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: 5 Output: 5 Explanation: Here are the non-negative integers <= 5 with their corresponding binary representations: 0 : 0 1 : 1 2 : 10 3 : 11 4 : 100 5 : 101 Among 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;
}
};