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Formatted question description: https://leetcode.ca/all/1794.html

1794. Count Pairs of Equal Substrings With Minimum Difference

Level

Medium

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 i-th character and ends at the j-th character (inclusive) is equal to the substring of secondString that starts at the a-th character and ends at the b-th 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 * 10^5
  • Both strings consist only of lowercase English letters.

Solution

First, find the minimum possible value j - a. If the equal substrings have lengths greater than 1, then after selecting shorter equal substrings, j - a can be reduced, so to obtain the minimum possible value j - a, always consider substrings with length 1. Find each character’s first occurrence in firstString and each character’s last occurrence in lastString, loop over all characters and find the minimum possible value j - a.

Next, count the number of pairs (j, a) that have the minimum possible value j - a. The number of pairs (j, a) is equal to the number of quadruples (i, j, a, b) since i == j and a == b.

Finally, return the count.

  • class Solution {
    public int countQuadruples(String firstString, String secondString) {
        Map<Character, Integer> map1 = new HashMap<Character, Integer>();
        Map<Character, Integer> map2 = new HashMap<Character, Integer>();
        int length1 = firstString.length(), length2 = secondString.length();
        for (int i = length1 - 1; i >= 0; i--) {
            char c = firstString.charAt(i);
            map1.put(c, i);
        }
        for (int i = 0; i < length2; i++) {
            char c = secondString.charAt(i);
            map2.put(c, i);
        }
        int minDifference = Integer.MAX_VALUE;
        Set<Character> keySet = map1.keySet();
        for (char c : keySet) {
            int index1 = map1.get(c);
            if (map2.containsKey(c)) {
                int index2 = map2.get(c);
                int difference = index1 - index2;
                minDifference = Math.min(minDifference, difference);
            }
        }
        if (minDifference == Integer.MAX_VALUE)
            return 0;
        int quadruples = 0;
        for (char c : keySet) {
            int index1 = map1.get(c);
            if (map2.containsKey(c)) {
                int index2 = map2.get(c);
                int difference = index1 - index2;
                if (difference == minDifference)
                    quadruples++;
            }
        }
        return quadruples;
    }
}

############

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;
    }
}

  • 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



  • 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;
    }
};

  • 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
}

  • function countQuadruples(firstString: string, secondString: string): number {
    const last: number[] = new Array(26).fill(0);
    for (let i = 0; i < secondString.length; ++i) {
        last[secondString.charCodeAt(i) - 97] = i + 1;
    }
    let [ans, mi] = [0, Infinity];
    for (let i = 0; i < firstString.length; ++i) {
        const j = last[firstString.charCodeAt(i) - 97];
        if (j) {
            const t = i - j;
            if (mi > t) {
                mi = t;
                ans = 1;
            } else if (mi === t) {
                ++ans;
            }
        }
    }
    return ans;
}

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