Given a positive integer, print the next smallest and the next largest number that have the same number of 1 bits in their binary representation.
Example1:
Input: num = 2 (0b10)
Output: [4, 1] ([0b100, 0b1])
Example2:
Input: num = 1
Output: [2, -1]
Note:
1 <= num <= 2147483647- If there is no next smallest or next largest number, output -1.
class Solution:
def findClosedNumbers(self, num: int) -> List[int]:
ans = [-1] * 2
dirs = (0, 1, 0)
for p in range(2):
a, b = dirs[p], dirs[p + 1]
x = num
for i in range(1, 31):
if (x >> i & 1) == a and (x >> (i - 1) & 1) == b:
x ^= 1 << i
x ^= 1 << (i - 1)
j, k = 0, i - 2
while j < k:
while j < k and (x >> j & 1) == b:
j += 1
while j < k and (x >> k & 1) == a:
k -= 1
if j < k:
x ^= 1 << j
x ^= 1 << k
ans[p] = x
break
return ansclass Solution {
public int[] findClosedNumbers(int num) {
int[] ans = {-1, -1};
int[] dirs = {0, 1, 0};
for (int p = 0; p < 2; ++p) {
int a = dirs[p], b = dirs[p + 1];
int x = num;
for (int i = 1; i < 31; ++i) {
if ((x >> i & 1) == a && (x >> (i - 1) & 1) == b) {
x ^= 1 << i;
x ^= 1 << (i - 1);
int j = 0, k = i - 2;
while (j < k) {
while (j < k && (x >> j & 1) == b) {
++j;
}
while (j < k && (x >> k & 1) == a) {
--k;
}
if (j < k) {
x ^= 1 << j;
x ^= 1 << k;
}
}
ans[p] = x;
break;
}
}
}
return ans;
}
}class Solution {
public:
vector<int> findClosedNumbers(int num) {
vector<int> ans(2, -1);
int dirs[3] = {0, 1, 0};
for (int p = 0; p < 2; ++p) {
int a = dirs[p], b = dirs[p + 1];
int x = num;
for (int i = 1; i < 31; ++i) {
if ((x >> i & 1) == a && (x >> (i - 1) & 1) == b) {
x ^= 1 << i;
x ^= 1 << (i - 1);
int j = 0, k = i - 2;
while (j < k) {
while (j < k && (x >> j & 1) == b) {
++j;
}
while (j < k && (x >> k & 1) == a) {
--k;
}
if (j < k) {
x ^= 1 << j;
x ^= 1 << k;
}
}
ans[p] = x;
break;
}
}
}
return ans;
}
};func findClosedNumbers(num int) []int {
ans := []int{-1, -1}
dirs := [3]int{0, 1, 0}
for p := 0; p < 2; p++ {
a, b := dirs[p], dirs[p+1]
x := num
for i := 1; i < 31; i++ {
if x>>i&1 == a && x>>(i-1)&1 == b {
x ^= 1 << i
x ^= 1 << (i - 1)
j, k := 0, i-2
for j < k {
for j < k && x>>j&1 == b {
j++
}
for j < k && x>>k&1 == a {
k--
}
if j < k {
x ^= 1 << j
x ^= 1 << k
}
}
ans[p] = x
break
}
}
}
return ans
}function findClosedNumbers(num: number): number[] {
const ans: number[] = [-1, -1];
const dirs: number[] = [0, 1, 0];
for (let p = 0; p < 2; ++p) {
const [a, b] = [dirs[p], dirs[p + 1]];
let x = num;
for (let i = 1; i < 31; ++i) {
if (((x >> i) & 1) === a && ((x >> (i - 1)) & 1) === b) {
x ^= 1 << i;
x ^= 1 << (i - 1);
let [j, k] = [0, i - 2];
while (j < k) {
while (j < k && ((x >> j) & 1) === b) {
++j;
}
while (j < k && ((x >> k) & 1) === a) {
--k;
}
if (j < k) {
x ^= 1 << j;
x ^= 1 << k;
}
}
ans[p] = x;
break;
}
}
}
return ans;
}