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Zanzabum
4 years ago
13

What is 1 and 3/4 in grams?

Computers and Technology
2 answers:
Mice21 [21]4 years ago
8 0
<span> 1 and 3/4 in grams is 
</span>1.75grams
DanielleElmas [232]4 years ago
6 0
1 and 3/4 in grams equal 1.75 grams.
Hope this helps! :)
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Assume the existence of an UNSORTED ARRAY of n characters. You are to trace the CS111Sort algorithm (as described here) to reord
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Answer:

class Main {

  public static void main(String[] args) {

      char arr[] = {'T','E','D','R','W','B','S','V','A'};

      int n = arr.length;

      System.out.println("Selection Sort:");

      System.out.println("Iteration\tArray\tComparisons");

      long comp1 = selectionSort(arr);

      System.out.println("Total comparisons: "+comp1);

      System.out.println("\nInsertion Sort:");

      System.out.println("Iteration\tArray\tComparisons");

      long comp2 = insertionSort(arr);

      System.out.println("Total Comparisons: "+comp2);

      System.out.println("\nOverall Total Comparisons: "+(comp1+comp2));

  }

  static long selectionSort(char arr[]) {

      // applies selection sort for n/2 elements

      // returns number of comparisons

      int n = arr.length;

      long comparisons = 0;

 

      // One by one move boundary of unsorted subarray

      for (int i = n-1; i>=n-n/2; i--) {

              // Find the minimum element in unsorted array

              int max_idx = i;

              for (int j = i-1; j>=0; j--) {

                      // there is a comparison everytime this loop returns

                      comparisons++;

                      if (arr[j] > arr[max_idx])

                              max_idx = j;

              }

              // Swap the found minimum element with the first

              // element

              char temp = arr[max_idx];

              arr[max_idx] = arr[i];

              arr[i] = temp;

              System.out.print(n-1-i+"\t");

              printArray(arr);

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      }

     

      return comparisons;

  }

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      // applies insertion sort for n/2 elements

      // returns number of comparisons

      int n = arr.length;

      n = n-n/2;   // sort only the first n/2 elements

      long comparisons = 0;

      for (int i = 1; i < n; ++i) {

          char key = arr[i];

          int j = i - 1;

          /* Move elements of arr[0..i-1], that are

                  greater than key, to one position ahead

                  of their current position */

          while (j >= 0) {

              // there is a comparison everytime this loop runs

              comparisons++;

              if (arr[j] > key) {

                  arr[j + 1] = arr[j];

              } else {

                  break;

              }

              j--;

          }

          arr[j + 1] = key;

          System.out.print(i-1+"\t");

          printArray(arr);

          System.out.println("\t"+comparisons);

      }

      return comparisons;

  }  

  static void printArray(char arr[]) {

      for (int i=0; i<arr.length; i++)

          System.out.print(arr[i]+" ");

  }

}

Explanation:

Explanation is in the answer.

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