From 2f30a4c03782529d3882862b1e60b3e873c43d38 Mon Sep 17 00:00:00 2001
From: Samuel Ortion <samuel@ortion.fr>
Date: Mon, 27 Oct 2025 20:15:33 +0100
Subject: [PATCH] solve: 136. Single Number

C solution with XOR operator
---
 c/136.single-number/136.single-number.org | 65 +++++++++++++++++++++++
 1 file changed, 65 insertions(+)
 create mode 100644 c/136.single-number/136.single-number.org

diff --git a/c/136.single-number/136.single-number.org b/c/136.single-number/136.single-number.org
new file mode 100644
index 0000000..7766a61
--- /dev/null
+++ b/c/136.single-number/136.single-number.org
@@ -0,0 +1,65 @@
+#+title: 136 Single Number
+
+Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.
+
+You must implement a solution with a linear runtime complexity and use only constant extra space.
+
+
+
+Example 1:
+
+Input: nums = [2,2,1]
+
+Output: 1
+
+Example 2:
+
+Input: nums = [4,1,2,1,2]
+
+Output: 4
+
+Example 3:
+
+Input: nums = [1]
+
+Output: 1
+
+
+
+Constraints:
+
+    1 <= nums.length <= 3 * 104
+    -3 * 104 <= nums[i] <= 3 * 104
+    Each element in the array appears twice except for one element which appears only once.
+
+* Solution
+
+#+name: solution
+#+begin_src C
+int xor(int a, int b) {
+   return a ^ b;
+}
+
+int singleNumber(int* nums, int numsSize) {
+    int number = 0;
+    for (int i=0; i < numsSize; i++) {
+        number = xor(number, nums[i]);
+    }
+    return number;
+}
+#+end_src
+
+#+RESULTS: solution
+
+#+begin_src C :noweb yes
+#include <stdio.h>
+
+<<solution>>
+
+int numbers[3] = {1, 1, 2};
+int single = singleNumber(numbers, 3);
+printf("%d\n", single);
+#+end_src
+
+#+RESULTS:
+: -2