1208. Get Equal Substrings Within Budget
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 <= 106sandtconsist 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 $s$.
Python3
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
Java
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;
}
}
C++
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;
}
};
Go
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
}
TypeScript
function equalSubstring(s: string, t: string, maxCost: number): number {
const n = s.length;
const f = Array(n + 1).fill(0);
for (let i = 0; i < n; i++) {
f[i + 1] = f[i] + Math.abs(s.charCodeAt(i) - t.charCodeAt(i));
}
const check = (x: number): boolean => {
for (let i = 0; i + x - 1 < n; i++) {
if (f[i + x] - f[i] <= maxCost) {
return true;
}
}
return false;
};
let l = 0,
r = n;
while (l < r) {
const mid = (l + r + 1) >> 1;
if (check(mid)) {
l = mid;
} else {
r = mid - 1;
}
}
return l;
}
Solution 2: Two Pointers
We can maintain two pointers $l$ and $r$, initially $l = r = 0$; maintain a variable $\textit{cost}$, which represents the sum of the absolute values of the ASCII code differences in the index interval $[l,..r]$. In each step, we move $r$ to the right by one position, then update $\textit{cost} = \textit{cost} + |s[r] - t[r]|$. If $\textit{cost} \gt \textit{maxCost}$, then we loop to move $l$ to the right by one position, and decrease the value of $\textit{cost}$, until $\textit{cost} \leq \textit{maxCost}$. Then we update the answer, that is, $\textit{ans} = \max(\textit{ans}, r - l + 1)$.
Finally, return the answer.
The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.
Python3
class Solution:
def equalSubstring(self, s: str, t: str, maxCost: int) -> int:
n = len(s)
ans = cost = l = 0
for r in range(n):
cost += abs(ord(s[r]) - ord(t[r]))
while cost > maxCost:
cost -= abs(ord(s[l]) - ord(t[l]))
l += 1
ans = max(ans, r - l + 1)
return ans
Java
class Solution {
public int equalSubstring(String s, String t, int maxCost) {
int n = s.length();
int ans = 0, cost = 0;
for (int l = 0, r = 0; r < n; ++r) {
cost += Math.abs(s.charAt(r) - t.charAt(r));
while (cost > maxCost) {
cost -= Math.abs(s.charAt(l) - t.charAt(l));
++l;
}
ans = Math.max(ans, r - l + 1);
}
return ans;
}
}
C++
class Solution {
public:
int equalSubstring(string s, string t, int maxCost) {
int n = s.length();
int ans = 0, cost = 0;
for (int l = 0, r = 0; r < n; ++r) {
cost += abs(s[r] - t[r]);
while (cost > maxCost) {
cost -= abs(s[l] - t[l]);
++l;
}
ans = max(ans, r - l + 1);
}
return ans;
}
};
Go
func equalSubstring(s string, t string, maxCost int) (ans int) {
var cost, l int
for r := range s {
cost += abs(int(s[r]) - int(t[r]))
for ; cost > maxCost; l++ {
cost -= abs(int(s[l]) - int(t[l]))
}
ans = max(ans, r-l+1)
}
return
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
TypeScript
function equalSubstring(s: string, t: string, maxCost: number): number {
const getCost = (i: number) => Math.abs(s[i].charCodeAt(0) - t[i].charCodeAt(0));
const n = s.length;
let ans = 0,
cost = 0;
for (let l = 0, r = 0; r < n; ++r) {
cost += getCost(r);
while (cost > maxCost) {
cost -= getCost(l++);
}
ans = Math.max(ans, r - l + 1);
}
return ans;
}
Solution 3: Another Way of Using Two Pointers
In Solution 2, the interval maintained by the two pointers may become shorter or longer. Since the problem only requires the maximum length, we can maintain a monotonically increasing interval.
Specifically, we use two pointers $l$ and $r$ to point to the left and right endpoints of the interval, initially $l = r = 0$. In each step, we move $r$ to the right by one position, then update $\textit{cost} = \textit{cost} + |s[r] - t[r]|$. If $\textit{cost} \gt \textit{maxCost}$, then we move $l$ to the right by one position, and decrease the value of $\textit{cost}$.
Finally, return $n - l$.
The time complexity is $O(n)$, and the space complexity is $O(1)$. Where $n$ is the length of the string $s$.
Python3
class Solution:
def equalSubstring(self, s: str, t: str, maxCost: int) -> int:
cost = l = 0
for a, b in zip(s, t):
cost += abs(ord(a) - ord(b))
if cost > maxCost:
cost -= abs(ord(s[l]) - ord(t[l]))
l += 1
return len(s) - l
Java
class Solution {
public int equalSubstring(String s, String t, int maxCost) {
int n = s.length();
int cost = 0, l = 0;
for (int r = 0; r < n; ++r) {
cost += Math.abs(s.charAt(r) - t.charAt(r));
if (cost > maxCost) {
cost -= Math.abs(s.charAt(l) - t.charAt(l));
++l;
}
}
return n - l;
}
}
C++
class Solution {
public:
int equalSubstring(string s, string t, int maxCost) {
int n = s.length();
int cost = 0, l = 0;
for (int r = 0; r < n; ++r) {
cost += abs(s[r] - t[r]);
if (cost > maxCost) {
cost -= abs(s[l] - t[l]);
++l;
}
}
return n - l;
}
};
Go
func equalSubstring(s string, t string, maxCost int) int {
n := len(s)
var cost, l int
for r := range s {
cost += abs(int(s[r]) - int(t[r]))
if cost > maxCost {
cost -= abs(int(s[l]) - int(t[l]))
l++
}
}
return n - l
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
TypeScript
function equalSubstring(s: string, t: string, maxCost: number): number {
const getCost = (i: number) => Math.abs(s[i].charCodeAt(0) - t[i].charCodeAt(0));
const n = s.length;
let cost = 0;
let l = 0;
for (let r = 0; r < n; ++r) {
cost += getCost(r);
if (cost > maxCost) {
cost -= getCost(l++);
}
}
return n - l;
}