Description#
You are given two strings s and t of the same length and an integer maxCost.
You want to change s to t. Changing the ith character of s to ith character of t costs |s[i] - t[i]| (i.e., the absolute difference between the ASCII values of the characters).
Return the maximum length of a substring of s that can be changed to be the same as the corresponding substring of t with a cost less than or equal to maxCost. If there is no substring from s that can be changed to its corresponding substring from t, return 0.
Example 1:
Input: s = "abcd", t = "bcdf", maxCost = 3
Output: 3
Explanation: "abc" of s can change to "bcd".
That costs 3, so the maximum length is 3.
Example 2:
Input: s = "abcd", t = "cdef", maxCost = 3
Output: 1
Explanation: Each character in s costs 2 to change to character in t, so the maximum length is 1.
Example 3:
Input: s = "abcd", t = "acde", maxCost = 0
Output: 1
Explanation: You cannot make any change, so the maximum length is 1.
Constraints:
1 <= s.length <= 105t.length == s.length0 <= maxCost <= 106s and t consist of only lowercase English letters.
Solutions#
Solution 1: Prefix Sum + Binary Search#
We can create an array $f$ of length $n + 1$, where $f[i]$ represents the sum of the absolute differences of ASCII values between the first $i$ characters of string $s$ and the first $i$ characters of string $t$. Thus, we can calculate the sum of the absolute differences of ASCII values from the $i$-th character to the $j$-th character of string $s$ by $f[j + 1] - f[i]$, where $0 \leq i \leq j < n$.
Note that the length has monotonicity, i.e., if there exists a substring of length $x$ that satisfies the condition, then a substring of length $x - 1$ must also satisfy the condition. Therefore, we can use binary search to find the maximum length.
We define a function $check(x)$, which indicates whether there exists a substring of length $x$ that satisfies the condition. In this function, we only need to enumerate all substrings of length $x$ and check whether they satisfy the condition. If there exists a substring that satisfies the condition, the function returns true, otherwise it returns false.
Next, we define the left boundary $l$ of binary search as $0$ and the right boundary $r$ as $n$. In each step, we let $mid = \lfloor \frac{l + r + 1}{2} \rfloor$. If the return value of $check(mid)$ is true, we update the left boundary to $mid$, otherwise we update the right boundary to $mid - 1$. After the binary search, the left boundary we get is the answer.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of string $s$.
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| class Solution:
def equalSubstring(self, s: str, t: str, maxCost: int) -> int:
def check(x):
for i in range(n):
j = i + mid - 1
if j < n and f[j + 1] - f[i] <= maxCost:
return True
return False
n = len(s)
f = list(accumulate((abs(ord(a) - ord(b)) for a, b in zip(s, t)), initial=0))
l, r = 0, n
while l < r:
mid = (l + r + 1) >> 1
if check(mid):
l = mid
else:
r = mid - 1
return l
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| class Solution {
private int maxCost;
private int[] f;
private int n;
public int equalSubstring(String s, String t, int maxCost) {
n = s.length();
f = new int[n + 1];
this.maxCost = maxCost;
for (int i = 0; i < n; ++i) {
int x = Math.abs(s.charAt(i) - t.charAt(i));
f[i + 1] = f[i] + x;
}
int l = 0, r = n;
while (l < r) {
int mid = (l + r + 1) >>> 1;
if (check(mid)) {
l = mid;
} else {
r = mid - 1;
}
}
return l;
}
private boolean check(int x) {
for (int i = 0; i + x - 1 < n; ++i) {
int j = i + x - 1;
if (f[j + 1] - f[i] <= maxCost) {
return true;
}
}
return false;
}
}
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| class Solution {
public:
int equalSubstring(string s, string t, int maxCost) {
int n = s.size();
int f[n + 1];
f[0] = 0;
for (int i = 0; i < n; ++i) {
f[i + 1] = f[i] + abs(s[i] - t[i]);
}
auto check = [&](int x) -> bool {
for (int i = 0; i + x - 1 < n; ++i) {
int j = i + x - 1;
if (f[j + 1] - f[i] <= maxCost) {
return true;
}
}
return false;
};
int l = 0, r = n;
while (l < r) {
int mid = (l + r + 1) >> 1;
if (check(mid)) {
l = mid;
} else {
r = mid - 1;
}
}
return l;
}
};
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| func equalSubstring(s string, t string, maxCost int) int {
n := len(s)
f := make([]int, n+1)
for i, a := range s {
f[i+1] = f[i] + abs(int(a)-int(t[i]))
}
check := func(x int) bool {
for i := 0; i+x-1 < n; i++ {
if f[i+x]-f[i] <= maxCost {
return true
}
}
return false
}
l, r := 0, n
for l < r {
mid := (l + r + 1) >> 1
if check(mid) {
l = mid
} else {
r = mid - 1
}
}
return l
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
|
Solution 2: Two Pointers#
We can maintain two pointers $j$ and $i$, initially $i = j = 0$; maintain a variable $sum$, representing the sum of the absolute differences of ASCII values in the index interval $[i,..j]$. In each step, we move $i$ to the right by one position, then update $sum = sum + |s[i] - t[i]|$. If $sum \gt maxCost$, then we move the pointer $j$ to the right in a loop, and continuously reduce the value of $sum$ during the moving process until $sum \leq maxCost$. Then we update the answer, i.e., $ans = \max(ans, i - j + 1)$.
Finally, return the answer.
The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is the length of string $s$.
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| class Solution:
def equalSubstring(self, s: str, t: str, maxCost: int) -> int:
n = len(s)
sum = j = 0
ans = 0
for i in range(n):
sum += abs(ord(s[i]) - ord(t[i]))
while sum > maxCost:
sum -= abs(ord(s[j]) - ord(t[j]))
j += 1
ans = max(ans, i - j + 1)
return ans
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| class Solution {
public int equalSubstring(String s, String t, int maxCost) {
int n = s.length();
int sum = 0;
int ans = 0;
for (int i = 0, j = 0; i < n; ++i) {
sum += Math.abs(s.charAt(i) - t.charAt(i));
while (sum > maxCost) {
sum -= Math.abs(s.charAt(j) - t.charAt(j));
++j;
}
ans = Math.max(ans, i - j + 1);
}
return ans;
}
}
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| class Solution {
public:
int equalSubstring(string s, string t, int maxCost) {
int n = s.size();
int ans = 0, sum = 0;
for (int i = 0, j = 0; i < n; ++i) {
sum += abs(s[i] - t[i]);
while (sum > maxCost) {
sum -= abs(s[j] - t[j]);
++j;
}
ans = max(ans, i - j + 1);
}
return ans;
}
};
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| func equalSubstring(s string, t string, maxCost int) (ans int) {
var sum, j int
for i := range s {
sum += abs(int(s[i]) - int(t[i]))
for ; sum > maxCost; j++ {
sum -= abs(int(s[j]) - int(t[j]))
}
if ans < i-j+1 {
ans = i - j + 1
}
}
return
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
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