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-rw-r--r--closest_prime_num_range.c61
1 files changed, 61 insertions, 0 deletions
diff --git a/closest_prime_num_range.c b/closest_prime_num_range.c
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+// 2523. Closest Prime Numbers in Range
+#include "leetcode.h"
+
+/*
+ * given two positive integers 'left' and 'right', find the two integers 'num1'
+ * and 'num2' such that 'left <= num1 < num2 <= right', both 'num1' and 'num2'
+ * are prime numbers, and 'num2 - num1' is the minimum amongst all other pairs
+ * satisfying the above conditions. return the positive integer array 'ans =
+ * [nums1, nums2]'. if there are multiple pairs satisfying these conditions,
+ * return the one with the smallest 'num1' value. if no such numbers exist,
+ * return '[-1, -1]'.
+ */
+
+int *closestPrimes(int left, int right, int *returnSize) {
+ int *ans = (int *)malloc(2 * sizeof(int));
+ *returnSize = 2;
+ ans[0] = -1;
+ ans[1] = -1;
+
+ bool *sieve = (bool *)malloc((right + 1) * sizeof(bool));
+ memset(sieve, true, (right + 1) * sizeof(bool));
+ sieve[0] = sieve[1] = false;
+ for (int i = 2; i * i <= right; i++)
+ if (sieve[i])
+ for (int j = i * i; j <= right; j += i)
+ sieve[j] = false;
+ int *prime = (int *)malloc((right - left + 1) * sizeof(int));
+ int idx = 0;
+ for (int i = left; i <= right; i++)
+ if (sieve[i])
+ prime[idx++] = i;
+ free(sieve);
+ if (idx < 2) {
+ free(prime);
+ return ans;
+ }
+ int min = INT_MAX;
+ for (int i = 1; i < idx; i++) {
+ int diff = prime[i] - prime[i - 1];
+ if (min > diff) {
+ min = diff;
+ ans[0] = prime[i - 1];
+ ans[1] = prime[i];
+ }
+ }
+ free(prime);
+ return ans;
+}
+
+int main() {
+ int rs1, *cp1 = closestPrimes(10, 19, &rs1);
+ int rs2, *cp2 = closestPrimes(4, 6, &rs2);
+ for (int i = 0; i < rs1; i++)
+ printf("%d ", cp1[i]); // expect: 11 13
+ printf("\n");
+ for (int i = 0; i < rs2; i++)
+ printf("%d ", cp2[i]); // expect: -1 -1
+ printf("\n");
+ free(cp1);
+ free(cp2);
+}