → 5
sum67([1, 1, 6, 7, 2]) → 4

public int sum67(int[] nums) {
      int before6 = 0;
      int sum = 0;
      int count6 = 0;
      int count7 = 0;
      for (int i = 0; i < nums.length; i++) {
          if (nums[i] != 6 && count6 == count7) {
              sum += nums[i];
          }
          if (nums[i] == 6 && count6 == count7) {
              before6 += sum;
              count6++;
          }
          if (nums[i] == 7 && count6 != 0) {
              count7++;
              if (count6 == count7) {
                  sum = 0;
              }
          }
      }
      return before6 + sum;
  }

has22
Given an array of ints, return true if the array contains a 2 next to a 2 somewhere.
has22([1, 2, 2]) → true
has22([1, 2, 1, 2]) → false
has22([2, 1, 2]) → false

public boolean has22(int[] nums) {
      int index2 = 0;
      for (int i = 0; i < nums.length - 1; i++) {
          if (nums[i] == 2 && nums[i + 1] == 2) {
              return true;
          }
      }
      return false;
  }

lucky13
Given an array of ints, return true if the array contains no 1’s and no 3’s.
lucky13([0, 2, 4]) → true
lucky13([1, 2, 3]) → false
lucky13([1, 2, 4]) → false

public boolean lucky13(int[] nums) {
       for (int i = 0; i < nums.length; i++) {
           if (nums[i] == 1 || nums[i] == 3) {
               return false;
           }
       }
       return true;
   }

sum28
Given an array of ints, return true if the sum of all the 2’s in the array is exactly 8.
sum28([2, 3, 2, 2, 4, 2]) → true
sum28([2, 3, 2, 2, 4, 2, 2]) → false
sum28([1, 2, 3, 4]) → false

public boolean sum28(int[] nums) {
       int count8 = 0;
       for (int i = 0; i < nums.length; i++) {
           if (nums[i] == 2) {
               count8++;
           }
       }
       if (count8 == 4) return true;
       return false;
   }

more14
Given an array of ints, return true if the number of 1’s is greater than the number of 4’s
more14([1, 4, 1]) → true
more14([1, 4, 1, 4]) → false
more14([1, 1]) → true

public boolean more14(int[] nums) {
      int count1 = 0;
      int count4 = 0;
      for (int i = 0; i < nums.length; i++) {
          if (nums[i] == 1) count1++;
          if (nums[i] == 4) count4++;
      }
      if (count1 > count4) return true;
      return false;
  }

fizzArray
Given a number n, create and return a new int array of length n, containing the numbers 0, 1, 2, … n-1. The given n may be 0, in which case just return a length 0 array. You do not need a separate if-statement for the length-0 case; the for-loop should naturally execute 0 times in that case, so it just works.
fizzArray(4) → [0, 1, 2, 3]
fizzArray(1) → [0]
fizzArray(10) → [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

public int[] fizzArray(int n) {
      int[] nums = new int[n];
      for(int i = 0; i < nums.length; i++){
          nums[i] = i;
      }
      return nums;
  }

only14
Given an array of ints, return true if every element is a 1 or a 4.
only14([1, 4, 1, 4]) → true
only14([1, 4, 2, 4]) → false
only14([1, 1]) → true

public boolean only14(int[] nums) {
       for (int i = 0; i < nums.length; i++) {
           if (nums[i] != 1 && nums[i] != 4) return false;
       }
       return true;
   }

fizzArray2
Given a number n, create and return a new string array of length n, containing the strings “0”, “1” “2” .. through n-1. N may be 0, in which case just return a length 0 array. Note: String.valueOf(xxx) will make the String form of most types.
fizzArray2(4) → [“0”, “1”, “2”, “3”]
fizzArray2(10) → [“0”, “1”, “2”, “3”, “4”, “5”, “6”, “7”, “8”, “9”]
fizzArray2(2) → [“0”, “1”]

public String[] fizzArray2(int n) {
       String[] nums = new String[n];
       String result = "";
       for (int i = 0; i < n; i++) {
           result += i;
           nums[i] = result.valueOf(i);
       }
       return nums;
   }

no14
Given an array of ints, return true if it contains no 1’s or it contains no 4’s.
no14([1, 2, 3]) → true
no14([1, 2, 3, 4]) → false
no14([2, 3, 4]) → true

public boolean no14(int[] nums) {
      int count1 = 0;
      int count4 = 0;
      for (int i = 0; i < nums.length; i++) {
          if (nums[i] == 1) count1++;
          if (nums[i] == 4) count4++;
      }
      if (count1 == 0 && count4 > 0) return true;
      if (count1 > 0 && count4 == 0) return true;
      if (nums.length <= 1) return true;
      return false;
  }

isEverywhere
We’ll say that a value is “everywhere” in an array if for every pair of adjacent elements in the array, at least one of the pair is that value. Return true if the given value is everywhere in the array.
isEverywhere([1, 2, 1, 3], 1) → true
isEverywhere([1, 2, 1, 3], 2) → false
isEverywhere([1, 2, 1, 3, 4], 1) → false

public boolean isEverywhere(int[] nums, int val) {
      for (int i = 0; i < nums.length - 1; i++) {
          if (nums[i] != val && nums[i + 1] != val) return false;
      }
      return true;
  }

either24
Given an array of ints, return true if the array contains a 2 next to a 2 or a 4 next to a 4, but not both.
either24([1, 2, 2]) → true
either24([4, 4, 1]) → true
either24([4, 4, 1, 2, 2]) → false

public boolean either24(int[] nums) {
       int count2 = 0;
       int count4 = 0;
       for (int i = 0; i < nums.length - 1; i++) {
           if (nums[i] == 2 && nums[i + 1] == 2) count2++;
           if (nums[i] == 4 && nums[i + 1] == 4) count4++;
       }
       if (count2 == count4) return false;
       return true;
   }

matchUp
Given arrays nums1 and nums2 of the same length, for every element in nums1, consider the corresponding element in nums2 (at the same index). Return the count of the number of times that the two elements differ by 2 or less, but are not equal.
matchUp([1, 2, 3], [2, 3, 10]) → 2
matchUp([1, 2, 3], [2, 3, 5]) → 3
matchUp([1, 2, 3], [2, 3, 3]) → 2

public int matchUp(int[] nums1, int[] nums2) {
       int count = 0;
       for (int i = 0; i < nums1.length; i++) {
           if (Math.abs(nums1[i] - nums2[i]) <= 2 && nums1[i] - nums2[i] != 0) count++;
       }
       return count;
   }

has77
Given an array of ints, return true if the array contains two 7’s next to each other, or there are two 7’s separated by one element, such as with {7, 1, 7}.
has77([1, 7, 7]) → true
has77([1, 7, 1, 7]) → true
has77([1, 7, 1, 1, 7]) → false

public boolean has77(int[] nums) {
 for (int i = 0; i < nums.length - 2; i++) {
 if ((nums[i] == 7 && nums[i + 1] == 7) || (nums[i] == 7 && nums[i + 2] == 7)) return true;
 if (nums[i + 1] == 7 && nums[i + 2] == 7) return true;
 }
 return false;
}

has12
Given an array of ints, return true if there is a 1 in the array with a 2 somewhere later in the array.
has12([1, 3, 2]) → true
has12([3, 1, 2]) → true
has12([3, 1, 4, 5, 2]) → true

public boolean has12(int[] nums) {
      int count1 = 0;
      for (int i = 0; i < nums.length; i++) {
          if (nums[i] == 1) count1++;
          if (nums[i] == 2 && count1 > 0) return true;
      }
      return false;
  }

modThree
Given an array of ints, return true if the array contains either 3 even or 3 odd values all next to each other.
modThree([2, 1, 3, 5]) → true
modThree([2, 1, 2, 5]) → false
modThree([2, 4, 2, 5]) → true

public boolean modThree(int[] nums) {
      for (int i = 0; i < nums.length - 2; i++) {
          if (nums[i] % 2 == 0 && nums[i + 1] % 2 == 0 && nums[i + 2] % 2 == 0) return true;
          if (nums[i] % 2 != 0 && nums[i + 1] % 2 != 0 && nums[i + 2] % 2 != 0) return true;
      }
      return false;
  }

haveThree
Given an array of ints, return true if the value 3 appears in the array exactly 3 times, and no 3’s are next to each other.
haveThree([3, 1, 3, 1, 3]) → true
haveThree([3, 1, 3, 3]) → false
haveThree([3, 4, 3, 3, 4]) → false

public boolean haveThree(int[] nums) {
       int count = 0;
       for (int i = 0; i < nums.length - 1; i++) {
           if (nums[i] == 3 && nums[i + 1] != 3) {
               count++;
           }
       }
       if (nums.length > 1 && nums[nums.length - 1] == 3 && nums[nums.length - 2] != 3) {
           count++;
       }
       if (count == 3) {
           return true;
       }
       return false;
   }

twoTwo
Given an array of ints, return true if every 2 that appears in the array is next to another 2.
twoTwo([4, 2, 2, 3]) → true
twoTwo([2, 2, 4]) → true
twoTwo([2, 2, 4, 2]) → false

public boolean twoTwo(int[] nums) {
    boolean found = false;
    int count = 0;
    if (nums.length == 2 && nums[0] + nums[1] == 4) return true;
    if (nums.length == 1 && nums[0] == 2) return false;
    for (int i = 0; i < nums.length - 1; i++) {
        if (nums[i] == 2) count++;
        if (nums[i] == 2 && nums[i + 1] == 2) found = true;
        if (nums.length > 1 && nums[nums.length - 2] != 2 && nums[nums.length - 1] == 2) {
            return false;
        }
    }
    if (count == 0 || (found == true && count % 2 == 0)) return true;

    return false;
}

sameEnds
Return true if the group of N numbers at the start and end of the array are the same. For example, with {5, 6, 45, 99, 13, 5, 6}, the ends are the same for n=0 and n=2, and false for n=1 and n=3. You may assume that n is in the range 0..nums.length inclusive.
sameEnds([5, 6, 45, 99, 13, 5, 6], 1) → false
sameEnds([5, 6, 45, 99, 13, 5, 6], 2) → true
sameEnds([5, 6, 45, 99, 13, 5, 6], 3) → false

public boolean sameEnds(int[] nums, int len) {
      for (int i = 0; i < len; i++) {
          if (nums[i] != nums[nums.length - len + i]) return false;
      }
      return true;
  }

tripleUp
Return true if the array contains, somewhere, three increasing adjacent numbers like …. 4, 5, 6, … or 23, 24, 25.
tripleUp([1, 4, 5, 6, 2]) → true
tripleUp([1, 2, 3]) → true
tripleUp([1, 2, 4]) → false

public boolean tripleUp(int[] nums) {
    for (int i = 0; i < nums.length - 2; i++) {
        if (nums.length >= 3) {
            if (nums[i] - nums[i + 1] == -1 && nums[i + 1] - nums[i + 2] == -1) return true;
        }
    }
    return false;
}

fizzArray3
Given start and end numbers, return a new array containing the sequence of integers from start up to but not including end, so start=5 and end=10 yields {5, 6, 7, 8, 9}. The end number will be greater or equal to the start number. Note that a length-0 array is valid.
fizzArray3(5, 10) → [5, 6, 7, 8, 9]
fizzArray3(11, 18) → [11, 12, 13, 14, 15, 16, 17]
fizzArray3(1, 3) → [1, 2]

public int[] fizzArray3(int start, int end) {
       int[] nums = new int[end-start];
       for(int i = 0; i<nums.length; i++){
           nums[i] = start;
           start++;
       }
       return nums;
   }

shiftLeft
Return an array that is “left shifted” by one — so {6, 2, 5, 3} returns {2, 5, 3, 6}. You may modify and return the given array, or return a new array.
shiftLeft([6, 2, 5, 3]) → [2, 5, 3, 6]
shiftLeft([1, 2]) → [2, 1]
shiftLeft([1]) → [1]

public int[] shiftLeft(int[] nums) {
      int[] nums1 = new int[nums.length];
      if (nums.length == 0) return nums;
      for (int i = 0; i < nums.length - 1; i++) {
          int a = i;
          nums1[i] = nums[a + 1];
      }
      nums1[nums1.length - 1] = nums[0];
      return nums1;
  }

tenRun
For each multiple of 10 in the given array, change all the values following it to be that multiple of 10, until encountering another multiple of 10. So {2, 10, 3, 4, 20, 5} yields {2, 10, 10, 10, 20, 20}.
tenRun([2, 10, 3, 4, 20, 5]) → [2, 10, 10, 10, 20, 20]
tenRun([10, 1, 20, 2]) → [10, 10, 20, 20]
tenRun([10, 1, 9, 20]) → [10, 10, 10, 20]

public int[] tenRun(int[] nums) {
        int[] nums1 = new int [nums.length];
        int multiTen = 0;
        boolean found = false;
        for(int i = 0; i < nums.length; i++){
            nums1[i] = nums[i];
            if(nums[i]%10 == 0){
                found = true;
                multiTen = nums[i];
            }
            if(found == true) nums1[i] = multiTen;
        }
        return nums1;
    }

pre4
Given a non-empty array of ints, return a new array containing the elements from the original array that come before the first 4 in the original array. The original array will contain at least one 4. Note that it is valid in java to create an array of length 0.
pre4([1, 2, 4, 1]) → [1, 2]
pre4([3, 1, 4]) → [3, 1]
pre4([1, 4, 4]) → [1]

public int[] pre4(int[] nums) {
    int count = 0;
    for(int i = 0; i<nums.length; i++){
        if(nums[i] != 4)  count++;
        if (nums[i] == 4){
            int[] res = Arrays.copyOf(nums, count);
            return res;
        }
    }
    return nums;
}

post4
Given a non-empty array of ints, return a new array containing the elements from the original array that come after the last 4 in the original array. The original array will contain at least one 4. Note that it is valid in java to create an array of length 0.
post4([2, 4, 1, 2]) → [1, 2]
post4([4, 1, 4, 2]) → [2]
post4([4, 4, 1, 2, 3]) → [1, 2, 3]

public int[] post4(int[] nums) {
    int count = nums.length;
    for(int i = nums.length-1; i >= 0; i--){
        if(nums[i] != 4) count--;
        if(nums[i] == 4){
            int[] res = Arrays.copyOfRange(nums, count, nums.length);
            return res;
        }
    }
    return nums;
}

notAlone
We’ll say that an element in an array is “alone” if there are values before and after it, and those values are different from it. Return a version of the given array where every instance of the given value which is alone is replaced by whichever value to its left or right is larger.
notAlone([1, 2, 3], 2) → [1, 3, 3]
notAlone([1, 2, 3, 2, 5, 2], 2) → [1, 3, 3, 5, 5, 2]
notAlone([3, 4], 3) → [3, 4]

public int[] notAlone(int[] nums, int val) {
    for(int i = 1; i< nums.length-1; i++){
        if( nums[i] == val){
            if(nums[i-1] != nums[i] && nums[i+1] != nums[i]){
                nums[i] = Math.max(nums[i-1], nums[i+1]);
            }
        }
    }
    return nums;
}

zeroFront
Return an array that contains the exact same numbers as the given array, but rearranged so that all the zeros are grouped at the start of the array. The order of the non-zero numbers does not matter. So {1, 0, 0, 1} becomes {0 ,0, 1, 1}. You may modify and return the given array or make a new array.
zeroFront([1, 0, 0, 1]) → [0, 0, 1, 1]
zeroFront([0, 1, 1, 0, 1]) → [0, 0, 1, 1, 1]
zeroFront([1, 0]) → [0, 1]

public int[] zeroFront(int[] nums) {
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 0) {
                count++;
                nums[i] = nums[count - 1];
                nums[count - 1] = 0;
            }
        }
        return nums;
    }

withoutTen
Return a version of the given array where all the 10’s have been removed. The remaining elements should shift left towards the start of the array as needed, and the empty spaces a the end of the array should be 0. So {1, 10, 10, 2} yields {1, 2, 0, 0}. You may modify and return the given array or make a new array.
withoutTen([1, 10, 10, 2]) → [1, 2, 0, 0]
withoutTen([10, 2, 10]) → [2, 0, 0]
withoutTen([1, 99, 10]) → [1, 99, 0]

public int[] withoutTen(int[] nums) {
       int count = 0;
       int[] res = new int[nums.length];
       for (int i = 0; i < nums.length; i++) {
           if (nums[i] != 10) {
               res[count] = nums[i];
               count++;
           }
       }
       return res;
   }

zeroMax
Return a version of the given array where each zero value in the array is replaced by the largest odd value to the right of the zero in the array. If there is no odd value to the right of the zero, leave the zero as a zero.
zeroMax([0, 5, 0, 3]) → [5, 5, 3, 3]
zeroMax([0, 4, 0, 3]) → [3, 4, 3, 3]
zeroMax([0, 1, 0]) → [1, 1, 0]

public int[] zeroMax(int[] nums) {
    int pos0 = 0;
    int maxOdd = 0;
    for(int i = 0; i<nums.length; i++){
        if(nums[i] == 0){
            pos0 = i;
            while(pos0 < nums.length){
                if(maxOdd < nums[pos0] && nums[pos0]%2 != 0) {
                    maxOdd = nums[pos0];
                }
                pos0++;
            }
            nums[i] = maxOdd;
            maxOdd = 0;
        }
    }
    return nums;
}

evenOdd
Return an array that contains the exact same numbers as the given array, but rearranged so that all the even numbers come before all the odd numbers. Other than that, the numbers can be in any order. You may modify and return the given array, or make a new array.
evenOdd([1, 0, 1, 0, 0, 1, 1]) → [0, 0, 0, 1, 1, 1, 1]
evenOdd([3, 3, 2]) → [2, 3, 3]
evenOdd([2, 2, 2]) → [2, 2, 2]

public int[] evenOdd(int[] nums) {
        int count = 0;
        int temp = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] % 2 == 0) {
                temp = nums[count];
                nums[count] = nums[i];
                nums[i] = temp;
                count++;
            }
        }
        return nums;
    }

fizzBuzz
This is slightly more difficult version of the famous FizzBuzz problem which is sometimes given as a first problem for job interviews. (See also: FizzBuzz Code.) Consider the series of numbers beginning at start and running up to but not including end, so for example start=1 and end=5 gives the series 1, 2, 3, 4. Return a new String[] array containing the string form of these numbers, except for multiples of 3, use “Fizz” instead of the number, for multiples of 5 use “Buzz”, and for multiples of both 3 and 5 use “FizzBuzz”. In Java, String.valueOf(xxx) will make the String form of an int or other type. This version is a little more complicated than the usual version since you have to allocate and index into an array instead of just printing, and we vary the start/end instead of just always doing 1..100.
fizzBuzz(1, 6) → [“1”, “2”, “Fizz”, “4”, “Buzz”]
fizzBuzz(1, 8) → [“1”, “2”, “Fizz”, “4”, “Buzz”, “Fizz”, “7”]
fizzBuzz(1, 11) → [“1”, “2”, “Fizz”, “4”, “Buzz”, “Fizz”, “7”, “8”, “Fizz”, “Buzz”]

public String[] fizzBuzz(int start, int end) {
       String result = "";
       String[] nums = new String[end - start];
       for (int i = 0; i < nums.length; i++) {
           if (start % 3 == 0 && start % 5 != 0) nums[i] = "Fizz";
           if (start % 5 == 0 && start % 3 != 0) nums[i] = "Buzz";
           if (start % 5 == 0 && start % 3 == 0) nums[i] = "FizzBuzz";
           if (start % 3 != 0 && start % 5 != 0) {
               result += start;
               nums[i] = result.valueOf(start);
           }
           start++;
       }
       return nums;
   }

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