2523. Closest Prime Numbers in Range
Description
Given two positive integers left and right, find the two integers num1 and num2 such that:
left <= num1 < num2 <= right.- Both
num1andnum2are prime numbers. num2 - num1is the minimum amongst all other pairs satisfying the above conditions.
Return the positive integer array ans = [num1, num2]. If there are multiple pairs satisfying these conditions, return the one with the smallest num1 value. If no such numbers exist, return [-1, -1].
Example 1:
Input: left = 10, right = 19 Output: [11,13] Explanation: The prime numbers between 10 and 19 are 11, 13, 17, and 19. The closest gap between any pair is 2, which can be achieved by [11,13] or [17,19]. Since 11 is smaller than 17, we return the first pair.
Example 2:
Input: left = 4, right = 6 Output: [-1,-1] Explanation: There exists only one prime number in the given range, so the conditions cannot be satisfied.
Constraints:
1 <= left <= right <= 106
Solutions
Solution 1: Linear Sieve
For the given range $[\textit{left}, \textit{right}]$, we can use the linear sieve method to find all prime numbers. Then, we traverse the prime numbers in ascending order to find the pair of adjacent prime numbers with the smallest difference, which will be the answer.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n = \textit{right}$.
Python3
class Solution:
def closestPrimes(self, left: int, right: int) -> List[int]:
cnt = 0
st = [False] * (right + 1)
prime = [0] * (right + 1)
for i in range(2, right + 1):
if not st[i]:
prime[cnt] = i
cnt += 1
j = 0
while prime[j] <= right // i:
st[prime[j] * i] = 1
if i % prime[j] == 0:
break
j += 1
p = [v for v in prime[:cnt] if left <= v <= right]
mi = inf
ans = [-1, -1]
for a, b in pairwise(p):
if (d := b - a) < mi:
mi = d
ans = [a, b]
return ans
Java
class Solution {
public int[] closestPrimes(int left, int right) {
int cnt = 0;
boolean[] st = new boolean[right + 1];
int[] prime = new int[right + 1];
for (int i = 2; i <= right; ++i) {
if (!st[i]) {
prime[cnt++] = i;
}
for (int j = 0; prime[j] <= right / i; ++j) {
st[prime[j] * i] = true;
if (i % prime[j] == 0) {
break;
}
}
}
int i = -1, j = -1;
for (int k = 0; k < cnt; ++k) {
if (prime[k] >= left && prime[k] <= right) {
if (i == -1) {
i = k;
}
j = k;
}
}
int[] ans = new int[] {-1, -1};
if (i == j || i == -1) {
return ans;
}
int mi = 1 << 30;
for (int k = i; k < j; ++k) {
int d = prime[k + 1] - prime[k];
if (d < mi) {
mi = d;
ans[0] = prime[k];
ans[1] = prime[k + 1];
}
}
return ans;
}
}
C++
class Solution {
public:
vector<int> closestPrimes(int left, int right) {
int cnt = 0;
bool st[right + 1];
memset(st, 0, sizeof st);
int prime[right + 1];
for (int i = 2; i <= right; ++i) {
if (!st[i]) {
prime[cnt++] = i;
}
for (int j = 0; prime[j] <= right / i; ++j) {
st[prime[j] * i] = true;
if (i % prime[j] == 0) {
break;
}
}
}
int i = -1, j = -1;
for (int k = 0; k < cnt; ++k) {
if (prime[k] >= left && prime[k] <= right) {
if (i == -1) {
i = k;
}
j = k;
}
}
vector<int> ans = {-1, -1};
if (i == j || i == -1) return ans;
int mi = 1 << 30;
for (int k = i; k < j; ++k) {
int d = prime[k + 1] - prime[k];
if (d < mi) {
mi = d;
ans[0] = prime[k];
ans[1] = prime[k + 1];
}
}
return ans;
}
};
Go
func closestPrimes(left int, right int) []int {
cnt := 0
st := make([]bool, right+1)
prime := make([]int, right+1)
for i := 2; i <= right; i++ {
if !st[i] {
prime[cnt] = i
cnt++
}
for j := 0; prime[j] <= right/i; j++ {
st[prime[j]*i] = true
if i%prime[j] == 0 {
break
}
}
}
i, j := -1, -1
for k := 0; k < cnt; k++ {
if prime[k] >= left && prime[k] <= right {
if i == -1 {
i = k
}
j = k
}
}
ans := []int{-1, -1}
if i == j || i == -1 {
return ans
}
mi := 1 << 30
for k := i; k < j; k++ {
d := prime[k+1] - prime[k]
if d < mi {
mi = d
ans[0], ans[1] = prime[k], prime[k+1]
}
}
return ans
}