Description
Given an integer n, return the decimal value of the binary string formed by concatenating the binary representations of 1 to n in order, modulo 109 + 7.
Example 1:
Input: n = 1
Output: 1
Explanation: "1" in binary corresponds to the decimal value 1.
Example 2:
Input: n = 3
Output: 27
Explanation: In binary, 1, 2, and 3 corresponds to "1", "10", and "11".
After concatenating them, we have "11011", which corresponds to the decimal value 27.
Example 3:
Input: n = 12
Output: 505379714
Explanation: The concatenation results in "1101110010111011110001001101010111100".
The decimal value of that is 118505380540.
After modulo 109 + 7, the result is 505379714.
Constraints:
Solutions
Solution 1: Bit Manipulation
By observing the pattern of number concatenation, we can find that when concatenating to the i-th number, the result ans formed by concatenating the previous i-1 numbers is actually shifted to the left by a certain number of bits, and then i is added. The number of bits shifted is the number of binary digits in i.
The time complexity is O(n), where n is the given integer. The space complexity is O(1).
PythonJavaC++GoTypeScriptRustJavaScriptC#
class Solution:
def concatenatedBinary(self, n: int) -> int:
mod = 10**9 + 7
ans = 0
for i in range(1, n + 1):
ans = (ans << i.bit_length() | i) % mod
return ans(code-box)
class Solution {
public int concatenatedBinary(int n) {
final int mod = (int) 1e9 + 7;
long ans = 0;
for (int i = 1; i <= n; ++i) {
ans = (ans << (32 - Integer.numberOfLeadingZeros(i)) | i) % mod;
}
return (int) ans;
}
}(code-box)
class Solution {
public:
int concatenatedBinary(int n) {
const int mod = 1e9 + 7;
long long ans = 0;
for (int i = 1; i <= n; ++i) {
ans = (ans << (32 - __builtin_clz(i)) | i) % mod;
}
return ans;
}
};(code-box)
func concatenatedBinary(n int) (ans int) {
const mod = 1e9 + 7
for i := 1; i <= n; i++ {
ans = (ans<<bits.Len(uint(i)) | i) % mod
}
return
}(code-box)
function concatenatedBinary(n: number): number {
const mod = 1_000_000_007;
let ans = 0;
for (let i = 1; i <= n; i++) {
ans = (((ans * (1 << (32 - Math.clz32(i)))) % mod) + i) % mod;
}
return ans;
}(code-box)
impl Solution {
pub fn concatenated_binary(n: i32) -> i32 {
let mod_: i64 = 1_000_000_007;
let mut ans: i64 = 0;
for i in 1..=n as i64 {
let bit_length: u32 = 64 - i.leading_zeros() as u32;
ans = ((ans << bit_length) | i) % mod_;
}
ans as i32
}
}(code-box)
/**
* @param {number} n
* @return {number}
*/
var concatenatedBinary = function (n) {
const mod = 1_000_000_007;
let ans = 0;
for (let i = 1; i <= n; i++) {
ans = (((ans * (1 << (32 - Math.clz32(i)))) % mod) + i) % mod;
}
return ans;
};(code-box)
public class Solution {
public int ConcatenatedBinary(int n) {
const int mod = 1000000007;
long ans = 0;
for (int i = 1; i <= n; ++i) {
int bitLength = 32 - System.Numerics.BitOperations.LeadingZeroCount((uint)i);
ans = ((ans << bitLength) | i) % mod;
}
return (int)ans;
}
}(code-box)
Solution 2: Bit Manipulation (Optimization)
In Solution 1, we need to calculate the number of binary digits of i each time, which adds some extra computation. We can use a variable shift to record the current number of bits to shift. When i is a power of 2, shift needs to be incremented by 1.
The time complexity is O(n), where n is the given integer. The space complexity is O(1).
PythonJavaC++GoTypeScriptRustJavaScriptC#
class Solution:
def concatenatedBinary(self, n: int) -> int:
mod = 10**9 + 7
ans = shift = 0
for i in range(1, n + 1):
if (i & (i - 1)) == 0:
shift += 1
ans = (ans << shift | i) % mod
return ans(code-box)
class Solution {
public int concatenatedBinary(int n) {
final int mod = (int) 1e9 + 7;
long ans = 0;
int shift = 0;
for (int i = 1; i <= n; ++i) {
if ((i & (i - 1)) == 0) {
++shift;
}
ans = (ans << shift | i) % mod;
}
return (int) ans;
}
}(code-box)
class Solution {
public:
int concatenatedBinary(int n) {
const int mod = 1e9 + 7;
long ans = 0;
int shift = 0;
for (int i = 1; i <= n; ++i) {
if ((i & (i - 1)) == 0) {
++shift;
}
ans = (ans << shift | i) % mod;
}
return ans;
}
};(code-box)
func concatenatedBinary(n int) (ans int) {
const mod = 1e9 + 7
shift := 0
for i := 1; i <= n; i++ {
if i&(i-1) == 0 {
shift++
}
ans = (ans<<shift | i) % mod
}
return
}(code-box)
function concatenatedBinary(n: number): number {
const mod = 1_000_000_007;
let ans = 0;
let shift = 0;
for (let i = 1; i <= n; i++) {
if ((i & (i - 1)) === 0) {
shift++;
}
ans = (((ans * (1 << shift)) % mod) + i) % mod;
}
return ans;
}(code-box)
impl Solution {
pub fn concatenated_binary(n: i32) -> i32 {
let mod_: i64 = 1_000_000_007;
let mut ans: i64 = 0;
let mut shift: u32 = 0;
for i in 1..=n as i64 {
if (i & (i - 1)) == 0 {
shift += 1;
}
ans = ((ans << shift) | i) % mod_;
}
ans as i32
}
}(code-box)
/**
* @param {number} n
* @return {number}
*/
var concatenatedBinary = function (n) {
const mod = 1_000_000_007;
let ans = 0;
let shift = 0;
for (let i = 1; i <= n; i++) {
if ((i & (i - 1)) === 0) {
shift++;
}
ans = (((ans * (1 << shift)) % mod) + i) % mod;
}
return ans;
};(code-box)
public class Solution {
public int ConcatenatedBinary(int n) {
const int mod = 1000000007;
long ans = 0;
int shift = 0;
for (int i = 1; i <= n; ++i) {
if ((i & (i - 1)) == 0) {
++shift;
}
ans = ((ans << shift) | i) % mod;
}
return (int)ans;
}
}(code-box)
