Welcome to Subscribe On Youtube

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;
    }
    
    

All Problems

All Solutions