README_EN.md

1794. Count Pairs of Equal Substrings With Minimum Difference

中文文档

Description

You are given two strings firstString and secondString that are 0-indexed and consist only of lowercase English letters. Count the number of index quadruples (i,j,a,b) that satisfy the following conditions:

• 0 <= i <= j < firstString.length
• 0 <= a <= b < secondString.length
• The substring of firstString that starts at the ith character and ends at the jth character (inclusive) is equal to the substring of secondString that starts at the ath character and ends at the bth character (inclusive).
• j - a is the minimum possible value among all quadruples that satisfy the previous conditions.

Return the number of such quadruples.

 

Example 1:

Input: firstString = "abcd", secondString = "bccda"
Output: 1
Explanation: The quadruple (0,0,4,4) is the only one that satisfies all the conditions and minimizes j - a.

Example 2:

Input: firstString = "ab", secondString = "cd"
Output: 0
Explanation: There are no quadruples satisfying all the conditions.

 

Constraints:

• 1 <= firstString.length, secondString.length <= 2 * 105
• Both strings consist only of lowercase English letters.

Solutions

Python3

class Solution:
    def countQuadruples(self, firstString: str, secondString: str) -> int:
        last = {c: i for i, c in enumerate(secondString)}
        ans, mi = 0, inf
        for i, c in enumerate(firstString):
            if c in last:
                t = i - last[c]
                if mi > t:
                    mi = t
                    ans = 1
                elif mi == t:
                    ans += 1
        return ans

Java

class Solution {
    public int countQuadruples(String firstString, String secondString) {
        int[] last = new int[26];
        for (int i = 0; i < secondString.length(); ++i) {
            last[secondString.charAt(i) - 'a'] = i + 1;
        }
        int ans = 0, mi = 1 << 30;
        for (int i = 0; i < firstString.length(); ++i) {
            int j = last[firstString.charAt(i) - 'a'];
            if (j > 0) {
                int t = i - j;
                if (mi > t) {
                    mi = t;
                    ans = 1;
                } else if (mi == t) {
                    ++ans;
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int countQuadruples(string firstString, string secondString) {
        int last[26] = {0};
        for (int i = 0; i < secondString.size(); ++i) {
            last[secondString[i] - 'a'] = i + 1;
        }
        int ans = 0, mi = 1 << 30;
        for (int i = 0; i < firstString.size(); ++i) {
            int j = last[firstString[i] - 'a'];
            if (j) {
                int t = i - j;
                if (mi > t) {
                    mi = t;
                    ans = 1;
                } else if (mi == t) {
                    ++ans;
                }
            }
        }
        return ans;
    }
};

Go

func countQuadruples(firstString string, secondString string) (ans int) {
	last := [26]int{}
	for i, c := range secondString {
		last[c-'a'] = i + 1
	}
	mi := 1 << 30
	for i, c := range firstString {
		j := last[c-'a']
		if j > 0 {
			t := i - j
			if mi > t {
				mi = t
				ans = 1
			} else if mi == t {
				ans++
			}
		}
	}
	return
}

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