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damaskus [11]
3 years ago
13

1. Write a recursive method to determine if a character is in a list of characters in O(logN) time. Mathematically prove (as we

did in class) that T(N) = O(logN). You can assume that this list is sorted lexicographically.
2. Write a function that determines if a string has the same number of 0’s and 1’s using a stack. The function must run in O(N) time. You can assume there already exists a stack class and can just use it
Computers and Technology
1 answer:
sergeinik [125]3 years ago
5 0

Answer:

1)

public class BinarySearch {

// Returns index of x if it is present in arr[l..

// r], else return -1

int binarySearch(char arr[], int l, int r, char x)

{

if (r >= l) {

int mid = l + (r - l) / 2;

 

// If the element is present at the

// middle itself

if (arr[mid] == x)

return mid;

 

// If element is smaller than mid, then

// it can only be present in left subarray

if (arr[mid] > x)

return binarySearch(arr, l, mid - 1, x);

 

// Else the element can only be present

// in right subarray

return binarySearch(arr, mid + 1, r, x);

}

 

// We reach here when element is not present

// in array

return -1;

}

 

// Driver method to test above

public static void main(String args[])

{

BinarySearch ob = new BinarySearch();

char arr[] = {'a','c','e','f','g','h'};

int n = arr.length;

char x = 'g';

int result = ob.binarySearch(arr, 0, n - 1, x);

if (result == -1)

System.out.println("Character not present");

else

System.out.println("Character found at index " + result);

}

}

2)

import java.io.*;

import java.util.*;

public class Test{

public static boolean checksame(String s)

{

 

Stack<Integer> stack= new Stack<Integer>();

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

{

if(s.charAt(i)=='0')

{

if(stack.empty()==false)

{

if((Integer)stack.peek()==1)

stack.pop();

else

stack.push(0);

}

else

{

stack.push(0);

}

}

else if(s.charAt(i)=='1')

{

if(stack.empty()==false)

{

if((Integer)stack.peek()==0)

stack.pop();

 

else

stack.push(1);

}

else

{

stack.push(1);

}

 

}

}

return stack.empty();

}

public static void main(String []args){

System.out.println(checksame("a0w1220"));

}

}

Explanation:

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Write a function which the counts the number of odd numbers and even numbers currently in the stack and prints the results.
Lerok [7]

Answer:

See the code below and the algorithm explanation on the figure.

Explanation:

The explanation in order to get the answer is given on the figure below.

Solving this problem with C. The program is given below:

#include <stdio.h>

int main(void) {

   int n, Even=0, Odd=0, Zeros=0;  

   for (;;) {

       printf("\nEnter the value the value that you want to check(remember just integers): ");

       //IF we input a non-numeric character the code end;

       if (scanf("%d", &n) != 1) break;

       if (n == 0) {

           Zeros++;

       }

       else {

           if (n % 2) {

               Even++;

           }

           else {

               Odd++;

           }

       }

   }  

   printf("for this case we have %d even, %d odd, and %d zero values.", Even, Odd, Zeros);

   return 0;

}

5 0
3 years ago
Analyze the following code. public class Test { public static void main(String[] args) { double radius; final double PI= 3.15169
sergiy2304 [10]

Answer:

The output of the given code as follows:

Output:

Area is: 12.60676

Explanation:

In the given code some information is missing so, the correct code to this question can be described as follows:

Program:

public class Test //defining class  

{

   public static void main(String[] args)//defining the main method

   {

       double radius= 2; //defining double variable radius

       final double PI= 3.15169; //defining double variable PI

       double area = radius * radius * PI; //defining double variable area that calculates values

       System.out.println("Area is: " + area); //print values

   }  

}

Explanation:

  • In the given java code a class "Test" is defined, in which a double variable "radius" is defined, which holds a value, that is 2.
  • In the next step, a double constant variable, that is PI is defined, that holds a value, that is "3.15169".
  • Then another double variable area is defined, that calculates the area value, and prints its value.
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What is the TSA motto
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Explanation:

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