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Minimum Array Sum

Solve the Minimum Array Sum problem using dynamic programming by tracking states and operations to minimize the sum of an array.

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Medium · State transition dynamic programming

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Answer-first summary

Solve the Minimum Array Sum problem using dynamic programming by tracking states and operations to minimize the sum of an array.

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The Minimum Array Sum problem asks you to minimize the sum of an array by applying up to two types of operations. The challenge lies in managing dynamic states and transitions as you track the remaining operations and their effects on the array's sum.

Problem Statement

You are given an integer array nums and three integers k, op1, and op2. You can apply two types of operations on the array elements: The first operation (op1) allows you to reduce an element by 1, while the second operation (op2) increases an element by 1. Both operations can be used at most once per index, but you can perform these operations up to k times in total across the entire array. The goal is to find the minimum possible sum of the array after performing at most k operations.

Return the minimum sum you can achieve by applying the operations as described above. Note that both operations can be applied to the same index, but at most once each, and you must keep track of how the operations interact to minimize the result.

Examples

Example 1

Input: nums = [2,8,3,19,3], k = 3, op1 = 1, op2 = 1

Output: 23

Example 2

Input: nums = [2,4,3], k = 3, op1 = 2, op2 = 1

Output: 3

Constraints

  • 1 <= nums.length <= 100
  • 0 <= nums[i] <= 105
  • 0 <= k <= 105
  • 0 <= op1, op2 <= nums.length

Solution Approach

State Transition Dynamic Programming

Start by representing the problem using dynamic programming (DP). You can define a DP table where each entry corresponds to the minimum sum achievable by performing a certain number of operations on a certain subset of the array. The key idea is to calculate the effect of each operation step by step, tracking the state of the array as you progress.

Tracking Progress and Remaining Operations

At each step, keep track of how many operations have been used and how many operations remain. For each element in the array, decide whether to apply op1 or op2 based on the current state and the remaining allowed operations. This tracking ensures that you are always working toward the optimal solution by carefully considering the impact of each operation.

Greedy Decisions within DP

Incorporate greedy strategies within the DP solution by prioritizing the operations that result in the largest reduction of the array sum. Since both operations can affect the same index, a careful balance of op1 and op2 is essential to minimizing the final sum, taking into account how many operations are left to perform.

Complexity Analysis

Metric Value
Time Depends on the final approach
Space Depends on the final approach

The time and space complexity depend on the final dynamic programming approach and how the operations are tracked. Generally, expect a time complexity of O(n * k) where n is the length of the array and k is the maximum number of operations, due to the DP table's size and state transitions. Space complexity is similarly affected by the DP state storage, which may require O(n * k) space for large inputs.

What Interviewers Usually Probe

  • Look for understanding of dynamic programming and state transitions.
  • Candidates should be able to explain how tracking operations optimizes the problem-solving approach.
  • Focus on how the operations are managed and how the dynamic state affects decision-making.

Common Pitfalls or Variants

Common pitfalls

  • Failing to track the correct number of operations left and overusing one type of operation.
  • Not considering the dynamic nature of the operations, which can lead to suboptimal results.
  • Overlooking edge cases where no operations are possible or the array length is very small.

Follow-up variants

  • Modify the problem to only allow a single operation (either op1 or op2).
  • Increase the constraints to allow for larger arrays or more operations.
  • Limit the number of times each operation can be applied to each element.

FAQ

What is the core approach to solving the Minimum Array Sum problem?

The problem is best tackled with dynamic programming, using state transitions to track operations applied to each element and minimize the final sum.

How does dynamic programming help in minimizing the array sum?

Dynamic programming allows you to track operations across different states, ensuring that each decision to apply op1 or op2 is optimal for the final result.

Can I apply both operations to the same element?

Yes, both operations can be applied to the same index, but only once for each operation, so choosing the right operation is key to minimizing the array sum.

What is the time complexity of the solution?

The time complexity of the solution is typically O(n * k), where n is the length of the array and k is the maximum number of operations.

How do I handle edge cases where no operations are allowed?

If no operations are allowed, the solution should simply return the sum of the array as is, without any modifications.

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Solution

Solution 1: Dynamic Programming

For convenience, we denote the given $k$ as $d$.

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class Solution:
    def minArraySum(self, nums: List[int], d: int, op1: int, op2: int) -> int:
        n = len(nums)
        f = [[[inf] * (op2 + 1) for _ in range(op1 + 1)] for _ in range(n + 1)]
        f[0][0][0] = 0
        for i, x in enumerate(nums, 1):
            for j in range(op1 + 1):
                for k in range(op2 + 1):
                    f[i][j][k] = f[i - 1][j][k] + x
                    if j > 0:
                        f[i][j][k] = min(f[i][j][k], f[i - 1][j - 1][k] + (x + 1) // 2)
                    if k > 0 and x >= d:
                        f[i][j][k] = min(f[i][j][k], f[i - 1][j][k - 1] + (x - d))
                    if j > 0 and k > 0:
                        y = (x + 1) // 2
                        if y >= d:
                            f[i][j][k] = min(f[i][j][k], f[i - 1][j - 1][k - 1] + y - d)
                        if x >= d:
                            f[i][j][k] = min(
                                f[i][j][k], f[i - 1][j - 1][k - 1] + (x - d + 1) // 2
                            )
        ans = inf
        for j in range(op1 + 1):
            for k in range(op2 + 1):
                ans = min(ans, f[n][j][k])
        return ans
Minimum Array Sum Solution: State transition dynamic programming | LeetCode #3366 Medium