# 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 :**

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Input:19Output: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 :183Output :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.

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++

`// 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;` `}` |

## Python3

`# Python3 program to find if bit-pattern` `# of a number has increasing value of` `# continuous-1 or not.` `# Returns true if n has increasing count of` `# continuous-1 else false` `def` `findContinuous1(n):` ` ` ` ` `# store the bit-pattern of n into` ` ` `# bit bitset- bp` ` ` `bp ` `=` `list` `(` `bin` `(n))` ` ` ` ` `bits ` `=` `len` `(bp)` ` ` ` ` `# set prev_count = 0 and curr_count = 0.` ` ` `prev_count ` `=` `0` ` ` `curr_count ` `=` `0` ` ` ` ` `i ` `=` `0` ` ` `while` `(i < bits):` ` ` `if` `(bp[i] ` `=` `=` `'1'` `):` ` ` `# increment current count of continuous-1` ` ` `curr_count ` `+` `=` `1` ` ` `i ` `+` `=` `1` ` ` ` ` `# traverse all continuous-0` ` ` `elif` `(bp[i ` `-` `1` `] ` `=` `=` `'0'` `):` ` ` `i ` `+` `=` `1` ` ` `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 ` `+` `=` `1` ` ` `prev_count ` `=` `curr_count` ` ` `curr_count ` `=` `0` ` ` ` ` `# check for last sequence of continuous-1` ` ` `if` `(prev_count > curr_count ` `and` `(curr_count !` `=` `0` `)):` ` ` `return` `0` ` ` ` ` `return` `1` ` ` `# Driver code` `n ` `=` `179` `if` `(findContinuous1(n)):` ` ` `print` `( ` `"Yes"` `)` `else` `:` ` ` `print` `( ` `"No"` `)` `# This code is contributed by SHUBHAMSINGH10` |

## Javascript

`<script>` `// JavaScript program to find if bit-pattern` `// of a number has increasing value of` `// continuous-1 or not.` `function` `dec2bin(dec) {` ` ` `return` `(dec >>> 0).toString(2);` `}` `// Returns true if n has increasing count of` `// continuous-1 else false` `function` `findContinuous1(n){` ` ` `// store the bit-pattern of n into` ` ` `// bit bitset- bp` ` ` `let bp = dec2bin(n)` ` ` `let bits = bp.length` ` ` ` ` `// set prev_count = 0 and curr_count = 0.` ` ` `let prev_count = 0` ` ` `let curr_count = 0` ` ` ` ` `let i = 0` ` ` `while` `(i < bits){` ` ` `if` `(bp[i] == ` `'1'` `){` ` ` `// increment current count of continuous-1` ` ` `curr_count += 1;` ` ` `i += 1;` ` ` `}` ` ` `// traverse all continuous-0` ` ` `else` `if` `(bp[i - 1] == ` `'0'` `){` ` ` `i += 1` ` ` `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 += 1` ` ` `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` `n = 179` `if` `(findContinuous1(n))` ` ` `document.write( ` `"Yes"` `)` `else` ` ` `document.write( ` `"No"` `)` `</script>` |

**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 the implementation.

## C++

`// 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;` `}` |

## Java

`// Java program to check if counts of` `// consecutive 1s are increasing order.` `import` `java .io.*;` `class` `GFG {` ` ` ` ` `// Returns true if n has counts of` ` ` `// consecutive 1's are increasing` ` ` `// order.` ` ` `static` `boolean` `areSetBitsIncreasing(` `int` `n)` ` ` `{` ` ` ` ` `// Initialize previous count` ` ` `int` `prev_count = Integer.MAX_VALUE;` ` ` ` ` `// 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` ` ` `static` `public` `void` `main (String[] args)` ` ` `{` ` ` `int` `n = ` `10` `;` ` ` ` ` `if` `(areSetBitsIncreasing(n))` ` ` `System.out.println(` `"Yes"` `);` ` ` `else` ` ` `System.out.println(` `"No"` `);` ` ` `}` `}` `// This code is contributed by anuj_67.` |

## Python3

`# Python3 program to check if counts of` `# consecutive 1s are increasing order.` `import` `sys` `# Returns true if n has counts of` `# consecutive 1's are increasing order.` `def` `areSetBitsIncreasing(n):` ` ` `# Initialize previous count` ` ` `prev_count ` `=` `sys.maxsize` ` ` `# 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` `and` `n ` `%` `2` `=` `=` `0` `):` ` ` `n ` `=` `int` `(n` `/` `2` `)` ` ` `# Count current set bits` ` ` `curr_count ` `=` `1` ` ` `while` `(n > ` `0` `and` `n ` `%` `2` `=` `=` `1` `):` ` ` ` ` `n ` `=` `n` `/` `2` ` ` `curr_count ` `+` `=` `1` ` ` ` ` `# Compare current with previous` ` ` `# and update previous.` ` ` `if` `(curr_count >` `=` `prev_count):` ` ` `return` `False` ` ` `prev_count ` `=` `curr_count` ` ` `return` `True` `# Driver code` `n ` `=` `10` `if` `(areSetBitsIncreasing(n)):` ` ` `print` `(` `"Yes"` `)` `else` `:` ` ` `print` `(` `"No"` `)` ` ` `# This code is contributed by Smitha` |

## C#

`// C# program to check if counts of` `// consecutive 1s are increasing order.` `using` `System;` `class` `GFG {` ` ` ` ` `// Returns true if n has counts of` ` ` `// consecutive 1's are increasing` ` ` `// order.` ` ` `static` `bool` `areSetBitsIncreasing(` `int` `n)` ` ` `{` ` ` ` ` `// Initialize previous count` ` ` `int` `prev_count = ` `int` `.MaxValue;` ` ` ` ` `// 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` ` ` `static` `public` `void` `Main ()` ` ` `{` ` ` `int` `n = 10;` ` ` ` ` `if` `(areSetBitsIncreasing(n))` ` ` `Console.WriteLine(` `"Yes"` `);` ` ` `else` ` ` `Console.WriteLine(` `"No"` `);` ` ` `}` `}` `// This code is contributed by anuj_67.` |

## PHP

`<?php` `// PHP program to check if` `// counts of consecutive` `// 1s are increasing order.` `// Returns true if n has` `// counts of consecutive` `// 1's are increasing order.` `function` `areSetBitsIncreasing( ` `$n` `)` `{` ` ` `// Initialize previous count` ` ` `$prev_count` `= PHP_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` `= ` `$n` `/ 2;` ` ` `// Count current set bits` ` ` `$curr_count` `= 1;` ` ` `while` `(` `$n` `> 0 ` `and` `$n` `% 2 == 1)` ` ` `{` ` ` `$n` `= ` `$n` `/ 2;` ` ` `$curr_count` `++;` ` ` `}` ` ` `// Compare current with previous` ` ` `// and update previous.` ` ` `if` `(` `$curr_count` `>= ` `$prev_count` `)` ` ` `return` `false;` ` ` `$prev_count` `= ` `$curr_count` `;` ` ` `}` ` ` `return` `true;` `}` `// Driver code` `$n` `= 10;` `if` `(areSetBitsIncreasing(` `$n` `))` ` ` `echo` `"Yes"` `;` `else` ` ` `echo` `"No"` `;` `// This code is contributed by anuj_67` `?>` |

## Javascript

`<script>` `// Javascript program to check if counts of` `// consecutive 1s are increasing order.` `// Returns true if n has counts of` `// consecutive 1's are increasing` `// order.` `function` `areSetBitsIncreasing(n)` `{` ` ` ` ` `// Initialize previous count` ` ` `var` `prev_count = Number.MAX_VALUE;` ` ` `// 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 = parseInt(n / 2);` ` ` `// Count current set bits` ` ` `var` `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` `var` `n = 10;` `if` `(areSetBitsIncreasing(n))` ` ` `document.write(` `"Yes"` `);` `else` ` ` `document.write(` `"No"` `);` `// This code is contributed by Rajput-Ji` `</script>` |

**Output :**

No

**Time Complexity: **O(log_{2}n)

**Auxiliary Space: **O(1)

This article is contributed by **Shivam Pradhan (anuj_charm)**. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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