leetcode

commit 60f3439c8377ef8b99b5055b471840d53ab200a0
parent 48f4961febc7d03056cd826682e4b3277c7b20ca
Author: Jack Sangdahl <jack@pngu.org>
Date:   Sat, 27 Sep 2025 20:37:04 -0600

461. hamming distance

Diffstat:
A
hamming_distance.c
 | 
22
++++++++++++++++++++++
A
hamming_distance.py
 | 
27
+++++++++++++++++++++++++++
A
largest_perimeter_triangle.c
 | 
24
++++++++++++++++++++++++
A
largest_perimeter_triangle.py
 | 
26
++++++++++++++++++++++++++

4 files changed, 99 insertions(+), 0 deletions(-)
diff --git a/hamming_distance.c b/hamming_distance.c
@@ -0,0 +1,22 @@
+// 461. Hamming Distance
+#include "leetcode.h"
+
+/*
+ * the hamming distance between two integers is the number of positions at which
+ * the corresponding bits are different. given two integers 'x' and 'y', return
+ * the hamming distance between them.
+ */
+
+int hammingDistance(int x, int y) {
+  int ans = 0, n = x ^ y;
+  while (n) {
+    ++ans;
+    n &= n - 1;
+  }
+  return ans;
+}
+
+int main() {
+  printf("%d\n", hammingDistance(1, 4)); // expect: 2
+  printf("%d\n", hammingDistance(3, 1)); // expect: 1
+}
diff --git a/hamming_distance.py b/hamming_distance.py
@@ -0,0 +1,27 @@
+# 461. Hamming Distance
+
+"""
+the hamming distance between two integers is the number of positions at which
+the corresponding bits are different. given two integers 'x' and 'y', return
+the hamming distance between them.
+"""
+
+
+class Solution(object):
+    def hammingDistance(self, x, y):
+        """
+        :type x: int
+        :type y: int
+        :rtype: int
+        """
+        ans, n = 0, x ^ y
+        while n:
+            ans += 1
+            n &= n - 1
+        return ans
+
+
+if __name__ == "__main__":
+    obj = Solution()
+    print(obj.hammingDistance(1, 4))
+    print(obj.hammingDistance(3, 1))
diff --git a/largest_perimeter_triangle.c b/largest_perimeter_triangle.c
@@ -0,0 +1,24 @@
+// 976. Largest Perimeter Triangle
+#include "leetcode.h"
+
+/*
+ * given an integer array 'nums', return the largest perimeter of a triangle
+ * with a non-zero area, formed from three of these lengths. if it is impossible
+ * to form any triangle of a non-zero area, return 0.
+ */
+
+int cmp(const void *a, const void *b) { return (*(int *)a - *(int *)b); }
+
+int largestPerimeter(int *nums, int numsSize) {
+  qsort(nums, numsSize, sizeof(int), cmp);
+  for (int i = numsSize - 3; i >= 0; i--)
+    if (nums[i] < nums[i - 1] + nums[i - 2])
+      return nums[i] + nums[i] + nums[i - 2];
+  return 0;
+}
+
+int main() {
+  int n1[] = {2, 1, 2}, n2[] = {3, 2, 3, 4};
+  printf("%d\n", largestPerimeter(n1, ARRAY_SIZE(n1))); // expect: 5
+  printf("%d\n", largestPerimeter(n2, ARRAY_SIZE(n2))); // expect: 0
+}
diff --git a/largest_perimeter_triangle.py b/largest_perimeter_triangle.py
@@ -0,0 +1,26 @@
+# 976. Largest Perimeter Triangle
+
+"""
+given an integer array 'nums', return the largest perimeter of a triangle
+with a non-zero area, formed from three of these lengths. if it is impossible
+to form any triangle of a non-zero area, return 0.
+"""
+
+
+class Solution(object):
+    def largestPerimeter(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        nums = sorted(nums)[::-1]
+        for i in range(len(nums) - 2):
+            if nums[i] < nums[i + 1] + nums[i + 2]:
+                return nums[i] + nums[i + 1] + nums[i + 2]
+        return 0
+
+
+if __name__ == "__main__":
+    obj = Solution()
+    print(obj.largestPerimeter(nums=[2, 1, 2]))
+    print(obj.largestPerimeter(nums=[3, 2, 3, 4]))