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+# 3689. Maximum Total Subarray Value I
+
+[LeetCode Link](https://leetcode.com/problems/maximum-total-subarray-value-i/)
+
+Difficulty: Medium
+Topics: Array, Greedy
+Acceptance Rate: 67.9%
+
+## Hints
+
+### Hint 1
+
+Notice the unusual freedom in this problem: subarrays may overlap **and** you can pick the exact same subarray more than once. Whenever a problem lets you reuse the same choice repeatedly, ask yourself: "What is the single best choice, and what stops me from just taking it every time?"
+
+### Hint 2
+
+Think about what bounds the **value** of a subarray, `max - min`. For any subarray `nums[l..r]`, its maximum can never exceed the maximum of the whole array, and its minimum can never go below the minimum of the whole array. So how large can a single subarray's value possibly get, and which subarray achieves that bound?
+
+### Hint 3
+
+The whole array `nums[0..n-1]` always contains the global maximum and the global minimum, so its value is exactly `max(nums) - min(nums)` — and no subarray can beat that. Since you must pick exactly `k` subarrays and you are allowed to repeat, the optimal move is to pick that single best subarray all `k` times. The answer is simply `k * (max(nums) - min(nums))`.
+
+## Approach
+
+This problem looks intimidating because of the "choose `k` subarrays" framing, but the constraints hide a one-line greedy result.
+
+Step 1 — Bound a single subarray's value. For any subarray, `max(subarray) <= max(nums)` and `min(subarray) >= min(nums)`. Therefore:
+
+```
+value(subarray) = max(subarray) - min(subarray) <= max(nums) - min(nums)
+```
+
+Step 2 — Show the bound is attainable. The full array is itself a valid subarray, and it contains both the global max and the global min, so its value equals exactly `max(nums) - min(nums)`. No subarray can do better.
+
+Step 3 — Use repetition. We must choose exactly `k` subarrays, overlaps are allowed, and duplicates are allowed. Since every chosen subarray contributes at most `max(nums) - min(nums)`, the best total is achieved by choosing the full array (or any subarray spanning both extremes) all `k` times:
+
+```
+answer = k * (max(nums) - min(nums))
+```
+
+Let's verify with the examples:
+
+- `nums = [1,3,2], k = 2`: max = 3, min = 1, diff = 2, answer = `2 * 2 = 4`. ✅
+- `nums = [4,2,5,1], k = 3`: max = 5, min = 1, diff = 4, answer = `3 * 4 = 12`. ✅
+
+So the algorithm is: scan once to find the global maximum and minimum, then multiply their difference by `k`. The hard part is the realization, not the coding — be honest with yourself about whether you saw *why* the whole array dominates every other choice.
+
+A subtle but important detail: `k` can be up to `10^5` and the difference up to `10^9`, so the product can reach `10^14`. That overflows 32-bit integers, so make sure the result is computed in 64-bit arithmetic.
+
+## Complexity Analysis
+
+Time Complexity: O(n) — a single pass to find the minimum and maximum.
+Space Complexity: O(1) — only a couple of scalar variables.
+
+## Edge Cases
+
+- **Single element array (`n == 1`)**: max equals min, so the difference is 0 and the answer is 0 regardless of `k`. The greedy formula handles this naturally.
+- **All equal elements**: every subarray has value 0, so the total is 0. Again handled automatically.
+- **Large `k` and large values**: the product `k * (max - min)` can exceed the 32-bit integer range, so the computation must use 64-bit integers to avoid overflow.
+- **`nums[i] = 0`**: zeros are valid; the minimum may legitimately be 0, which is fine.
diff --git a/problems/3689-maximum-total-subarray-value-i/problem.md b/problems/3689-maximum-total-subarray-value-i/problem.md
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+---
+number: "3689"
+frontend_id: "3689"
+title: "Maximum Total Subarray Value I"
+slug: "maximum-total-subarray-value-i"
+difficulty: "Medium"
+topics:
+  - "Array"
+  - "Greedy"
+acceptance_rate: 6787.1
+is_premium: false
+created_at: "2026-06-09T04:43:36.205822+00:00"
+fetched_at: "2026-06-09T04:43:36.205822+00:00"
+link: "https://leetcode.com/problems/maximum-total-subarray-value-i/"
+date: "2026-06-09"
+---
+
+# 3689. Maximum Total Subarray Value I
+
+You are given an integer array `nums` of length `n` and an integer `k`.
+
+You need to choose **exactly** `k` non-empty subarrays `nums[l..r]` of `nums`. Subarrays may overlap, and the exact same subarray (same `l` and `r`) **can** be chosen more than once.
+
+The **value** of a subarray `nums[l..r]` is defined as: `max(nums[l..r]) - min(nums[l..r])`.
+
+The **total value** is the sum of the **values** of all chosen subarrays.
+
+Return the **maximum** possible total value you can achieve.
+
+ 
+
+**Example 1:**
+
+**Input:** nums = [1,3,2], k = 2
+
+**Output:** 4
+
+**Explanation:**
+
+One optimal approach is:
+
+  * Choose `nums[0..1] = [1, 3]`. The maximum is 3 and the minimum is 1, giving a value of `3 - 1 = 2`.
+  * Choose `nums[0..2] = [1, 3, 2]`. The maximum is still 3 and the minimum is still 1, so the value is also `3 - 1 = 2`.
+
+
+
+Adding these gives `2 + 2 = 4`.
+
+**Example 2:**
+
+**Input:** nums = [4,2,5,1], k = 3
+
+**Output:** 12
+
+**Explanation:**
+
+One optimal approach is:
+
+  * Choose `nums[0..3] = [4, 2, 5, 1]`. The maximum is 5 and the minimum is 1, giving a value of `5 - 1 = 4`.
+  * Choose `nums[0..3] = [4, 2, 5, 1]`. The maximum is 5 and the minimum is 1, so the value is also `4`.
+  * Choose `nums[2..3] = [5, 1]`. The maximum is 5 and the minimum is 1, so the value is again `4`.
+
+
+
+Adding these gives `4 + 4 + 4 = 12`.
+
+ 
+
+**Constraints:**
+
+  * `1 <= n == nums.length <= 5 * 10​​​​​​​4`
+  * `0 <= nums[i] <= 109`
+  * `1 <= k <= 105`
diff --git a/problems/3689-maximum-total-subarray-value-i/solution_daily_20260609.go b/problems/3689-maximum-total-subarray-value-i/solution_daily_20260609.go
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+package main
+
+// Maximum Total Subarray Value I
+//
+// The value of any subarray is max(subarray) - min(subarray), which can never
+// exceed max(nums) - min(nums). The full array attains that bound, and since we
+// may choose the same subarray repeatedly, the best total is to pick that single
+// best subarray all k times. Thus the answer is k * (max(nums) - min(nums)).
+//
+// We use int64 for the result because k (up to 1e5) times the difference (up to
+// 1e9) can reach 1e14, which overflows 32-bit integers.
+//
+// Time: O(n), Space: O(1).
+func maxTotalValue(nums []int, k int) int64 {
+	mn, mx := nums[0], nums[0]
+	for _, v := range nums {
+		if v < mn {
+			mn = v
+		}
+		if v > mx {
+			mx = v
+		}
+	}
+	return int64(k) * int64(mx-mn)
+}
diff --git a/problems/3689-maximum-total-subarray-value-i/solution_daily_20260609_test.go b/problems/3689-maximum-total-subarray-value-i/solution_daily_20260609_test.go
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index 0000000..5d0fb9a
--- /dev/null
+++ b/problems/3689-maximum-total-subarray-value-i/solution_daily_20260609_test.go
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+package main
+
+import "testing"
+
+func TestMaxTotalValue(t *testing.T) {
+	tests := []struct {
+		name     string
+		nums     []int
+		k        int
+		expected int64
+	}{
+		{"example 1: [1,3,2] k=2", []int{1, 3, 2}, 2, 4},
+		{"example 2: [4,2,5,1] k=3", []int{4, 2, 5, 1}, 3, 12},
+		{"edge case: single element", []int{7}, 5, 0},
+		{"edge case: all equal elements", []int{3, 3, 3, 3}, 4, 0},
+		{"edge case: k=1 takes whole-array span", []int{10, 0, 5}, 1, 10},
+		{"edge case: large product needs int64", []int{0, 1000000000}, 100000, 100000000000000},
+		{"edge case: contains zero as min", []int{0, 2, 9, 4}, 3, 27},
+	}
+
+	for _, tt := range tests {
+		t.Run(tt.name, func(t *testing.T) {
+			result := maxTotalValue(tt.nums, tt.k)
+			if result != tt.expected {
+				t.Errorf("maxTotalValue(%v, %d) = %d, want %d", tt.nums, tt.k, result, tt.expected)
+			}
+		})
+	}
+}