questions A (01-matrix - avoid-flood)

backend/data/questions/append-k-integers-with-minimal-sum.yaml

@@ -0,0 +1,191 @@
+title: Append K Integers With Minimal Sum
+slug: append-k-integers-with-minimal-sum
+difficulty: medium
+leetcode_id: 2195
+leetcode_url: https://leetcode.com/problems/append-k-integers-with-minimal-sum/
+categories:
+  - arrays
+  - math
+  - sorting
+patterns:
+  - greedy
+
+description: |
+  You are given an integer array `nums` and an integer `k`. Append `k` **unique positive** integers that do **not** appear in `nums` to `nums` such that the resulting total sum is **minimum**.
+
+  Return *the sum of the* `k` *integers appended to* `nums`.
+
+constraints: |
+  - `1 <= nums.length <= 10^5`
+  - `1 <= nums[i] <= 10^9`
+  - `1 <= k <= 10^8`
+
+examples:
+  - input: "nums = [1,4,25,10,25], k = 2"
+    output: "5"
+    explanation: "The two unique positive integers that do not appear in nums which we append are 2 and 3. The sum of the two integers appended is 2 + 3 = 5."
+  - input: "nums = [5,6], k = 6"
+    output: "25"
+    explanation: "The six unique positive integers that do not appear in nums which we append are 1, 2, 3, 4, 7, and 8. The sum is 1 + 2 + 3 + 4 + 7 + 8 = 25."
+
+explanation:
+  intuition: |
+    To minimise the sum, we want to pick the **smallest possible positive integers** that aren't already in `nums`. Imagine counting from 1 upward: 1, 2, 3, 4, ... and skipping any number that already exists in the array.
+
+    The key insight is that we don't need to iterate through each number one by one. Instead, we can use the **arithmetic series formula** to calculate sums of consecutive integers in bulk: the sum of integers from `1` to `n` is `n * (n + 1) / 2`.
+
+    Think of it like filling gaps: if `nums` contains some numbers that "block" certain positions, we need to count how many integers we can pick before hitting a blocker, calculate that sum efficiently, then jump past the blocker and continue.
+
+    By sorting `nums` and processing gaps between consecutive elements, we can quickly determine how many "free" integers exist in each range and compute their sum using the formula.
+
+  approach: |
+    We solve this using a **Greedy Gap-Filling Approach**:
+
+    **Step 1: Sort and deduplicate the array**
+
+    - Sort `nums` to process elements in order
+    - Remove duplicates since they don't affect which integers are "taken"
+
+    &nbsp;
+
+    **Step 2: Initialise tracking variables**
+
+    - `result`: Accumulates the sum of chosen integers (starts at `0`)
+    - `prev`: Tracks the last integer we've considered (starts at `0`, meaning we begin from `1`)
+
+    &nbsp;
+
+    **Step 3: Process each number in the sorted array**
+
+    - For each number `num` in the sorted array, calculate the gap: how many integers exist between `prev + 1` and `num - 1` inclusive
+    - The count of available integers in this gap is `num - prev - 1`
+    - If the gap has more integers than we still need (`k`), we only take `k` of them
+    - Use the arithmetic series formula to add the sum of the integers we take: sum from `prev + 1` to `prev + take` is `take * (2 * prev + take + 1) / 2`
+    - Subtract the taken count from `k` and update `prev = num`
+    - If `k` reaches `0`, we're done
+
+    &nbsp;
+
+    **Step 4: Handle remaining integers after the array**
+
+    - If `k > 0` after processing all elements, we need `k` more integers starting from `prev + 1`
+    - Add the sum of integers from `prev + 1` to `prev + k` using the arithmetic series formula
+
+    &nbsp;
+
+    **Step 5: Return the result**
+
+    - Return the accumulated `result` sum
+
+  common_pitfalls:
+    - title: Iterating One-by-One
+      description: |
+        With `k` up to `10^8`, iterating through each integer individually and checking membership would be far too slow.
+
+        For example, if `nums = [10^9]` and `k = 10^8`, you'd need to check and sum 100 million integers one at a time. This results in **O(k)** operations which causes TLE.
+
+        Instead, use the arithmetic series formula `n * (n + 1) / 2` to calculate sums of ranges in **O(1)** time.
+      wrong_approach: "Loop through integers 1, 2, 3, ... checking each"
+      correct_approach: "Calculate range sums using arithmetic series formula"
+
+    - title: Not Handling Duplicates
+      description: |
+        The input array can contain duplicate values (e.g., `[1,4,25,10,25]`). If you don't deduplicate, you might incorrectly count the same "blocked" position multiple times.
+
+        For instance, with `nums = [2, 2, 2]`, the integer `2` only blocks one position, not three. Deduplicating ensures each blocked position is counted exactly once.
+      wrong_approach: "Process array with duplicates"
+      correct_approach: "Convert to set or deduplicate after sorting"
+
+    - title: Integer Overflow
+      description: |
+        The sum of `k` integers (where `k` can be `10^8`) starting from 1 is approximately `k * k / 2`, which can exceed `10^16`. In languages with fixed-size integers, this can cause overflow.
+
+        Python handles arbitrary precision integers automatically, but in other languages you'd need to use 64-bit integers (`long long` in C++, `Long` in Java).
+      wrong_approach: "Use 32-bit integers for sum calculation"
+      correct_approach: "Use 64-bit integers or language with arbitrary precision"
+
+    - title: Off-by-One Errors in Gap Calculation
+      description: |
+        When calculating the number of integers between `prev` and `num`, it's easy to make off-by-one mistakes.
+
+        The count of integers from `a` to `b` inclusive is `b - a + 1`. The gap between `prev` (exclusive) and `num` (exclusive) contains `num - prev - 1` integers.
+
+        Example: between `prev = 2` and `num = 5`, the available integers are `3, 4` — that's `5 - 2 - 1 = 2` integers.
+      wrong_approach: "Miscounting gap size"
+      correct_approach: "Gap from prev to num (exclusive both) is num - prev - 1"
+
+  key_takeaways:
+    - "**Arithmetic series formula**: Sum from 1 to n is `n * (n + 1) / 2`. This transforms O(n) iteration into O(1) calculation"
+    - "**Gap-filling strategy**: When filling positions with constraints, sort the constraints and process gaps between them"
+    - "**Greedy correctness**: Taking the smallest available integers first always yields the minimum sum — no need to consider alternatives"
+    - "**Handle large ranges**: When k or values can be very large, look for mathematical formulas to avoid iteration"
+
+  time_complexity: "O(n log n). Sorting dominates the complexity; processing the sorted array is O(n)."
+  space_complexity: "O(n). We store the deduplicated sorted array. Can be O(1) extra space if we sort in-place and handle duplicates during iteration."
+
+solutions:
+  - approach_name: Greedy with Arithmetic Series
+    is_optimal: true
+    code: |
+      def min_sum(nums: list[int], k: int) -> int:
+          # Sort and deduplicate to process gaps in order
+          nums = sorted(set(nums))
+          result = 0
+          prev = 0  # Last integer we've accounted for (start before 1)
+
+          for num in nums:
+              # How many integers are available between prev and num?
+              gap = num - prev - 1
+
+              if gap > 0:
+                  # Take at most k integers from this gap
+                  take = min(gap, k)
+                  # Sum of integers from (prev + 1) to (prev + take)
+                  # Using formula: sum = take * (first + last) / 2
+                  first = prev + 1
+                  last = prev + take
+                  result += take * (first + last) // 2
+                  k -= take
+
+                  if k == 0:
+                      return result
+
+              prev = num
+
+          # Still need more integers after the last element in nums
+          if k > 0:
+              first = prev + 1
+              last = prev + k
+              result += k * (first + last) // 2
+
+          return result
+    explanation: |
+      **Time Complexity:** O(n log n) — Sorting the array dominates; iteration is O(n).
+
+      **Space Complexity:** O(n) — Creating a sorted set of unique elements.
+
+      We process gaps between consecutive elements in the sorted array, using the arithmetic series formula to efficiently sum ranges of consecutive integers. This avoids iterating through potentially billions of integers.
+
+  - approach_name: Brute Force (TLE)
+    is_optimal: false
+    code: |
+      def min_sum(nums: list[int], k: int) -> int:
+          # Convert to set for O(1) lookup
+          num_set = set(nums)
+          result = 0
+          current = 1
+
+          # Find k integers not in nums
+          while k > 0:
+              if current not in num_set:
+                  result += current
+                  k -= 1
+              current += 1
+
+          return result
+    explanation: |
+      **Time Complexity:** O(k + n) — We iterate up to k + max(nums) times in the worst case.
+
+      **Space Complexity:** O(n) — Storing nums in a set.
+
+      This straightforward approach iterates through positive integers one by one, adding those not in nums. While correct, it's far too slow when k is large (up to 10^8). Included to illustrate why the arithmetic series approach is necessary.