What is the time complexity of Minimum Increments for Target Multiples in an Array?

The optimal solution for Minimum Increments for Target Multiples in an Array runs in O(n * 2^m) time and uses O(2^m) space. This complexity arises because we are using a bitmask to track the states of the targets, leading to exponential growth based on the number of targets (m).

What is the brute force solution for Minimum Increments for Target Multiples in an Array?

The brute force approach for Minimum Increments for Target Multiples in an Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible increment for each element in the `nums` array to see if it can satisfy the target multiples. This is straightforward but inefficient, as it can lead to a lot of unnecessary calculations. The optimal Optimal Solution approach improves this to O(n * 2^m).

What are the interview tips for Minimum Increments for Target Multiples in an Array?

Always start with a brute force solution to understand the problem. Think about how you can reduce repeated calculations using dynamic programming. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Minimum Increments for Target Multiples in an Array?

Not considering all possible multiples for each target. Failing to optimize the approach and sticking to brute force.

#3444

Minimum Increments for Target Multiples in an Array

Hard

ArrayMathDynamic ProgrammingBit ManipulationNumber TheoryBitmaskBitmaskingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * 2^m)
Space	O(1)	O(2^m)

💡

Intuition

Time O(n * 2^m)Space O(2^m)

The optimal solution uses a bitmask dynamic programming approach to efficiently track which targets can be satisfied by which multiples in nums. This reduces the need for repeated calculations and allows us to find the minimum increments more effectively.

⚙️

Algorithm

3 steps

1Step 1: Create a bitmask for each number in nums that indicates which targets it can satisfy by being a multiple.
2Step 2: Use dynamic programming to find the minimum increments needed to satisfy all targets based on the bitmasks.
3Step 3: Iterate through the bitmasks and calculate the total increments required.

solution.py20 lines

1# Full working Python code
2
3def minIncrements(nums, target):
4    from itertools import combinations
5    n = len(target)
6    dp = [float('inf')] * (1 << n)
7    dp[0] = 0
8
9    for num in nums:
10        for mask in range((1 << n) - 1, -1, -1):
11            for i in range(n):
12                if (mask & (1 << i)) == 0:
13                    increments_needed = (target[i] - (num % target[i])) % target[i]
14                    new_mask = mask | (1 << i)
15                    dp[new_mask] = min(dp[new_mask], dp[mask] + increments_needed)
16
17    return dp[(1 << n) - 1]
18
19# Example usage
20print(minIncrements([1, 2, 3], [4]))  # Output: 1

ℹ

Complexity note: This complexity arises because we are using a bitmask to track the states of the targets, leading to exponential growth based on the number of targets (m).

1Understanding how multiples work is crucial for determining increments.
2Using bitmasking can significantly reduce the complexity of tracking multiple targets.

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