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authorJack Sangdahl <[email protected]>2025-07-28 20:58:11 -0600
committerJack Sangdahl <[email protected]>2025-07-28 20:58:11 -0600
commit64ccda3aff7e186e03bdc963a8ab110d2bc01e19 (patch)
treed3aba8d067fefe1072a97fc774759a73e7a03e2d
parent59613ec9d3a899b43dd9c788cb1c0d14719eff76 (diff)
downloadleetcode-64ccda3aff7e186e03bdc963a8ab110d2bc01e19.tar.gz
leetcode-64ccda3aff7e186e03bdc963a8ab110d2bc01e19.zip
2411. smallest subarrays with maximum bitwise or
-rw-r--r--smallest_subarr_max_or.c41
-rw-r--r--smallest_subarr_max_or.py35
2 files changed, 76 insertions, 0 deletions
diff --git a/smallest_subarr_max_or.c b/smallest_subarr_max_or.c
new file mode 100644
index 0000000..99c4a1f
--- /dev/null
+++ b/smallest_subarr_max_or.c
@@ -0,0 +1,41 @@
+// 2411. Smallest Subarrays With Maximum Bitwise OR
+#include "leetcode.h"
+
+/*
+ * you are given a 0-indexed array 'nums' of length 'n', consisting of
+ * non-negative integers. for each index 'i' from 0 to 'n - 1', you must
+ * determine the size of the minimum sized non-empty subarray of 'nums' starting
+ * at 'i' that has the maximum possible bitwise or. the bitwosee or of an array
+ * is the bitwise or of all the numbers in it. return an integer array 'answer'
+ * of size 'n' where 'answer[i]' is the length of the minimum sized subarray
+ * starting at 'i' with maximum bitwise or. a subarray is a contiguous non-empty
+ * sequence of elements within an array.
+ */
+
+int *smallestSubarrays(int *nums, int numsSize, int *returnSize) {
+ int prev[30], *ans = (int *)malloc(numsSize * sizeof(int));
+ memset(ans, 1, numsSize * sizeof(int));
+ *returnSize = numsSize;
+ for (int i = numsSize - 1; i >= 0; --i)
+ for (int j = 0; j < 30; ++j) {
+ if (nums[i] & (1 << j))
+ prev[j] = i;
+ ans[i] = fmax(ans[i], prev[j] - i + 1);
+ }
+ return ans;
+}
+
+int main() {
+ int n1[] = {1, 0, 2, 1, 3}, rs1;
+ int n2[] = {1, 2}, rs2;
+ int *sb1 = smallestSubarrays(n1, ARRAY_SIZE(n1), &rs1);
+ int *sb2 = smallestSubarrays(n2, ARRAY_SIZE(n2), &rs2);
+ for (int i = 0; i < rs1; i++)
+ printf("%d ", sb1[i]); // expect: 3 3 2 2 1
+ printf("\n");
+ for (int i = 0; i < rs2; i++)
+ printf("%d ", sb2[i]); // expect: 2 1
+ printf("\n");
+ free(sb1);
+ free(sb2);
+}
diff --git a/smallest_subarr_max_or.py b/smallest_subarr_max_or.py
new file mode 100644
index 0000000..a0506fc
--- /dev/null
+++ b/smallest_subarr_max_or.py
@@ -0,0 +1,35 @@
+# 2411. Smallest Subarrays With Maximum Bitwise OR
+
+"""
+you are given a 0-indexed array 'nums' of length 'n', consisting of
+non-negative integers. for each index 'i' from 0 to 'n - 1', you must
+determine the size of the minimum sized non-empty subarray of 'nums' starting
+at 'i' that has the maximum possible bitwise or. the bitwosee or of an array
+is the bitwise or of all the numbers in it. return an integer array 'answer'
+of size 'n' where 'answer[i]' is the length of the minimum sized subarray
+starting at 'i' with maximum bitwise or. a subarray is a contiguous non-empty
+sequence of elements within an array.
+"""
+
+
+class Solution(object):
+ def smallestSubarrays(self, nums):
+ """
+ :type nums: List[int]
+ :rtype: List[int]
+ """
+ prev = [0] * 32
+ n = len(nums)
+ ans = [0] * n
+ for i in range(n - 1, -1, -1):
+ for j in range(32):
+ if nums[i] & (1 << j):
+ prev[j] = i
+ ans[i] = max(1, max(prev) - i + 1)
+ return ans
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.smallestSubarrays(nums=[1, 0, 2, 1, 3]))
+ print(obj.smallestSubarrays(nums=[1, 2]))