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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.
1#include <stdio.h>
2#include <string.h>
3
4int findIntegers(int num) {
5 int dp[31] = {1, 2};
6 for(int i = 2; i < 31; i++) {
7 dp[i] = dp[i - 1] + dp[i - 2];
8 }
9 int prev_bit = 0, result = 0;
10
11 for (int i = 29; i >= 0; i--) {
12 if ((num & (1 << i)) != 0) {
13 result += dp[i];
14 if (prev_bit == 1) {
15 return result;
16 }
17 prev_bit = 1;
18 } else {
19 prev_bit = 0;
20 }
21 }
22 return result + 1;
23}
24
25int main() {
26 int n = 5;
27 printf("%d\n", findIntegers(n));
28 return 0;
29}The C solution utilizes a dynamic programming array dp to store the number of valid integers for each bit length. As we iterate over each bit in n, we decide whether to add the number of valid binary numbers of that length. We keep track of consecutive 1s and adjust our result accordingly. The final result is incremented to account for the number itself if valid.
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.