How do you solve Get Equal Substrings Within Budget on LeetCode?

Get Equal Substrings Within Budget (LeetCode #1208) 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: Using a sliding window allows us to efficiently manage the cost without recalculating for every substring.

What is the brute force solution for Get Equal Substrings Within Budget?

The brute force approach for Get Equal Substrings Within Budget is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible substring of `s` and calculating the cost to change it to the corresponding substring in `t`. This is simple but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Get Equal Substrings Within Budget?

Get Equal Substrings Within Budget uses the following concepts: String, Binary Search, Sliding Window, Prefix Sum. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Get Equal Substrings Within Budget?

Always clarify the problem constraints and edge cases before jumping into coding. Think about the efficiency of your solution and whether a brute-force approach is feasible given the input size. Practice explaining your thought process as you code; this helps in interviews.

What are common mistakes when solving Get Equal Substrings Within Budget?

Forgetting to reset the left pointer when the cost exceeds maxCost. Not considering edge cases where the strings are already equal or maxCost is zero.

#1208

Get Equal Substrings Within Budget

Medium

StringBinary SearchSliding WindowPrefix SumSliding WindowTwo Pointers

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)

The optimal solution uses a sliding window technique to efficiently find the longest valid substring. By maintaining a window that expands and contracts based on the cost, we can achieve linear time complexity.

⚙️

Algorithm

5 steps

1Step 1: Initialize two pointers (left and right) and a variable to keep track of the total cost.
2Step 2: Expand the right pointer to include characters until the total cost exceeds maxCost.
3Step 3: If the cost exceeds maxCost, move the left pointer to reduce the cost.
4Step 4: Update the maximum length of the substring whenever the cost is within the allowed budget.
5Step 5: Return the maximum length found.

solution.py13 lines

1# Full working Python code
2
3def equalSubstring(s, t, maxCost):
4    left = 0
5    total_cost = 0
6    max_length = 0
7    for right in range(len(s)):
8        total_cost += abs(ord(s[right]) - ord(t[right]))
9        while total_cost > maxCost:
10            total_cost -= abs(ord(s[left]) - ord(t[left]))
11            left += 1
12        max_length = max(max_length, right - left + 1)
13    return max_length

ℹ

Complexity note: This complexity is linear because each character is processed at most twice (once by the right pointer and once by the left pointer), leading to efficient traversal of the string.

1Using a sliding window allows us to efficiently manage the cost without recalculating for every substring.
2The absolute difference in ASCII values directly translates to the cost of changing characters, making calculations straightforward.

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