# Check if bits of a number has count of consecutive set bits in increasing order

Given a integer n > 0, the task is to find whether in the bit pattern of integer count of continuous 1’s are in increasing from left to right.
Examples:

```Input:19
Output:Yes
Explanation: Bit-pattern of 19 = 10011,
Counts of continuous 1's from left to right
are 1, 2 which are in increasing order.

Input  : 183
Output : yes
Explanation: Bit-pattern of 183 = 10110111,
Counts of continuous 1's from left to right
are 1, 2, 3 which are in increasing order.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to store binary representation of given number into a string, then traverse from left to right and count the number of continuous 1’s. For every encounter of 0 check the value of previous count of continuous 1’s to that of current value, if the value of previous count is greater than the value of current count then return False, Else when string ends return True.

```// C++ program to find if bit-pattern
// of a number has increasing value of
// continuous-1 or not.
#include<bits/stdc++.h>
using namespace std;

// Returns true if n has increasing count of
// continuous-1 else false
bool findContinuous1(int n)
{
const int bits = 8*sizeof(int);

// store the bit-pattern of n into
// bit bitset- bp
string bp = bitset <bits>(n).to_string();

// set prev_count = 0 and curr_count = 0.
int prev_count = 0, curr_count = 0;

int i = 0;
while (i < bits)
{
if (bp[i] == '1')
{
// increment current count of continuous-1
curr_count++;
i++;
}

// traverse all continuous-0
else if (bp[i-1] == '0')
{
i++;
curr_count = 0;
continue;
}

// check  prev_count and curr_count
// on encounter of first zero after
// continuous-1s
else
{
if (curr_count < prev_count)
return 0;
i++;
prev_count=curr_count;
curr_count = 0;
}
}

// check for last sequence of continuous-1
if (prev_count > curr_count && (curr_count != 0))
return 0;

return 1;
}

// Driver code
int main()
{
int n = 179;
if (findContinuous1(n))
cout << "Yes";
else
cout << "No";

return 0;
}
```

Output:

```Yes
```

An efficient solution is to use decimal to binary conversion loop that divides number by 2 and take remainder as bit. This loop finds bits from right to left. So we check if right to left is in decreasing order or not.

Below is C++ implementation.

```// C++ program to check if counts of consecutive
// 1s are increasing order.
#include<bits/stdc++.h>
using namespace std;

// Returns true if n has counts of consecutive
// 1's are increasing order.
bool areSetBitsIncreasing(int n)
{
// Initialize previous count
int prev_count = INT_MAX;

// We traverse bits from right to left
// and check if counts are decreasing
// order.
while (n > 0)
{
// Ignore 0s until we reach a set bit.
while (n > 0 && n % 2 == 0)
n = n/2;

// Count current set bits
int curr_count = 1;
while (n > 0 && n % 2 == 1)
{
n = n/2;
curr_count++;
}

// Compare current with previous and
// update previous.
if (curr_count >= prev_count)
return false;
prev_count = curr_count;
}

return true;
}

// Driver code
int main()
{
int n = 10;
if (areSetBitsIncreasing(n))
cout << "Yes";
else
cout << "No";

return 0;
}
```

Output:

```Yes
```

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