4.13 Implementing 2D Array Algorithms

1. Exam Points

• Implement 2D array algorithms using selection and loop statements:
  • determine a minimum or maximum value
  • compute a sum or average
  • determine if at least one element has a particular property
  • determine if all elements of the 2D array have a particular property.
  • determine the number of elements in the 2D array
  • access all consecutive pairs of elements
  • determine the presence or absence of duplicate elements
  • shift or rotate elements in a row left or right or in a column up or down
  • reverse the order of the elements in a row or column

2. Knowledge Points

(1) Implementing 2D Array Algorithms

• There are standard algorithms that utilize 2D array traversals to:
  • determine a minimum or maximum value of all the elements or for a designated row, column, or other subsection.
  • compute a sum or average of all the elements or for a designated row, column, or other subsection.
  • determine if at least one element has a particular property in the entire 2D array or for a designated row, column, or other subsection.
  • determine if all elements of the 2D array or a designated row, column, or other subsection have a particular property.
  • determine the number of elements in the 2D array or in a designated row, column, or other subsection having a particular property.
  • access all consecutive pairs of elements.
  • determine the presence or absence of duplicate elements in the 2D array or in a designated row, column, or other subsection.
  • shift or rotate elements in a row left or right or in a column up or down
  • reverse the order of the elements in a row or column.

(2) Examples

• Example 1: compute a sum or average
  

  int[][] nums = { 
          { 1, 2, 3 }, 
          { 4, 5, 6 } 
  };
  
  double sum = 0;
  
  // i for row index, j for column index
  for (int i = 0; i < nums.length; i++) {
  
      for (int j = 0; j < nums[0].length; j++) {
          sum += nums[i][j];
      }
  }
  
  System.out.println("sum: " + sum);
  	
  int rows = nums.length;
  int cols = nums[0].length;
  int n = rows * cols;
  	
  double avg=sum/n;
  System.out.println("average: "+ avg);
  

• Example 2: find a maximum or minimum value
  

  int[][] nums = { 
      { 11, 2, 31 }, 
      { 42, 15, 26 } 
  };
  
  int max = nums[0][0]; // or Integer.MIN_VALUE
  
  for (int i = 0; i < nums.length; i++) { // i: row index
  
      for (int j = 0; j < nums[i].length; j++) { // j: column index
  
          if(nums[i][j] > max) { // if a greater value if found
              max = nums[i][j]; // update max
          }
      }
  }
  
  System.out.println("max: " + max);
  

• Example 3: Check if all the elements are odd
  

  int[][] nums = { 
      { 11, 13, 31 }, { 41, 15, 21 } 
  };
  
  // suppose all the elements are odd
  boolean allOdd = true; 
  
  // i: row index, j: column index
  for (int i = 0; i < nums.length; i++) {
  
      for (int j = 0; j < nums[0].length; j++) {
  
          if (nums[i][j] % 2 == 0) { // if an even number is found
              allOdd = false; // update allOdd
          }
      }
  }
  
  System.out.println("all odd: " + allOdd);
  

3. Exercises

• Practice
• Solution