How do you solve Minimum Cost to Make Two Binary Strings Equal on LeetCode?

Minimum Cost to Make Two Binary Strings Equal (LeetCode #3800) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Identifying mismatches is crucial for minimizing costs.

What is the time complexity of Minimum Cost to Make Two Binary Strings Equal?

The optimal solution for Minimum Cost to Make Two Binary Strings Equal runs in O(n) time and uses O(1) space. The complexity is linear since we only traverse the strings once to count mismatches.

What is the brute force solution for Minimum Cost to Make Two Binary Strings Equal?

The brute force approach for Minimum Cost to Make Two Binary Strings Equal is Brute Force, with O(n²) time complexity and O(1) space complexity. Try every possible combination of operations to make the strings equal. This method is inefficient as it explores all possibilities, leading to high time complexity. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Cost to Make Two Binary Strings Equal?

Minimum Cost to Make Two Binary Strings Equal uses the following concepts: String, Greedy. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Cost to Make Two Binary Strings Equal?

Clarify the problem requirements before diving into solutions. Discuss your thought process and potential edge cases. Practice explaining your code as you write it.

What are common mistakes when solving Minimum Cost to Make Two Binary Strings Equal?

Overlooking the cost of flipping versus swapping. Not counting mismatches correctly.

#3800

Minimum Cost to Make Two Binary Strings Equal

Medium

StringGreedyHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Count mismatches and resolve them using the cheapest operations. Pair opposite mismatches for efficient cost management.

⚙️

Algorithm

3 steps

1Step 1: Count mismatches a (0 in s, 1 in t) and b (1 in s, 0 in t).
2Step 2: Calculate pairs p = min(a, b) and resolve with cost min(swapCost, 2 * flipCost).
3Step 3: Handle remaining mismatches with flipCost.

solution.py9 lines

1def minCost(s, t, flipCost, swapCost, crossCost):
2    a = b = 0
3    for i in range(len(s)):
4        if s[i] != t[i]:
5            if s[i] == '0': a += 1
6            else: b += 1
7    p = min(a, b)
8    cost = p * min(swapCost, 2 * flipCost) + (a - p + b - p) * flipCost
9    return cost

ℹ

Complexity note: The complexity is linear since we only traverse the strings once to count mismatches.

1Identifying mismatches is crucial for minimizing costs.
2Pairing mismatches can lead to significant savings in operation costs.

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