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This approach utilizes dynamic programming to count the number of non-negative integers without consecutive ones. By breaking down the problem, we can avoid redundant work by storing solutions to subproblems.
Time Complexity: O(1) since we iterate over a fixed number of bits (30 bits for int).
Space Complexity: O(1) for storing a constant size dp array.
1class Solution:
2 def findIntegers(self, num: int) -> int:
3 dp = [0] * 31
4 dp[0] = 1
5 dp[1] = 2
6 for i in range(2, 31):
7 dp[i] = dp[i - 1] + dp[i - 2]
8
9 prev_bit = 0
10 result = 0
11
12 for i in range(29, -1, -1):
13 if (num & (1 << i)) != 0:
14 result += dp[i]
15 if prev_bit == 1:
16 return result
17 prev_bit = 1
18 else:
19 prev_bit = 0
20
21 return result + 1
22
23if __name__ == "__main__":
24 sol = Solution()
25 n = 5
26 print(sol.findIntegers(n))In Python, the solution similarly utilizes a dp list to manage the counts of valid integer representations without consecutive ones for binary integers up to 31-bit. This implementation iterates over each bit of num to calcualte the total valid integers.
In this approach, we tackle the problem recursively and use memoization to store and retrieve solutions to subproblems, thereby optimizing overlapping subproblem calculations.
Time Complexity: Generally O(log n), as the recursion iterates over each bit once.
Space Complexity: O(log n) due to the recursive call stack and memo storage.
1#include <iostream>
2#include <vector>
3#include <unordered_map>
4using namespace std;
5
6class Solution {
private:
unordered_map<int, int> memo;
int dfs(int pos, bool is_tight, bool prev_one, vector<int>& bits) {
if (pos == -1) return 1;
if (!is_tight && memo.count(pos * 2 + prev_one)) return memo[pos * 2 + prev_one];
int limit = is_tight ? bits[pos] : 1;
int res = 0;
for (int i = 0; i <= limit; ++i) {
if (prev_one && i == 1) continue;
res += dfs(pos - 1, is_tight && i == bits[pos], i == 1, bits);
}
if (!is_tight) memo[pos * 2 + prev_one] = res;
return res;
}
public:
int findIntegers(int num) {
vector<int> bits;
while (num) {
bits.push_back(num & 1);
num >>= 1;
}
return dfs(bits.size() - 1, true, false, bits);
}
};
int main() {
Solution sol;
int n = 5;
cout << sol.findIntegers(n) << endl;
return 0;
}This C++ solution implements depth-first search recursion, coupled with an unordered_map that serves as the memoization strategy. The function performs bitwise checks and combines overlaps using the memo to prevent redundant calculations.