Problem Description

You have an integer array nums and three integers k, op1, and op2. Your goal is to minimize the sum of all elements in the array by applying up to two types of operations.

Operation 1 (Division):

• Select an index i and divide nums[i] by 2, rounding up to the nearest whole number
• You can use this operation at most op1 times total
• Each index can receive this operation at most once

Operation 2 (Subtraction):

• Select an index i where nums[i] ≥ k and subtract k from nums[i]
• You can use this operation at most op2 times total
• Each index can receive this operation at most once

Key Constraints:

• Both operations can be applied to the same index, but each operation can only be used once per index
• Operation 2 can only be applied if the element is at least k

The task is to find the minimum possible sum of all elements after optimally applying these operations.

Example scenarios:

• If you apply Operation 1 to an element with value 7, it becomes ⌈7/2⌉ = 4
• If you apply Operation 2 to an element with value 10 and k = 3, it becomes 10 - 3 = 7
• You could apply both operations to the same element: first Operation 1 on 10 to get 5, then Operation 2 (if k ≤ 5) to get 5 - k

The solution uses dynamic programming where f[i][j][k] represents the minimum sum achievable for the first i elements using exactly j Operation 1's and exactly k Operation 2's. The algorithm considers all possible ways to apply operations to each element, including using no operation, using one operation, or using both operations in either order.

Intuition

The key insight is that we need to make optimal decisions about which operations to apply to which elements. Since we have limited operations (op1 and op2), we can't just greedily apply operations to the largest elements. We need to consider all possibilities systematically.

Think about this problem as making a sequence of decisions: for each element in the array, we need to decide:

1. Do we apply no operation?
2. Do we apply only Operation 1 (divide by 2)?
3. Do we apply only Operation 2 (subtract k)?
4. Do we apply both operations? If so, in which order?

This screams dynamic programming because:

• We have overlapping subproblems (the minimum sum for a prefix of the array with certain operations used)
• We have optimal substructure (the best way to handle n elements depends on the best way to handle n-1 elements)

The state representation becomes natural: f[i][j][k] = minimum sum for the first i elements using j Operation 1's and k Operation 2's. This captures everything we need to know about our progress through the array.

For each new element x, we explore all valid transitions:

• No operation: Just add x to the previous state
• Only Operation 1: Transform x to ⌈(x+1)/2⌉ and use one Operation 1
• Only Operation 2: If x ≥ d, transform x to x - d and use one Operation 2
• Both operations: Here's where it gets interesting - the order matters!
  • Operation 1 first: x becomes ⌈(x+1)/2⌉, then if this result ≥ d, subtract d
  • Operation 2 first: If x ≥ d, subtract d to get x - d, then divide by 2 to get ⌈(x-d+1)/2⌉

The reason order matters is that division rounds up, so ⌈(x+1)/2⌉ - d might be different from ⌈(x-d+1)/2⌉. We need to try both orders to ensure we find the minimum.

By building up the DP table element by element, considering all possible operation combinations, we guarantee finding the optimal solution. The final answer is the minimum value among all f[n][j][k] where j ≤ op1 and k ≤ op2.

Learn more about Dynamic Programming patterns.

Solution Approach

We implement a 3D dynamic programming solution where f[i][j][k] represents the minimum sum of the first i numbers using j operations of type 1 and k operations of type 2.

Initialization:

• Create a 3D DP array with dimensions (n+1) × (op1+1) × (op2+1) initialized to infinity
• Set f[0][0][0] = 0 as the base case (0 elements with 0 operations has sum 0)
• For convenience, denote the given k parameter as d to avoid confusion with the loop variable

State Transitions: For each element x = nums[i-1] (using 1-based indexing for the DP array), we consider five possible scenarios:

1. No operation applied:

   f[i][j][k] = f[i-1][j][k] + x

   Simply add the current element to the previous state.

2. Only Operation 1 (divide by 2): If j > 0, we can use one Operation 1:

   f[i][j][k] = min(f[i][j][k], f[i-1][j-1][k] + (x + 1) // 2)

   The expression (x + 1) // 2 computes the ceiling of x/2.

3. Only Operation 2 (subtract d): If k > 0 and x >= d, we can use one Operation 2:

   f[i][j][k] = min(f[i][j][k], f[i-1][j][k-1] + (x - d))

4. Both operations - Operation 1 first: If j > 0 and k > 0, apply Operation 1 first to get y = (x + 1) // 2. Then if y >= d, we can apply Operation 2:

   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)

5. Both operations - Operation 2 first: If j > 0, k > 0, and x >= d, apply Operation 2 first to get x - d. Then apply Operation 1:

   if x >= d:
       f[i][j][k] = min(f[i][j][k], f[i-1][j-1][k-1] + (x - d + 1) // 2)

Computing the Final Answer: After processing all elements, the answer is the minimum value among all valid states:

ans = inf
for j in range(op1 + 1):
    for k in range(op2 + 1):
        ans = min(ans, f[n][j][k])

This considers all possible combinations of using at most op1 Operation 1's and at most op2 Operation 2's.

Time Complexity: O(n × op1 × op2) - we fill a 3D DP table with these dimensions.

Space Complexity: O(n × op1 × op2) - for storing the DP table.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Input: nums = [10, 7], k = 3, op1 = 1, op2 = 1

We'll build our DP table f[i][j][k] where:

• i represents elements processed (0 to 2)
• j represents Operation 1's used (0 to 1)
• k represents Operation 2's used (0 to 1)

Initialization:

• f[0][0][0] = 0 (base case: no elements, no operations)
• All other states = infinity

Processing first element (x = 10):

For i = 1, we consider all possible operations on element 10:

1. No operation: f[1][0][0] = f[0][0][0] + 10 = 0 + 10 = 10

2. Only Operation 1 (divide by 2):

   • f[1][1][0] = f[0][0][0] + ⌈10/2⌉ = 0 + 5 = 5

3. Only Operation 2 (subtract 3):

   • Since 10 ≥ 3: f[1][0][1] = f[0][0][0] + (10 - 3) = 0 + 7 = 7

4. Both operations - Operation 1 first:

   • Apply Op1: 10 → 5
   • Since 5 ≥ 3, apply Op2: 5 → 2
   • f[1][1][1] = f[0][0][0] + 2 = 0 + 2 = 2

5. Both operations - Operation 2 first:

   • Apply Op2: 10 → 7
   • Apply Op1: 7 → ⌈7/2⌉ = 4
   • f[1][1][1] = min(2, 0 + 4) = 2

After processing element 10:

• f[1][0][0] = 10 (no operations)
• f[1][1][0] = 5 (only Op1)
• f[1][0][1] = 7 (only Op2)
• f[1][1][1] = 2 (both operations)

Processing second element (x = 7):

For i = 2, we build on previous states:

1. From f[1][0][0] = 10 (no ops used yet):

   • No operation: f[2][0][0] = 10 + 7 = 17
   • Only Op1 on 7: f[2][1][0] = 10 + ⌈7/2⌉ = 10 + 4 = 14
   • Only Op2 on 7: f[2][0][1] = 10 + (7-3) = 10 + 4 = 14
   • Both on 7 (Op1 first): 7 → 4, 4 ≥ 3, so 4 → 1: f[2][1][1] = 10 + 1 = 11
   • Both on 7 (Op2 first): 7 → 4 → 2: f[2][1][1] = min(11, 10 + 2) = 11

2. From f[1][1][0] = 5 (Op1 used on first element):

   • No operation: f[2][1][0] = min(14, 5 + 7) = 12
   • Op2 on 7: f[2][1][1] = min(11, 5 + 4) = 9

3. From f[1][0][1] = 7 (Op2 used on first element):

   • No operation: f[2][0][1] = min(14, 7 + 7) = 14
   • Op1 on 7: f[2][1][1] = min(9, 7 + 4) = 9

4. From f[1][1][1] = 2 (both ops used on first element):

   • No operation: f[2][1][1] = min(9, 2 + 7) = 9

Final DP table for i=2:

• f[2][0][0] = 17 (no operations on either)
• f[2][1][0] = 12 (Op1 on one element)
• f[2][0][1] = 14 (Op2 on one element)
• f[2][1][1] = 9 (using both operations optimally)

Answer: The minimum sum is 9, achieved by:

• Applying both operations to element 10 (10 → 5 → 2)
• Not applying any operation to element 7
• Final array: [2, 7] with sum = 9

Solution Implementation

Time and Space Complexity

The time complexity is O(n × op1 × op2), where:

• n is the length of the input array nums
• op1 is the maximum number of operations of type 1 (halving operation)
• op2 is the maximum number of operations of type 2 (subtraction operation)

This is because the code uses three nested loops:

• The outer loop iterates through all elements in nums (n iterations)
• The middle loop iterates through all possible values of j from 0 to op1 (op1 + 1 iterations)
• The inner loop iterates through all possible values of k from 0 to op2 (op2 + 1 iterations)

Within the innermost loop, all operations (comparisons, arithmetic operations, and array accesses) take constant time O(1).

The space complexity is O(n × op1 × op2) due to the 3D dynamic programming array f which has dimensions (n + 1) × (op1 + 1) × (op2 + 1). This array stores the minimum sum achievable for each state defined by:

• The number of elements processed (up to n)
• The number of type 1 operations used (up to op1)
• The number of type 2 operations used (up to op2)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Order of Operations Assumption

A critical pitfall is assuming that one order of operations is always better than the other when both operations are applied to the same element. Many developers might think that always applying Operation 1 (division) first is optimal, or vice versa.

Why it's wrong: The optimal order depends on the specific values:

• For num = 15 and k = 5:

  • Op1 first: ⌈15/2⌉ = 8, then 8 - 5 = 3 (final: 3)
  • Op2 first: 15 - 5 = 10, then ⌈10/2⌉ = 5 (final: 5)
  • Op1 first is better here!

• For num = 11 and k = 5:

  • Op1 first: ⌈11/2⌉ = 6, then 6 - 5 = 1 (final: 1)
  • Op2 first: 11 - 5 = 6, then ⌈6/2⌉ = 3 (final: 3)
  • Op1 first is better again, but the difference varies!

Solution: Always check both orders in the DP transition:

# Check both orders when applying both operations
if j > 0 and k > 0:
    # Order 1: Division first, then subtraction
    value_after_op1 = (num + 1) // 2
    if value_after_op1 >= d:
        final_value = value_after_op1 - d
        dp[i][j][k] = min(dp[i][j][k], dp[i - 1][j - 1][k - 1] + final_value)
  
    # Order 2: Subtraction first, then division
    if num >= d:
        value_after_op2 = num - d
        final_value = (value_after_op2 + 1) // 2
        dp[i][j][k] = min(dp[i][j][k], dp[i - 1][j - 1][k - 1] + final_value)

2. Forgetting Feasibility Checks for Operation 2

Another common mistake is applying Operation 2 without checking if the element is at least k. This can lead to invalid states or negative values.

Why it's wrong: Operation 2 has a prerequisite: nums[i] >= k. Forgetting this check can cause:

• Invalid transitions in the DP table
• Negative values that shouldn't exist
• Incorrect minimum sum calculations

Solution: Always verify the condition before applying Operation 2:

# Single Operation 2
if k > 0 and num >= d:  # Check num >= d before applying
    value_after_op2 = num - d
    dp[i][j][k] = min(dp[i][j][k], dp[i - 1][j][k - 1] + value_after_op2)

# When applying both operations with Op1 first
value_after_op1 = (num + 1) // 2
if value_after_op1 >= d:  # Check the intermediate value
    final_value = value_after_op1 - d
    # ... update DP

3. Incorrect Ceiling Division Implementation

Using floor division when ceiling division is required is a subtle but critical error.

Why it's wrong: The problem specifies rounding UP to the nearest whole number. Using x // 2 gives floor division:

• For x = 7: x // 2 = 3 (wrong), but we need ⌈7/2⌉ = 4
• For x = 8: x // 2 = 4 (correct by coincidence)

Solution: Use the formula (x + 1) // 2 for ceiling division:

# Correct ceiling division
ceiling_div = (num + 1) // 2

# Alternative correct implementations:
import math
ceiling_div = math.ceil(num / 2)
# or
ceiling_div = -(-num // 2)  # Using the negative floor trick

4. Not Considering "Use Fewer Operations" Cases

Initializing the DP table incorrectly or not propagating states where fewer operations are used can miss optimal solutions.

Why it's wrong: Sometimes using fewer operations gives a better result. If we only consider states where exactly op1 and op2 operations are used, we miss cases where using fewer operations is optimal.

Solution: When finding the minimum, check all valid combinations:

# Check ALL combinations, not just dp[n][op1][op2]
min_sum = float('inf')
for j in range(op1 + 1):  # 0 to op1 inclusive
    for k in range(op2 + 1):  # 0 to op2 inclusive
        min_sum = min(min_sum, dp[n][j][k])