What is the time complexity of Minimum Deletion Cost to Make All Characters Equal?

The optimal solution for Minimum Deletion Cost to Make All Characters Equal runs in O(n) time and uses O(1) space. Single pass through the string results in linear complexity.

What is the brute force solution for Minimum Deletion Cost to Make All Characters Equal?

The brute force approach for Minimum Deletion Cost to Make All Characters Equal is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all possible characters to keep and calculate the cost of deleting others. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Deletion Cost to Make All Characters Equal?

Minimum Deletion Cost to Make All Characters Equal uses the following concepts: Array, Hash Table, String, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Deletion Cost to Make All Characters Equal?

Explain your thought process clearly. Optimize your solution step by step. Practice similar problems to build confidence.

What are common mistakes when solving Minimum Deletion Cost to Make All Characters Equal?

Forgetting to consider all characters. Incorrectly summing costs.

#3784

Minimum Deletion Cost to Make All Characters Equal

Medium

ArrayHash TableStringEnumerationHash 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)

Group characters and keep the one with the highest deletion cost. The rest will be deleted to minimize total cost.

⚙️

Algorithm

3 steps

1Step 1: Initialize variables to track the current character and total deletion cost.
2Step 2: Iterate through the string, summing costs of characters that are not the current character.
3Step 3: Update the minimum cost by comparing with the total cost for each character.

solution.py10 lines

1def minCost(s, cost):
2    total_cost = 0
3    max_cost = 0
4    current_char = s[0]
5    for i in range(len(s)):
6        if s[i] == current_char:
7            max_cost = max(max_cost, cost[i])
8        else:
9            total_cost += cost[i]
10    return total_cost + (sum(cost) - max_cost)

ℹ

Complexity note: Single pass through the string results in linear complexity.

1Keep the character with the highest deletion cost.
2Sum costs of characters to be deleted.

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