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"));
}
}
Step-by-step explanation: