Problem statement:

Given a positive integer n, find the number of non-negative integers less than or equal to n, whose binary representations do NOT contain consecutive ones.

Example 1:

Input: 5Output: 5Explanation: Here are the non-negative integers <= 5 with their corresponding binary representations:0 : 01 : 12 : 103 : 114 : 1005 : 101Among them, only integer 3 disobeys the rule (two consecutive ones) and the other 5 satisfy the rule. 

Note: 1 <= n <= 10^9

Solution:

This is the last question of leetcode weekly contest 34. It is a DP and definitely is hard problem. But, the DP formula is not difficult to figure it out. What we need to be careful is there is one requirement in the description: less than or equal  to n.  After we get the answer, we still need to do one more step.

Generally, this problem can be divided into several steps:

(1) Convert the original n into a binary representation string. Get the size of string to allocate memory for DP array.

(2)Do DP, like 198. House Robber, we should define two dp arrays:

• dp0[i]: the number of integers when current bit set to 0
• dp1[i]: the number of integers when current bit set to 1

(3)Any integer can not contain any consecutive ones.

• dp0[i] = dp1[i - 1] + dp0[i - 1]
• dp1[i] = dp0[i - 1]

(4) And do the last processing to find the integers which are less than or equal to n.

Time complexity is O(k), k is the bit count of n.

class Solution {public:    int findIntegers(int num) {        string str_num;        while(num){            str_num.push_back(num % 2 + ‘0‘);            num /= 2;        }        int size = str_num.size();        vector<int> dp0(size, 0);        vector<int> dp1(size, 0);        dp0[0] = 1;        dp1[0] = 1;        for(int i = 1; i < size; i++){            dp0[i] = dp0[i - 1] + dp1[i - 1];            dp1[i] = dp0[i - 1];        }        int cnt = dp0[size - 1] + dp1[size - 1];        for (int i = size - 2; i >= 0; i--) {            if (str_num[i] == ‘1‘ && str_num[i + 1] == ‘1‘) {                break;            }            if (str_num[i] == ‘0‘ && str_num[i + 1] == ‘0‘) {                cnt -= dp1[i];            }        }        return cnt;    }};

600. Non-negative Integers without Consecutive Ones