# 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 <= 105
• t.length == s.length
• 0 <= maxCost <= 106
• s 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 $s$.

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 $s$.

• 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;
}
}

• 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;
}
};

• 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

• 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
}

• 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;
}