Question

Pseudocode:

1. The first problem has multiple steps but is a little bit easier than the second.

   1. Define a module main

   2. In main, declare an array of Integers named goldFishPerChild with 4

      elements

   3. Set the values of the array to 6, 4, 3, 3 ( you may do both in 1 line)

   4. You tried to teach them responsibility, but your kids are terrible with

      pets. Each child killed two of its goldfish. This is a learning

      experience for them and for you.

   5. In main, write a loop to update the value of each element so that it is

      2 less ​by using the accumulator pattern​.

2. Define a function named linearSearchInStringArray

   1. Has String parameter searchTerm : value you are searching for

   2. Has String[] parameter inputArray : array you wish to search through

   3. Has Integer parameter arraySize : number of elements in the array

   4. Returns Boolean : True if search term is one of the values in the

      array, False if not.

   CLUE: use count-controlled loop to iterate through all elements CLUE: needs to use a decision structure each iteration
   CLUE: can simply return True if you find a match!
   CLUE: consider when you know you can return false.

   CLUE: drawing pictures can help

3. Define a function named evenCount

1. Has Integer[] parameter inputArray : array of elements you are

   counting through

2. Has Integer parameter arraySize : number of elements in array

3. Returns Integer : number of elements with even-numbered values in

   array

CLUE: This will require a loop and an extra variable to keep a running total

CLUE: this is a great opportunity to use the "%" operator. Look up "modulus" operator. This operator tells you the 'remainder' of a division operation.

A remainder of 0 means the number can be evenly divided...

Answer 1

1.

public static void main(String[] args) {
    int[] goldFishPerChild = {6, 4, 3, 3};
    for(int i=0; i<4; i++){
      goldFishPerChild[i] = goldFishPerChild[i] - 2;
}

2.

public static boolean linearSearchInStringArray(String searchTerm, String[] inputArray, int arraySize){
    for(int i=0; i<arraySize; i++){
      if(inputArray[i]==searchTerm){
        return true;
      }
    }
    return false;
}

3.

public static int evenCount(int[] inputArray, int arraySize){
    int count=0;
    for(int i=0; i<arraySize; i++){
      if(inputArray[i] % 2 == 0){
        count++;
      }
    }
    return count;
}

Driver function to test:

class Main {

  public static boolean linearSearchInStringArray(String searchTerm, String[] inputArray, int arraySize){
    for(int i=0; i<arraySize; i++){
      if(inputArray[i]==searchTerm){
        return true;
      }
    }
    return false;
  }

  public static int evenCount(int[] inputArray, int arraySize){
    int count=0;
    for(int i=0; i<arraySize; i++){
      if(inputArray[i] % 2 == 0){
        count++;
      }
    }
    return count;
  }

  public static void main(String[] args) {
    int[] goldFishPerChild = {6, 4, 3, 3};
    for(int i=0; i<4; i++){
      goldFishPerChild[i] = goldFishPerChild[i] - 2;
    }
    String[] str = {"Ronaldo", "Messi", "Kaka"};
    System.out.println("String exists?: "+linearSearchInStringArray("Messi",str,3));

    int[] num = {1, 2, 3, 4, 5, 6};
    System.out.println("Number of evens: "+evenCount(num,6));

  }
}

(Feel free to give an upvote or comment below for any doubts)