diff --git a/problems/3867-sum-of-gcd-of-formed-pairs/analysis.md b/problems/3867-sum-of-gcd-of-formed-pairs/analysis.md
new file mode 100644
index 0000000..fa20010
--- /dev/null
+++ b/problems/3867-sum-of-gcd-of-formed-pairs/analysis.md
@@ -0,0 +1,93 @@
+# 3867. Sum of GCD of Formed Pairs
+
+[LeetCode Link](https://leetcode.com/problems/sum-of-gcd-of-formed-pairs/)
+
+Difficulty: Medium
+Topics: Array, Math, Two Pointers, Sorting, Simulation, Number Theory
+Acceptance Rate: 69.6%
+
+## Hints
+
+### Hint 1
+
+The problem reads like a chain of separate steps: build one array, sort it, pair
+things up, then reduce. Don't try to be clever about merging the steps — this is a
+simulation problem. Handle each stage faithfully and the pieces fall into place. The
+only questions are: how do I build `prefixGcd` efficiently, and how do I pair the
+sorted elements without an expensive nested loop?
+
+### Hint 2
+
+For `prefixGcd`, notice that `mxi` is just a running (prefix) maximum. You can compute
+the whole array in one pass while carrying the max seen so far. For the pairing step,
+"smallest unpaired with largest unpaired" after sorting is the textbook signal for the
+**two-pointers** technique: one pointer at the front, one at the back, walking inward.
+
+### Hint 3
+
+Once `prefixGcd` is sorted, set `left = 0` and `right = n - 1`. While `left < right`,
+add `gcd(prefixGcd[left], prefixGcd[right])` to the answer, then move `left++` and
+`right--`. The loop condition `left < right` handles the odd-length case for free: the
+single middle element is exactly where the two pointers meet, so it is never paired and
+never counted. No special-casing required.
+
+## Approach
+
+The task is a direct simulation with two efficiency-sensitive stages.
+
+**Stage 1 — Build `prefixGcd`.** For each index `i`, `mxi` is the maximum of
+`nums[0..i]`, which is a prefix maximum. Keep a variable `mx` that we update to
+`max(mx, nums[i])` as we scan left to right, and set
+`prefixGcd[i] = gcd(nums[i], mx)`. This is a single `O(n)` pass (each `gcd` is
+`O(log V)`), avoiding any `O(n^2)` recomputation of the max.
+
+For example, with `nums = [3, 6, 2, 8]`:
+
+- `i=0`: `mx=3`, `gcd(3,3)=3`
+- `i=1`: `mx=6`, `gcd(6,6)=6`
+- `i=2`: `mx=6`, `gcd(2,6)=2`
+- `i=3`: `mx=8`, `gcd(8,8)=8`
+
+So `prefixGcd = [3, 6, 2, 8]`.
+
+**Stage 2 — Sort.** Sort `prefixGcd` in non-decreasing order: `[2, 3, 6, 8]`.
+
+**Stage 3 — Pair with two pointers.** The rule "pair the smallest unpaired with the
+largest unpaired, repeatedly" is exactly a two-pointer sweep from both ends of the
+sorted array. Start `left = 0`, `right = n - 1`:
+
+- `left=0, right=3`: `gcd(2, 8) = 2`
+- `left=1, right=2`: `gcd(3, 6) = 3`
+- `left=2, right=1`: `left < right` fails, stop.
+
+Sum `= 2 + 3 = 5`, matching the expected output.
+
+When `n` is odd, the two pointers eventually satisfy `left == right` on the middle
+element, and the `left < right` guard stops the loop before it is used — so the middle
+element is naturally ignored, as the problem requires.
+
+Computing `gcd` uses the Euclidean algorithm. Since values can be up to `10^9`, the sum
+of GCDs across up to ~`5 * 10^4` pairs can exceed a 32-bit range in the worst case
+(each GCD up to `10^9`), so accumulate the answer in a 64-bit integer to be safe.
+
+## Complexity Analysis
+
+Time Complexity: O(n log n) — the sort dominates; building `prefixGcd` is `O(n log V)`
+and the two-pointer pass is `O(n log V)`, where `V` is the maximum value.
+Space Complexity: O(n) — for the `prefixGcd` array (or O(1) auxiliary beyond it,
+ignoring the sort's internal usage).
+
+## Edge Cases
+
+- **Single element (`n == 1`):** `prefixGcd` has one element; no pair can be formed and
+  the answer is `0`. The two-pointer loop never executes since `left < right` is false
+  immediately.
+- **Odd length:** The middle element after sorting must be skipped. The `left < right`
+  condition handles this automatically — no explicit middle-index arithmetic needed.
+- **All equal values:** e.g. `[5, 5, 5, 5]` gives `prefixGcd = [5, 5, 5, 5]`, and each
+  pair contributes `gcd(5, 5) = 5`. A good sanity check that pairing/summing works.
+- **Large values (up to `10^9`) and many pairs:** Accumulate the sum in a 64-bit
+  integer to avoid overflow, since the total can exceed the 32-bit range.
+- **Strictly increasing input:** `mxi == nums[i]` at every index, so
+  `prefixGcd[i] = gcd(nums[i], nums[i]) = nums[i]`; a useful case to confirm the prefix
+  max is tracked correctly.
diff --git a/problems/3867-sum-of-gcd-of-formed-pairs/problem.md b/problems/3867-sum-of-gcd-of-formed-pairs/problem.md
new file mode 100644
index 0000000..0805f9d
--- /dev/null
+++ b/problems/3867-sum-of-gcd-of-formed-pairs/problem.md
@@ -0,0 +1,95 @@
+---
+number: "3867"
+frontend_id: "3867"
+title: "Sum of GCD of Formed Pairs"
+slug: "sum-of-gcd-of-formed-pairs"
+difficulty: "Medium"
+topics:
+  - "Array"
+  - "Math"
+  - "Two Pointers"
+  - "Sorting"
+  - "Simulation"
+  - "Number Theory"
+acceptance_rate: 6958.0
+is_premium: false
+created_at: "2026-07-16T03:45:53.859780+00:00"
+fetched_at: "2026-07-16T03:45:53.859780+00:00"
+link: "https://leetcode.com/problems/sum-of-gcd-of-formed-pairs/"
+date: "2026-07-16"
+---
+
+# 3867. Sum of GCD of Formed Pairs
+
+You are given an integer array `nums` of length `n`.
+
+Construct an array `prefixGcd` where for each index `i`:
+
+  * Let `mxi = max(nums[0], nums[1], ..., nums[i])`.
+  * `prefixGcd[i] = gcd(nums[i], mxi)`.
+
+
+
+After constructing `prefixGcd`:
+
+  * Sort `prefixGcd` in **non-decreasing** order.
+  * Form pairs by taking the **smallest unpaired** element and the **largest unpaired** element.
+  * Repeat this process until no more pairs can be formed.
+  * For each formed pair, **compute** the `gcd` of the two elements.
+  * If `n` is odd, the **middle** element in the `prefixGcd` array remains **unpaired** and should be ignored.
+
+
+
+Return an integer denoting the **sum of the GCD** values of all formed pairs.
+
+The term `gcd(a, b)` denotes the **greatest common divisor** of `a` and `b`. 
+
+ 
+
+**Example 1:**
+
+**Input:** nums = [2,6,4]
+
+**Output:** 2
+
+**Explanation:**
+
+Construct `prefixGcd`:
+
+`i` | `nums[i]` | `mxi` | `prefixGcd[i]`  
+---|---|---|---  
+0 | 2 | 2 | 2  
+1 | 6 | 6 | 6  
+2 | 4 | 6 | 2  
+  
+`prefixGcd = [2, 6, 2]`. After sorting, it forms `[2, 2, 6]`.
+
+Pair the smallest and largest elements: `gcd(2, 6) = 2`. The remaining middle element 2 is ignored. Thus, the sum is 2.
+
+**Example 2:**
+
+**Input:** nums = [3,6,2,8]
+
+**Output:** 5
+
+**Explanation:**
+
+Construct `prefixGcd`:
+
+`i` | `nums[i]` | `mxi` | `prefixGcd[i]`  
+---|---|---|---  
+0 | 3 | 3 | 3  
+1 | 6 | 6 | 6  
+2 | 2 | 6 | 2  
+3 | 8 | 8 | 8  
+  
+`prefixGcd = [3, 6, 2, 8]`. After sorting, it forms `[2, 3, 6, 8]`.
+
+Form pairs: `gcd(2, 8) = 2` and `gcd(3, 6) = 3`. Thus, the sum is `2 + 3 = 5`.
+
+ 
+
+**Constraints:**
+
+  * `1 <= n == nums.length <= 105`
+  * `1 <= nums[i] <= 10​​​​​​​9`
diff --git a/problems/3867-sum-of-gcd-of-formed-pairs/solution_daily_20260716.go b/problems/3867-sum-of-gcd-of-formed-pairs/solution_daily_20260716.go
new file mode 100644
index 0000000..0fb2205
--- /dev/null
+++ b/problems/3867-sum-of-gcd-of-formed-pairs/solution_daily_20260716.go
@@ -0,0 +1,48 @@
+package main
+
+import "sort"
+
+// Approach: Simulate the three stages directly.
+//  1. Build prefixGcd in one pass using a running prefix maximum:
+//     prefixGcd[i] = gcd(nums[i], max(nums[0..i])).
+//  2. Sort prefixGcd in non-decreasing order.
+//  3. Pair the smallest unpaired with the largest unpaired using two pointers
+//     (left at the front, right at the back) and sum gcd(left, right) for each
+//     pair. The `left < right` guard leaves the middle element unpaired when n
+//     is odd, exactly as required.
+//
+// The answer is accumulated in an int64 to avoid overflow, since values can be
+// up to 1e9 across many pairs.
+
+// gcd returns the greatest common divisor of a and b using the Euclidean
+// algorithm. Assumes non-negative inputs (constraints guarantee nums[i] >= 1).
+func gcd(a, b int) int {
+	for b != 0 {
+		a, b = b, a%b
+	}
+	return a
+}
+
+func sumGcd(nums []int) int64 {
+	n := len(nums)
+
+	// Stage 1: build prefixGcd with a running maximum.
+	prefixGcd := make([]int, n)
+	mx := 0
+	for i, v := range nums {
+		if v > mx {
+			mx = v
+		}
+		prefixGcd[i] = gcd(v, mx)
+	}
+
+	// Stage 2: sort in non-decreasing order.
+	sort.Ints(prefixGcd)
+
+	// Stage 3: two-pointer pairing from both ends.
+	var sum int64
+	for left, right := 0, n-1; left < right; left, right = left+1, right-1 {
+		sum += int64(gcd(prefixGcd[left], prefixGcd[right]))
+	}
+	return sum
+}
diff --git a/problems/3867-sum-of-gcd-of-formed-pairs/solution_daily_20260716_test.go b/problems/3867-sum-of-gcd-of-formed-pairs/solution_daily_20260716_test.go
new file mode 100644
index 0000000..41461a9
--- /dev/null
+++ b/problems/3867-sum-of-gcd-of-formed-pairs/solution_daily_20260716_test.go
@@ -0,0 +1,56 @@
+package main
+
+import "testing"
+
+func TestSumGcd(t *testing.T) {
+	tests := []struct {
+		name     string
+		nums     []int
+		expected int64
+	}{
+		{
+			name:     "example 1: [2,6,4] -> sorted [2,2,6], gcd(2,6)=2",
+			nums:     []int{2, 6, 4},
+			expected: 2,
+		},
+		{
+			name:     "example 2: [3,6,2,8] -> sorted [2,3,6,8], gcd(2,8)+gcd(3,6)=5",
+			nums:     []int{3, 6, 2, 8},
+			expected: 5,
+		},
+		{
+			name:     "edge case: single element has no pairs",
+			nums:     []int{7},
+			expected: 0,
+		},
+		{
+			name:     "edge case: two identical elements, gcd(5,5)=5",
+			nums:     []int{5, 5},
+			expected: 5,
+		},
+		{
+			name:     "edge case: odd length middle ignored",
+			nums:     []int{5, 5, 5, 5, 5},
+			expected: 10, // sorted [5,5,5,5,5]: gcd(5,5)+gcd(5,5)=10, middle skipped
+		},
+		{
+			name:     "edge case: strictly increasing keeps values as prefixGcd",
+			nums:     []int{1, 2, 3, 4},
+			expected: 2, // prefixGcd [1,2,3,4]: gcd(1,4)+gcd(2,3)=1+1=2
+		},
+		{
+			name:     "edge case: large values, no overflow",
+			nums:     []int{1000000000, 1000000000},
+			expected: 1000000000, // gcd(1e9,1e9)=1e9
+		},
+	}
+
+	for _, tt := range tests {
+		t.Run(tt.name, func(t *testing.T) {
+			result := sumGcd(tt.nums)
+			if result != tt.expected {
+				t.Errorf("sumGcd(%v) = %d, want %d", tt.nums, result, tt.expected)
+			}
+		})
+	}
+}