Question

1208. Get Equal Substrings Within Budget

中文文档

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 $s$.

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

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;
  }
};

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 $s$.

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

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;
  }
}

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;
  }
};

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
}