# Check whether consecutive decreasing active bits exist in a number

Given a positive integer, check whether the binary representation bits restrict the decreasing order of continuously active bits present in the right-to-left direction or not. For example, valid decreasing contiguous sequence.

``````
Example 1 :
------------
num = 398 (000110001110) Binary bits
First pair of consecutive set bits is
(111) length is 3
Second pair of consecutive set bits is
(11)  length is 2
Valid decreasing order (3 > 2 ) (Right to left)

Output : Yes

Example 2 :
------------
num = 623 (1001101111)
First pair of consecutive set bits is
(1111) length is 4
Second pair of consecutive set bits is
(11)  length is 2
Third pair of consecutive set bits is
(1)  length is 1

Valid decreasing order (4 > 2 > 1)

Output : Yes

Example 3 :
-----------
num = 46 (101110)
First pair of consecutive set bits is
(111) length is 3
Second pair of consecutive set bits is
(1)  length is 1
Valid decreasing order ( 2 > 1)
Output : Yes

Example 4 :
------------
num = 14 (1110)
First pair of consecutive set bits is
(111) length is 3

Any one pair of active bits are valid

Output : Yes
```
```

Some of case which is not valid to this problem their examples such as follows.

``````Example 1 :
27  (11011)
First pair of consecutive set bits is
(11) length is 2
Second pair of consecutive set bits is
(11)  length is 2
Valid decreasing order (2 > 2) Not true
Output : No

Example 2:
58 (111010)
First pair of consecutive set bits is
(1) length is 1
Second pair of consecutive set bits is
(111)  length is 3
Valid decreasing order (1 > 3) Not true

Output : No
``````

Here given code implementation process.

``````// C program
// Check whether consecutive decreasing active bits exist in a number
#include <stdio.h>

// Determine that given number have binary active bits in decreasing order or not
void decreasingActiveBit(int num)
{
if (num < 0)
{
return;
}
// Define some auxiliary variables
int count = 0;
int result = 0;
int temp = num;
// Execute loop until temp is gratore than zero and
// Consecutive active bit length is decreasing order
// in form of right to left
while (temp > 0 && result != -1)
{
if ((temp & 1) == 1)
{
// Count consecutive active bit length
count++;
}
else if (count > 0)
{
if (result == 0)
{
// Get first sequence of active bits
result = count;
}
else if (result > count)
{
// Get next small active bits pair length
result = count;
}
else
{
// When next active pair length is greater
result = -1;
}
count = 0;
}
// Shift by one bit to left
temp = temp >> 1;
}
if (result == 0 && count > 0)
{
// Whe only first pair exist
result = count;
}
else if (count >= result)
{
// When left side consecutive set bits length are not smaller
result = -1;
}
// Display given number
printf("\n Number : %d", num);
if (result == -1)
{
printf("\n No");
}
else
{
printf("\n Yes");
}
}
int main(int argc, char
const *argv[])
{
int num = 567;
// 567 (1000110111)
decreasingActiveBit(num);
num = 439;
// 439 (110110111)
decreasingActiveBit(num);
num = 7;
// 7 (111)
decreasingActiveBit(num);
num = 92;
// 92 (1011100)
decreasingActiveBit(num);
num = 29;
// 29 (11101)
decreasingActiveBit(num);
return 0;
}``````

#### Output

`````` Number : 567
Yes
Number : 439
No
Number : 7
Yes
Number : 92
Yes
Number : 29
No``````
``````/*
Java program
Check whether consecutive decreasing active bits exist in a number
*/
public class Sequence
{
// Determine that given number have binary active bits in decreasing order or not
public void decreasingActiveBit(int num)
{
if (num < 0)
{
return;
}
// Define some auxiliary variables
int count = 0;
int result = 0;
int temp = num;
// Execute loop until temp is gratore than zero and
// Consecutive active bit length is decreasing order
// in form of right to left
while (temp > 0 && result != -1)
{
if ((temp & 1) == 1)
{
// Count consecutive active bit length
count++;
}
else if (count > 0)
{
if (result == 0)
{
// Get first sequence of active bits
result = count;
}
else if (result > count)
{
// Get next small active bits pair length
result = count;
}
else
{
// When next active pair length is greater
result = -1;
}
count = 0;
}
// Shift by one bit to left
temp = temp >> 1;
}
if (result == 0 && count > 0)
{
// Whe only first pair exist
result = count;
}
else if (count >= result)
{
// When left side consecutive set bits length are not smaller
result = -1;
}
// Display given number
System.out.print("\n Number : " + num + "");
if (result == -1)
{
System.out.print("\n No");
}
else
{
System.out.print("\n Yes");
}
}
public static void main(String[] args)
{
int num = 567;
// 567 (1000110111)
num = 439;
// 439 (110110111)
num = 7;
// 7 (111)
num = 92;
// 92 (1011100)
num = 29;
// 29 (11101)
}
}``````

#### Output

`````` Number : 567
Yes
Number : 439
No
Number : 7
Yes
Number : 92
Yes
Number : 29
No``````
``````// Include header file
#include <iostream>
using namespace std;

/*
C++ program
Check whether consecutive decreasing active bits exist in a number
*/

class Sequence
{
public:
// Determine that given number have binary active bits in decreasing order or not
void decreasingActiveBit(int num)
{
if (num < 0)
{
return;
}
// Define some auxiliary variables
int count = 0;
int result = 0;
int temp = num;
// Execute loop until temp is gratore than zero and
// Consecutive active bit length is decreasing order
// in form of right to left
while (temp > 0 && result != -1)
{
if ((temp &1) == 1)
{
// Count consecutive active bit length
count++;
}
else if (count > 0)
{
if (result == 0)
{
// Get first sequence of active bits
result = count;
}
else if (result > count)
{
// Get next small active bits pair length
result = count;
}
else
{
// When next active pair length is greater
result = -1;
}
count = 0;
}
// Shift by one bit to left
temp = temp >> 1;
}
if (result == 0 && count > 0)
{
// Whe only first pair exist
result = count;
}
else if (count >= result)
{
// When left side consecutive set bits length are not smaller
result = -1;
}
// Display given number
cout << "\n Number : " << num << "";
if (result == -1)
{
cout << "\n No";
}
else
{
cout << "\n Yes";
}
}
};
int main()
{
int num = 567;
// 567 (1000110111)
num = 439;
// 439 (110110111)
num = 7;
// 7 (111)
num = 92;
// 92 (1011100)
num = 29;
// 29 (11101)
return 0;
}``````

#### Output

`````` Number : 567
Yes
Number : 439
No
Number : 7
Yes
Number : 92
Yes
Number : 29
No``````
``````// Include namespace system
using System;
/*
C# program
Check whether consecutive decreasing active bits exist in a number
*/
public class Sequence
{
// Determine that given number have binary active bits in decreasing order or not
public void decreasingActiveBit(int num)
{
if (num < 0)
{
return;
}
// Define some auxiliary variables
int count = 0;
int result = 0;
int temp = num;
// Execute loop until temp is gratore than zero and
// Consecutive active bit length is decreasing order
// in form of right to left
while (temp > 0 && result != -1)
{
if ((temp & 1) == 1)
{
// Count consecutive active bit length
count++;
}
else if (count > 0)
{
if (result == 0)
{
// Get first sequence of active bits
result = count;
}
else if (result > count)
{
// Get next small active bits pair length
result = count;
}
else
{
// When next active pair length is greater
result = -1;
}
count = 0;
}
// Shift by one bit to left
temp = temp >> 1;
}
if (result == 0 && count > 0)
{
// Whe only first pair exist
result = count;
}
else if (count >= result)
{
// When left side consecutive set bits length are not smaller
result = -1;
}
// Display given number
Console.Write("\n Number : " + num + "");
if (result == -1)
{
Console.Write("\n No");
}
else
{
Console.Write("\n Yes");
}
}
public static void Main(String[] args)
{
int num = 567;
// 567 (1000110111)
num = 439;
// 439 (110110111)
num = 7;
// 7 (111)
num = 92;
// 92 (1011100)
num = 29;
// 29 (11101)
}
}``````

#### Output

`````` Number : 567
Yes
Number : 439
No
Number : 7
Yes
Number : 92
Yes
Number : 29
No``````
``````<?php
/*
Php program
Check whether consecutive decreasing active bits exist in a number
*/
class Sequence
{
// Determine that given number have binary active bits in decreasing order or not
public	function decreasingActiveBit(\$num)
{
if (\$num < 0)
{
return;
}
// Define some auxiliary variables
\$count = 0;
\$result = 0;
\$temp = \$num;
// Execute loop until temp is gratore than zero and
// Consecutive active bit length is decreasing order
// in form of right to left
while (\$temp > 0 && \$result != -1)
{
if ((\$temp & 1) == 1)
{
// Count consecutive active bit length
\$count++;
}
else if (\$count > 0)
{
if (\$result == 0)
{
// Get first sequence of active bits
\$result = \$count;
}
else if (\$result > \$count)
{
// Get next small active bits pair length
\$result = \$count;
}
else
{
// When next active pair length is greater
\$result = -1;
}
\$count = 0;
}
// Shift by one bit to left
\$temp = \$temp >> 1;
}
if (\$result == 0 && \$count > 0)
{
// Whe only first pair exist
\$result = \$count;
}
else if (\$count >= \$result)
{
// When left side consecutive set bits length are not smaller
\$result = -1;
}
// Display given number
echo "\n Number : ". \$num ."";
if (\$result == -1)
{
echo "\n No";
}
else
{
echo "\n Yes";
}
}
}

function main()
{
\$num = 567;
// 567 (1000110111)
\$num = 439;
// 439 (110110111)
\$num = 7;
// 7 (111)
\$num = 92;
// 92 (1011100)
\$num = 29;
// 29 (11101)
}
main();``````

#### Output

`````` Number : 567
Yes
Number : 439
No
Number : 7
Yes
Number : 92
Yes
Number : 29
No``````
``````/*
Node Js program
Check whether consecutive decreasing active bits exist in a number
*/
class Sequence
{
// Determine that given number have binary active bits in decreasing order or not
decreasingActiveBit(num)
{
if (num < 0)
{
return;
}
// Define some auxiliary variables
var count = 0;
var result = 0;
var temp = num;
// Execute loop until temp is gratore than zero and
// Consecutive active bit length is decreasing order
// in form of right to left
while (temp > 0 && result != -1)
{
if ((temp & 1) == 1)
{
// Count consecutive active bit length
count++;
}
else if (count > 0)
{
if (result == 0)
{
// Get first sequence of active bits
result = count;
}
else if (result > count)
{
// Get next small active bits pair length
result = count;
}
else
{
// When next active pair length is greater
result = -1;
}
count = 0;
}
// Shift by one bit to left
temp = temp >> 1;
}
if (result == 0 && count > 0)
{
// Whe only first pair exist
result = count;
}
else if (count >= result)
{
// When left side consecutive set bits length are not smaller
result = -1;
}
// Display given number
process.stdout.write("\n Number : " + num + "");
if (result == -1)
{
process.stdout.write("\n No");
}
else
{
process.stdout.write("\n Yes");
}
}
}

function main()
{
var num = 567;
// 567 (1000110111)
num = 439;
// 439 (110110111)
num = 7;
// 7 (111)
num = 92;
// 92 (1011100)
num = 29;
// 29 (11101)
}
main();``````

#### Output

`````` Number : 567
Yes
Number : 439
No
Number : 7
Yes
Number : 92
Yes
Number : 29
No``````
``````#   Python 3 program
#   Check whether consecutive decreasing active bits exist in a number

class Sequence :
#  Determine that given number have binary active bits in decreasing order or not
def decreasingActiveBit(self, num) :
if (num < 0) :
return

#  Define some auxiliary variables
count = 0
result = 0
temp = num
#  Execute loop until temp is gratore than zero and
#  Consecutive active bit length is decreasing order
#  in form of right to left
while (temp > 0 and result != -1) :
if ((temp & 1) == 1) :
#  Count consecutive active bit length
count += 1

elif(count > 0) :
if (result == 0) :
#  Get first sequence of active bits
result = count

elif(result > count) :
#  Get next small active bits pair length
result = count
else :
#  When next active pair length is greater
result = -1

count = 0

#  Shift by one bit to left
temp = temp >> 1

if (result == 0 and count > 0) :
#  Whe only first pair exist
result = count

elif(count >= result) :
#  When left side consecutive set bits length are not smaller
result = -1

#  Display given number
print("\n Number : ", num ,"", end = "")
if (result == -1) :
print("\n No", end = "")
else :
print("\n Yes", end = "")

def main() :
num = 567
#  567 (1000110111)
num = 439
#  439 (110110111)
num = 7
#  7 (111)
num = 92
#  92 (1011100)
num = 29
#  29 (11101)

if __name__ == "__main__": main()``````

#### Output

`````` Number :  567
Yes
Number :  439
No
Number :  7
Yes
Number :  92
Yes
Number :  29
No``````
``````#   Ruby program
#   Check whether consecutive decreasing active bits exist in a number

class Sequence
#  Determine that given number have binary active bits in decreasing order or not
def decreasingActiveBit(num)
if (num < 0)
return
end

#  Define some auxiliary variables
count = 0
result = 0
temp = num
#  Execute loop until temp is gratore than zero and
#  Consecutive active bit length is decreasing order
#  in form of right to left
while (temp > 0 && result != -1)
if ((temp & 1) == 1)
#  Count consecutive active bit length
count += 1
elsif(count > 0)
if (result == 0)
#  Get first sequence of active bits
result = count
elsif(result > count)
#  Get next small active bits pair length
result = count
else
#  When next active pair length is greater
result = -1
end

count = 0
end

#  Shift by one bit to left
temp = temp >> 1
end

if (result == 0 && count > 0)
#  Whe only first pair exist
result = count
elsif(count >= result)
#  When left side consecutive set bits length are not smaller
result = -1
end

#  Display given number
print("\n Number : ", num ,"")
if (result == -1)
print("\n No")
else
print("\n Yes")
end

end

end

def main()
num = 567
#  567 (1000110111)
num = 439
#  439 (110110111)
num = 7
#  7 (111)
num = 92
#  92 (1011100)
num = 29
#  29 (11101)
end

main()``````

#### Output

`````` Number : 567
Yes
Number : 439
No
Number : 7
Yes
Number : 92
Yes
Number : 29
No``````
``````/*
Scala program
Check whether consecutive decreasing active bits exist in a number
*/
class Sequence
{
// Determine that given number have binary active bits in decreasing order or not
def decreasingActiveBit(num: Int): Unit = {
if (num < 0)
{
return;
}
// Define some auxiliary variables
var count: Int = 0;
var result: Int = 0;
var temp: Int = num;
// Execute loop until temp is gratore than zero and
// Consecutive active bit length is decreasing order
// in form of right to left
while (temp > 0 && result != -1)
{
if ((temp & 1) == 1)
{
// Count consecutive active bit length
count += 1;
}
else if (count > 0)
{
if (result == 0)
{
// Get first sequence of active bits
result = count;
}
else if (result > count)
{
// Get next small active bits pair length
result = count;
}
else
{
// When next active pair length is greater
result = -1;
}
count = 0;
}
// Shift by one bit to left
temp = temp >> 1;
}
if (result == 0 && count > 0)
{
// Whe only first pair exist
result = count;
}
else if (count >= result)
{
// When left side consecutive set bits length are not smaller
result = -1;
}
// Display given number
print("\n Number : " + num + "");
if (result == -1)
{
print("\n No");
}
else
{
print("\n Yes");
}
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: Sequence = new Sequence();
var num: Int = 567;
// 567 (1000110111)
num = 439;
// 439 (110110111)
num = 7;
// 7 (111)
num = 92;
// 92 (1011100)
num = 29;
// 29 (11101)
}
}``````

#### Output

`````` Number : 567
Yes
Number : 439
No
Number : 7
Yes
Number : 92
Yes
Number : 29
No``````
``````/*
Swift 4 program
Check whether consecutive decreasing active bits exist in a number
*/
class Sequence
{
// Determine that given number have binary active bits in decreasing order or not
func decreasingActiveBit(_ num: Int)
{
if (num < 0)
{
return;
}
// Define some auxiliary variables
var count: Int = 0;
var result: Int = 0;
var temp: Int = num;
// Execute loop until temp is gratore than zero and
// Consecutive active bit length is decreasing order
// in form of right to left
while (temp > 0 && result  != -1)
{
if ((temp & 1) == 1)
{
// Count consecutive active bit length
count += 1;
}
else if (count > 0)
{
if (result == 0)
{
// Get first sequence of active bits
result = count;
}
else if (result > count)
{
// Get next small active bits pair length
result = count;
}
else
{
// When next active pair length is greater
result = -1;
}
count = 0;
}
// Shift by one bit to left
temp = temp >> 1;
}
if (result == 0 && count > 0)
{
// Whe only first pair exist
result = count;
}
else if (count >= result)
{
// When left side consecutive set bits length are not smaller
result = -1;
}
// Display given number
print("\n Number : ", num ,"", terminator: "");
if (result == -1)
{
print("\n No", terminator: "");
}
else
{
print("\n Yes", terminator: "");
}
}
}
func main()
{
var num: Int = 567;
// 567 (1000110111)
num = 439;
// 439 (110110111)
num = 7;
// 7 (111)
num = 92;
// 92 (1011100)
num = 29;
// 29 (11101)
}
main();``````

#### Output

`````` Number :  567
Yes
Number :  439
No
Number :  7
Yes
Number :  92
Yes
Number :  29
No``````
``````/*
Kotlin program
Check whether consecutive decreasing active bits exist in a number
*/
class Sequence
{
// Determine that given number have binary active bits in decreasing order or not
fun decreasingActiveBit(num: Int): Unit
{
if (num < 0)
{
return;
}
// Define some auxiliary variables
var count: Int = 0;
var result: Int = 0;
var temp: Int = num;
// Execute loop until temp is gratore than zero and
// Consecutive active bit length is decreasing order
// in form of right to left
while (temp > 0 && result != -1)
{
if ((temp and 1) == 1)
{
// Count consecutive active bit length
count += 1;
}
else if (count > 0)
{
if (result == 0)
{
// Get first sequence of active bits
result = count;
}
else if (result > count)
{
// Get next small active bits pair length
result = count;
}
else
{
// When next active pair length is greater
result = -1;
}
count = 0;
}
// Shift by one bit to left
temp = temp shr 1;
}
if (result == 0 && count > 0)
{
// Whe only first pair exist
result = count;
}
else if (count >= result)
{
// When left side consecutive set bits length are not smaller
result = -1;
}
// Display given number
print("\n Number : " + num + "");
if (result == -1)
{
print("\n No");
}
else
{
print("\n Yes");
}
}
}
fun main(args: Array <String> ): Unit
{
var num: Int = 567;
// 567 (1000110111)
num = 439;
// 439 (110110111)
num = 7;
// 7 (111)
num = 92;
// 92 (1011100)
num = 29;
// 29 (11101)
}``````

#### Output

`````` Number : 567
Yes
Number : 439
No
Number : 7
Yes
Number : 92
Yes
Number : 29
No``````

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