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authorJack Sangdahl <jack@pngu.org>2024-09-13 18:23:35 -0600
committerJack Sangdahl <jack@pngu.org>2024-09-13 18:23:35 -0600
commit7d6ce37fbb409bc5c6675eeabb58166abe463aef (patch)
tree607fc6f4fffec8688b8b070368cff3dd807c2b83
parentf6175e306d3fb120d8f83804f57467fca72c9baf (diff)
downloadleetcode-7d6ce37fbb409bc5c6675eeabb58166abe463aef.tar.gz
leetcode-7d6ce37fbb409bc5c6675eeabb58166abe463aef.zip
2419. longest subarray with maximum bitwise and
-rw-r--r--longest_subarr_max_bitwise_and.c35
-rw-r--r--longest_subarr_max_bitwise_and.py27
2 files changed, 62 insertions, 0 deletions
diff --git a/longest_subarr_max_bitwise_and.c b/longest_subarr_max_bitwise_and.c
new file mode 100644
index 0000000..17b9155
--- /dev/null
+++ b/longest_subarr_max_bitwise_and.c
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+// 2419. Longest Subarray With Maximum Bitwise AND
+#include "leetcode.h"
+
+/*
+ * you are given an integer array 'nums' of size 'n'. consider a non-empty
+ * subarray from 'nums' that has the maximum possible bitwise and. in other
+ * words, let 'k' be the maximum value of the bitwise and of any subarray of
+ * 'nums'. then, only subarrays with a bitwise and equal to 'k' should be
+ * considered. return the length of the longest such subarray. the bitwise and
+ * of an array is the bitwise and of all the numbers in it. a subarray is a
+ * contiguous sequence of elements within an array
+ */
+
+int longestSubarray(int *nums, int numsSize) {
+ int max = 0, ans = 0;
+ for (int i = 0, j = 0; i < numsSize; i++) {
+ if (nums[i] > max) {
+ max = nums[i];
+ j = 1;
+ ans = 1;
+ } else if (nums[i] == max)
+ j++;
+ else
+ j = 0;
+ if (j > ans)
+ ans = j;
+ }
+ return ans;
+}
+
+int main() {
+ int n1[] = {1, 2, 3, 3, 2, 2}, n2[] = {1, 2, 3, 4};
+ printf("%d\n", longestSubarray(n1, ARRAY_SIZE(n1))); // expect: 2
+ printf("%d\n", longestSubarray(n2, ARRAY_SIZE(n2))); // expect: 1
+}
diff --git a/longest_subarr_max_bitwise_and.py b/longest_subarr_max_bitwise_and.py
new file mode 100644
index 0000000..95d3a36
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+++ b/longest_subarr_max_bitwise_and.py
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+# 2419. Longest Subarray With Maximum Bitwise AND
+from itertools import groupby
+
+"""
+you are given an integer array 'nums' of size 'n'. consider a non-empty
+subarray from 'nums' that has the maximum possible bitwise and. in other
+words, let 'k' be the maximum value of the bitwise and of any subarray of
+'nums'. then, only subarrays with a bitwise and equal to 'k' should be
+considered. return the length of the longest such subarray. the bitwise and
+of an array is the bitwise and of all the numbers in it. a subarray is a
+contiguous sequence of elements within an array
+"""
+
+
+class Solution(object):
+ def longestSubarray(self, nums):
+ """
+ :type nums: List[int]
+ :rtype: int
+ """
+ return max(len(list(i)) for n, i in groupby(nums) if n == max(nums))
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.longestSubarray(nums=[1, 2, 3, 3, 2, 2]))
+ print(obj.longestSubarray(nums=[1, 2, 3, 4]))