leetcode

commit 9e00d6582e7b3758610b8dab014180dc15171c00
parent 090091712aea5abfb0b7dfc17db3a00e6fc131b5
Author: Jack Sangdahl <jack@pngu.org>
Date:   Thu,  1 Jan 2026 19:45:06 -0700

961. n-repeated element in size 2n array

Diffstat:
A
n_repeat_elem_size_2n_arr.c
 | 
28
++++++++++++++++++++++++++++
A
n_repeat_elem_size_2n_arr.py
 | 
24
++++++++++++++++++++++++

2 files changed, 52 insertions(+), 0 deletions(-)
diff --git a/n_repeat_elem_size_2n_arr.c b/n_repeat_elem_size_2n_arr.c
@@ -0,0 +1,28 @@
+// 961. N-Repeated Element in Size 2N Array
+#include "leetcode.h"
+
+/*
+ * you are given an integer array 'nums' with the following properties:
+ * 'nums.length == 2 * n', 'nums' contains 'n + 1' unique elements, and exactly
+ * one element of 'nums' is repeated 'n' times. return the element that is
+ * repeated 'n' times.
+ */
+
+int repeatedNTimes(int *nums, int numsSize) {
+  bool hash[10001] = {false};
+  for (int i = 0; i < numsSize; ++i) {
+    if (hash[nums[i]])
+      return nums[i];
+    else
+      hash[nums[i]] = true;
+  }
+  return -1;
+}
+
+int main() {
+  int n1[] = {1, 2, 3, 3}, n2[] = {2, 1, 2, 5, 3, 2},
+      n3[] = {5, 1, 5, 2, 5, 3, 5, 4};
+  printf("%d\n", repeatedNTimes(n1, ARRAY_SIZE(n1))); // expect: 3
+  printf("%d\n", repeatedNTimes(n2, ARRAY_SIZE(n2))); // expect: 2
+  printf("%d\n", repeatedNTimes(n3, ARRAY_SIZE(n3))); // expect: 5
+}
diff --git a/n_repeat_elem_size_2n_arr.py b/n_repeat_elem_size_2n_arr.py
@@ -0,0 +1,24 @@
+# 961. N-Repeated Element in Size 2N Array
+
+"""
+you are given an integer array 'nums' with the following properties:
+'nums.length == 2  n', 'nums' contains 'n + 1' unique elements, and exactly
+one element of 'nums' is repeated 'n' times. return the element that is
+repeated 'n' times.
+"""
+
+
+class Solution(object):
+    def repeatedNTimes(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        return int((sum(nums) - sum(set(nums))) // (len(nums) // 2 - 1))
+
+
+if __name__ == "__main__":
+    obj = Solution()
+    print(obj.repeatedNTimes(nums=[1, 2, 3, 3]))
+    print(obj.repeatedNTimes(nums=[2, 1, 2, 5, 3, 2]))
+    print(obj.repeatedNTimes(nums=[5, 1, 5, 2, 5, 3, 5, 4]))