Using stacks for decimal to binary conversion is a more flexible method that you can use to convert any decimal number into its equivalent binary numbers. Here we show you how you can use stacks to perform this conversion and how it is used in all the major programming languages. These Decimal to Binary Using Stacks Question and Answer with FREE PDF will help you prepare for any exam including BCA, MCA, GATE, GRE, IES, PSC, UGC NET, DOEACC Exams at all levels – you just have to practice regularly.
Decimal to Binary Using Stacks Question
1. Express -15 as a 6-bit signed binary number.
a) 001111
b) 101111
c) 101110
d) 001110
Answer: b
2. Which of the following code snippet is used to convert decimal to binary numbers?
a)
public void convertBinary(int num)
{
int bin[] = new int[50];
int index = 0;
while(num > 0)
{
bin[index++] = num%2;
num = num/2;
}
for(int i = index-1;i >= 0;i--)
{
System.out.print(bin[i]);
}
}
b)
public void convertBinary(int num)
{
int bin[] = new int[50];
int index = 0;
while(num > 0)
{
bin[++index] = num%2;
num = num/2;
}
for(int i = index-1;i >= 0;i--)
{
System.out.print(bin[i]);
}
}
c)
public void convertBinary(int num)
{
int bin[] = new int[50];
int index = 0;
while(num > 0)
{
bin[index++] = num/2;
num = num%2;
}
for(int i = index-1;i >= 0;i--)
{
System.out.print(bin[i]);
}
}
d)
public void convertBinary(int num)
{
int bin[] = new int[50];
int index = 0;
while(num > 0)
{
bin[++index] = num/2;
num = num%2;
}
for(int i = index-1;i >= 0;i--)
{
System.out.print(bin[i]);
}
}
Answer:
public void convertBinary(int num)
{
int bin[] = new int[50];
int index = 0;
while(num > 0)
{
bin[index++] = num%2;
num = num/2;
}
for(int i = index-1;i >= 0;i--)
{
System.out.print(bin[i]);
}
}
3. Which is the predefined method available in Java to convert decimal to binary numbers?
a) toBinaryInteger(int)
b) toBinaryValue(int)
c) toBinaryNumber(int)
d) toBinaryString(int)
Answer: toBinaryString(int)
4. Using stacks, how to obtain the binary representation of the number?
a)
public void convertBinary(int num)
{
Stack<Integer> stack = new Stack<Integer>();
while (num != 0)
{
int digit = num / 2;
stack.push(digit);
num = num % 2;
}
System.out.print("nBinary representation is:");
while (!(stack.isEmpty() ))
{
System.out.print(stack.pop());
}
}
b)
public void convertBinary(int num)
{
Stack<Integer> stack = new Stack<Integer>();
while (num != 0)
{
int digit = num % 2;
stack.push(digit);
}
System.out.print("nBinary representation is:");
while (!(stack.isEmpty() ))
{
System.out.print(stack.pop());
}
}
c)
public void convertBinary(int num)
{
Stack<Integer> stack = new Stack<Integer>();
while (num != 0)
{
int digit = num % 2;
stack.push(digit);
num = num / 2;
}
System.out.print("nBinary representation is:");
while (!(stack.isEmpty() ))
{
System.out.print(stack.pop());
}
}
d)
public void convertBinary(int num)
{
Stack<Integer> stack = new Stack<Integer>();
while (num != 0)
{
int digit = num % 2;
stack.push(digit%2);
num = num / 2;
}
System.out.print("nBinary representation is:");
while (!(stack.isEmpty() ))
{
System.out.print(stack.pop());
}
}
Answer:
public void convertBinary(int num)
{
Stack<Integer> stack = new Stack<Integer>();
while (num != 0)
{
int digit = num % 2;
stack.push(digit);
num = num / 2;
}
System.out.print("nBinary representation is:");
while (!(stack.isEmpty() ))
{
System.out.print(stack.pop());
}
}
5. What is the time complexity for converting decimal to binary numbers?
a) O(1)
b) O(n)
c) O(logn)
d) O(nlogn)
Answer: O(logn)
6. Write a piece of code which returns true if the string contains balanced parenthesis, false otherwise.
a)
public boolean isBalanced(String exp)
{
int len = exp.length();
Stack<Integer> stk = new Stack<Integer>();
for(int i = 0; i < len; i++)
{
char ch = exp.charAt(i);
if (ch == '(')
stk.push(i);
else if (ch == ')')
{
if(stk.peek() == null)
{
return false;
}
stk.pop();
}
}
return true;
}
b)
public boolean isBalanced(String exp)
{
int len = exp.length();
Stack<Integer> stk = new Stack<Integer>();
for(int i = 0; i < len; i++)
{
char ch = exp.charAt(i);
if (ch == '(')
stk.push(i);
else if (ch == ')')
{
if(stk.peek() != null)
{
return true;
}
stk.pop();
}
}
return false;
}
c)
public boolean isBalanced(String exp)
{
int len = exp.length();
Stack<Integer> stk = new Stack<Integer>();
for(int i = 0; i < len; i++)
{
char ch = exp.charAt(i);
if (ch == ')')
stk.push(i);
else if (ch == '(')
{
if(stk.peek() == null)
{
return false;
}
stk.pop();
}
}
return true;
}
d)
public boolean isBalanced(String exp)
{
int len = exp.length();
Stack<Integer> stk = new Stack<Integer>();
for(int i = 0; i < len; i++)
{
char ch = exp.charAt(i);
if (ch == '(')
stk.push(i);
else if (ch == ')')
{
if(stk.peek() != null)
{
return false;
}
stk.pop();
}
}
return true;
}
Answer:
public boolean isBalanced(String exp)
{
int len = exp.length();
Stack<Integer> stk = new Stack<Integer>();
for(int i = 0; i < len; i++)
{
char ch = exp.charAt(i);
if (ch == '(')
stk.push(i);
else if (ch == ')')
{
if(stk.peek() == null)
{
return false;
}
stk.pop();
}
}
return true;
}
7. What is the time complexity of the following code?
public boolean isBalanced(String exp)
{
int len = exp.length();
Stack<Integer> stk = new Stack<Integer>();
for(int i = 0; i < len; i++)
{
char ch = exp.charAt(i);
if (ch == '(')
stk.push(i);
else if (ch == ')')
{
if(stk.peek() == null)
{
return false;
}
stk.pop();
}
}
return true;
}
a) O(logn)
b) O(n)
c) O(1)
d) O(nlogn)
Answer: O(n)
8. Which of the following program prints the index of every matching parenthesis?
a)
public void dispIndex(String exp)
{
Stack<Integer> stk = new Stack<Integer>();
for (int i = 0; i < len; i++)
{
char ch = exp.charAt(i);
if (ch == '(')
stk.push(i);
else if (ch == ')')
{
try
{
int p = stk.pop() + 1;
System.out.println("')' at index "+(i+1)+" matched with ')' at index "+p);
}
catch(Exception e)
{
System.out.println("')' at index "+(i+1)+" is unmatched");
}
}
}
while (!stk.isEmpty() )
System.out.println("'(' at index "+(stk.pop() +1)+" is unmatched");
}
b)
public void dispIndex(String exp)
{
Stack<Integer> stk = new Stack<Integer>();
for (int i = 0; i < len; i++)
{
char ch = exp.charAt(i);
if (ch == '(')
stk.push(i);
else if (ch == ')')
{
try
{
int p = stk.pop() + 1;
System.out.println("')' at index "+(i)+" matched with ')' at index "+p);
}
catch(Exception e)
{
System.out.println("')' at index "+(i)+" is unmatched");
}
}
}
while (!stk.isEmpty() )
System.out.println("'(' at index "+(stk.pop() +1)+" is unmatched");
}
c)
public void dispIndex(String exp)
{
Stack<Integer> stk = new Stack<Integer>();
for (int i = 0; i < len; i++)
{
char ch = exp.charAt(i);
if (ch == ')')
stk.push(i);
else if (ch == '(')
{
try
{
int p = stk.pop() +1;
System.out.println("')' at index "+(i+1)+" matched with ')' at index "+p);
}
catch(Exception e)
{
System.out.println("')' at index "+(i+1)+" is unmatched");
}
}
}
while (!stk.isEmpty() )
System.out.println("'(' at index "+(stk.pop() +1)+" is unmatched");
}
d)
public void dispIndex(String exp)
{
Stack<Integer> stk = new Stack<Integer>();
for (int i = 0; i < len; i++)
{
char ch = exp.charAt(i);
if (ch == ')')
stk.push(i);
else if (ch == '(')
{
try
{
int p = stk.pop();
System.out.println("')' at index "+(i+1)+" matched with ')' at index "+p);
}
catch(Exception e)
{
System.out.println("')' at index "+(i+1)+" is unmatched");
}
}
}
while (!stk.isEmpty() )
System.out.println("'(' at index "+(stk.pop() +1)+" is unmatched");
}
Answer:
public void dispIndex(String exp)
{
Stack<Integer> stk = new Stack<Integer>();
for (int i = 0; i < len; i++)
{
char ch = exp.charAt(i);
if (ch == '(')
stk.push(i);
else if (ch == ')')
{
try
{
int p = stk.pop() + 1;
System.out.println("')' at index "+(i+1)+" matched with ')' at index "+p);
}
catch(Exception e)
{
System.out.println("')' at index "+(i+1)+" is unmatched");
}
}
}
while (!stk.isEmpty() )
System.out.println("'(' at index "+(stk.pop() +1)+" is unmatched");
}
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Categories: Application of Stacks
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