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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.
1class Solution:
2 def __init__(self):
The Python solution mirrors the C++ solution by breaking down the integer into binary bits and evaluating recursively through depth-first search. It stores computed results in a dictionary to avoid re-calculating overlapping subproblems.