Leetcode 1208: Get Equal Substrings Within Budget
You are given two strings s and t of the same length, and an integer maxCost. You want to change string s to string t. The cost of changing the ith character of s to the ith character of t is the absolute difference between their ASCII values. Your task is to find the maximum length of a substring of s that can be changed to match the corresponding substring of t, with the total cost not exceeding maxCost.
Problem
Approach
Steps
Complexity
Input: You are given two strings s and t of the same length, and an integer maxCost.
Example: Input: s = "hello", t = "hxllo", maxCost = 2
Constraints:
• 1 <= s.length <= 10^5
• t.length == s.length
• 0 <= maxCost <= 10^6
• s and t consist of only lowercase English letters.
Output: Return the maximum length of a substring of s that can be changed to the corresponding substring of t with a total cost no greater than maxCost.
Example: Output: 3
Constraints:
Goal: The goal is to find the longest possible substring that can be converted from s to t such that the total cost of the transformation is within the allowed maxCost.
Steps:
• Calculate the cost of changing each character of s to the corresponding character of t.
• Use a sliding window approach to find the longest substring where the total change cost does not exceed maxCost.
Goal: The problem ensures the input strings are of the same length and constrained by the specified limits.
Steps:
• 1 <= s.length <= 10^5
• t.length == s.length
• 0 <= maxCost <= 10^6
• s and t consist of only lowercase English letters.
Assumptions:
• The input strings s and t are non-empty and consist of lowercase English letters.
• Input: Input: s = "hello", t = "hxllo", maxCost = 2
• Explanation: You can change the substring 'hel' of s to 'hxl' of t. The total cost is 2, which is within the maxCost.
• Input: Input: s = "abcde", t = "bbcde", maxCost = 1
• Explanation: You can change the substring 'ab' of s to 'bb' of t. The cost is 1, which is within the maxCost.
• Input: Input: s = "abc", t = "xyz", maxCost = 0
• Explanation: Since maxCost is 0, no changes can be made. The longest valid substring is 0.
Approach: We can use a sliding window technique to solve this problem. The idea is to keep track of the running cost of converting substrings of s to substrings of t and adjust the window size accordingly.
Observations:
• We need to track the cost of character changes between s and t for substrings.
• The sliding window will help in dynamically adjusting the substring length while ensuring the total cost stays within maxCost.
• The total cost of transforming any substring can be efficiently managed with a sliding window technique.
Steps:
• Calculate the absolute difference in ASCII values for each corresponding character in s and t.
• Use a sliding window to find the longest substring where the sum of character differences does not exceed maxCost.
• Return the length of this longest valid substring.
Empty Inputs:
• Input strings s and t should always have at least one character, as per the constraints.
Large Inputs:
• Ensure that the solution can handle inputs with lengths up to 10^5.
Special Values:
• Consider cases where maxCost is 0, which means no changes can be made.
Constraints:
• The lengths of s and t are guaranteed to be the same, so there is no need to check for different lengths.
int equalSubstring(string s, string t, int mxc) {
vector<int> nums(s.size(), 0);
for(int i = 0; i < s.size(); i++) {
nums[i] = abs(s[i] - t[i]);
}
int res = 0, j = 0;
for(int i = 0; i < nums.size(); i++) {
mxc -= nums[i];
while(mxc < 0) {
mxc += nums[j++];
}
res = max(res, i - j + 1);
}
return res;
}
1 : Function Definition
int equalSubstring(string s, string t, int mxc) {
Define the function `equalSubstring` that takes two strings `s`, `t`, and an integer `mxc`, which represents the maximum allowed difference.
2 : Vector Initialization
vector<int> nums(s.size(), 0);
Initialize a vector `nums` of the same size as `s`, filled with zeros. This vector will store the absolute differences between corresponding characters in `s` and `t`.
3 : Loop to Calculate Differences
for(int i = 0; i < s.size(); i++) {
Start a for loop to iterate over the characters in the strings `s` and `t`.
4 : Calculate Absolute Difference
nums[i] = abs(s[i] - t[i]);
For each character pair in `s` and `t`, compute the absolute difference between their ASCII values and store it in the `nums` vector.
5 : Variable Initialization
int res = 0, j = 0;
Initialize `res` to 0 to store the maximum length of the valid substring, and `j` to 0 as the starting index for the sliding window.
6 : Sliding Window Loop
for(int i = 0; i < nums.size(); i++) {
Start a for loop that will iterate over the `nums` vector to find the maximum substring with the allowed differences.
7 : Update Maximum Difference
mxc -= nums[i];
Subtract the current difference from `mxc` to track how much the allowed difference decreases as the window expands.
8 : Adjust Window for Validity
while(mxc < 0) {
Start a while loop to shrink the window from the left when the total difference exceeds the allowed `mxc`.
9 : Expand Window by Moving Left Pointer
mxc += nums[j++];
Increment `j` (move the left pointer) and add back the difference for the element that is being excluded from the window.
10 : Update Maximum Substring Length
res = max(res, i - j + 1);
Update `res` to store the maximum length of a valid substring found so far by comparing the current window length `i - j + 1`.
11 : Return Result
return res;
Return the final result `res`, which is the maximum length of a valid substring that can be transformed by no more than `mxc` changes.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n), where n is the length of the strings. This is because each character in the strings is processed at most twice.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the storage of the differences between corresponding characters in s and t.
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