How do you solve Minimum Array Sum on LeetCode?

Minimum Array Sum (LeetCode #3366) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * op1 * op2) time complexity and O(n * op1 * op2) space complexity. Key insight: Using both operations strategically can lead to significant reductions in the sum.

What is the brute force solution for Minimum Array Sum?

The brute force approach for Minimum Array Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will try every possible combination of applying the two operations on each element of the array. This will help us understand the impact of each operation but will be inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n * op1 * op2).

What algorithm or data structure is used to solve Minimum Array Sum?

Minimum Array Sum uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, State Space Exploration.

What are the interview tips for Minimum Array Sum?

Think about how to break the problem down into smaller subproblems. Always consider edge cases, such as when k is 0 or when op1 and op2 are both 0. Be prepared to explain your thought process and the trade-offs of different approaches.

What are common mistakes when solving Minimum Array Sum?

Not considering all combinations of operations. Failing to initialize the DP table correctly.

#3366

Minimum Array Sum

Medium

ArrayDynamic ProgrammingDynamic ProgrammingState Space Exploration

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * op1 * op2)
Space	O(1)	O(n * op1 * op2)

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Intuition

Time O(n * op1 * op2)Space O(n * op1 * op2)

The optimal approach uses dynamic programming to keep track of the minimum sum possible at each index with the remaining operations. This allows us to efficiently explore the state space without redundant calculations.

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Algorithm

3 steps

1Step 1: Initialize a 3D DP array where dp[i][j][l] represents the minimum sum possible using the first i elements with j operations of type 1 and l operations of type 2.
2Step 2: Iterate through each element and update the DP table based on whether we apply operation 1, operation 2, or neither.
3Step 3: Return the minimum value found in the last layer of the DP table.

solution.py13 lines

1def minArraySum(nums, k, op1, op2):
2    n = len(nums)
3    dp = [[[float('inf')] * (op2 + 1) for _ in range(op1 + 1)] for _ in range(n + 1)]
4    dp[0][0][0] = 0
5    for i in range(1, n + 1):
6        for j in range(op1 + 1):
7            for l in range(op2 + 1):
8                dp[i][j][l] = dp[i - 1][j][l] + nums[i - 1]
9                if j > 0:
10                    dp[i][j][l] = min(dp[i][j][l], dp[i - 1][j - 1][l] + (nums[i - 1] + 1) // 2)
11                if l > 0 and nums[i - 1] >= k:
12                    dp[i][j][l] = min(dp[i][j][l], dp[i - 1][j][l - 1] + nums[i - 1] - k)
13    return min(dp[n][j][l] for j in range(op1 + 1) for l in range(op2 + 1))

ℹ

Complexity note: The complexity is O(n * op1 * op2) because we are iterating through each element and considering all combinations of operations left, leading to a manageable state space.

1Using both operations strategically can lead to significant reductions in the sum.
2Dynamic programming allows us to efficiently explore the state space without redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.