How do you solve Minimum Cost to Partition a Binary String on LeetCode?

Minimum Cost to Partition a Binary String (LeetCode #3864) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Recursive partitioning minimizes costs effectively.

What is the time complexity of Minimum Cost to Partition a Binary String?

The optimal solution for Minimum Cost to Partition a Binary String runs in O(n) time and uses O(n) space. The complexity is linear due to memoization, which avoids redundant calculations for segments.

What is the brute force solution for Minimum Cost to Partition a Binary String?

The brute force approach for Minimum Cost to Partition a Binary String is Brute Force, with O(n²) time complexity and O(1) space complexity. We can evaluate every possible partition of the string and calculate the cost for each segment. This approach is straightforward but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Cost to Partition a Binary String?

Minimum Cost to Partition a Binary String uses the following concepts: String, Divide and Conquer, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Recursion.

What are the interview tips for Minimum Cost to Partition a Binary String?

Explain your thought process clearly. Discuss edge cases and their handling. Practice similar problems to build intuition.

What are common mistakes when solving Minimum Cost to Partition a Binary String?

Not considering even-length splits. Forgetting to memoize results.

#3864

Minimum Cost to Partition a Binary String

Hard

StringDivide and ConquerPrefix SumDynamic ProgrammingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Using a recursive approach with memoization, we can efficiently calculate the minimum cost by evaluating segments and considering splits only when the length is even.

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Algorithm

3 steps

1Step 1: Use a recursive function to calculate the cost of a segment.
2Step 2: If the segment's length is even, compute the cost for both halves and take the minimum.
3Step 3: Store results in a memoization table to avoid redundant calculations.

solution.py15 lines

1def minCost(s, encCost, flatCost):
2    memo = {}
3    def dfs(start, end):
4        if (start, end) in memo: return memo[(start, end)]
5        segment = s[start:end]
6        X = segment.count('1')
7        L = len(segment)
8        cost_no_split = flatCost if X == 0 else L * X * encCost
9        min_cost = cost_no_split
10        if (end - start) % 2 == 0:
11            mid = (start + end) // 2
12            min_cost = min(min_cost, dfs(start, mid) + dfs(mid, end))
13        memo[(start, end)] = min_cost
14        return min_cost
15    return dfs(0, len(s))

ℹ

Complexity note: The complexity is linear due to memoization, which avoids redundant calculations for segments.

1Recursive partitioning minimizes costs effectively.
2Memoization avoids redundant calculations.

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