README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3361.Shift%20Distance%20Between%20Two%20Strings/README_EN.md	1553	Biweekly Contest 144 Q2	Array String Prefix Sum

3361. Shift Distance Between Two Strings

中文文档

Description

You are given two strings s and t of the same length, and two integer arrays nextCost and previousCost.

In one operation, you can pick any index i of s, and perform either one of the following actions:

• Shift s[i] to the next letter in the alphabet. If s[i] == 'z', you should replace it with 'a'. This operation costs nextCost[j] where j is the index of s[i] in the alphabet.
• Shift s[i] to the previous letter in the alphabet. If s[i] == 'a', you should replace it with 'z'. This operation costs previousCost[j] where j is the index of s[i] in the alphabet.

The shift distance is the minimum total cost of operations required to transform s into t.

Return the shift distance from s to t.

 

Example 1:

Input: s = "abab", t = "baba", nextCost = [100,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], previousCost = [1,100,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]

Output: 2

Explanation:

• We choose index i = 0 and shift s[0] 25 times to the previous character for a total cost of 1.
• We choose index i = 1 and shift s[1] 25 times to the next character for a total cost of 0.
• We choose index i = 2 and shift s[2] 25 times to the previous character for a total cost of 1.
• We choose index i = 3 and shift s[3] 25 times to the next character for a total cost of 0.

Example 2:

Input: s = "leet", t = "code", nextCost = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1], previousCost = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]

Output: 31

Explanation:

• We choose index i = 0 and shift s[0] 9 times to the previous character for a total cost of 9.
• We choose index i = 1 and shift s[1] 10 times to the next character for a total cost of 10.
• We choose index i = 2 and shift s[2] 1 time to the previous character for a total cost of 1.
• We choose index i = 3 and shift s[3] 11 times to the next character for a total cost of 11.

 

Constraints:

• 1 <= s.length == t.length <= 105
• s and t consist only of lowercase English letters.
• nextCost.length == previousCost.length == 26
• 0 <= nextCost[i], previousCost[i] <= 109

Solutions

Solution 1

Python3

class Solution:
    def shiftDistance(
        self, s: str, t: str, nextCost: List[int], previousCost: List[int]
    ) -> int:
        m = 26
        s1 = [0] * (m << 1 | 1)
        s2 = [0] * (m << 1 | 1)
        for i in range(m << 1):
            s1[i + 1] = s1[i] + nextCost[i % m]
            s2[i + 1] = s2[i] + previousCost[(i + 1) % m]
        ans = 0
        for a, b in zip(s, t):
            x, y = ord(a) - ord("a"), ord(b) - ord("a")
            c1 = s1[y + m if y < x else y] - s1[x]
            c2 = s2[x + m if x < y else x] - s2[y]
            ans += min(c1, c2)
        return ans

Java

class Solution {
    public long shiftDistance(String s, String t, int[] nextCost, int[] previousCost) {
        int m = 26;
        long[] s1 = new long[(m << 1) + 1];
        long[] s2 = new long[(m << 1) + 1];
        for (int i = 0; i < (m << 1); i++) {
            s1[i + 1] = s1[i] + nextCost[i % m];
            s2[i + 1] = s2[i] + previousCost[(i + 1) % m];
        }
        long ans = 0;
        for (int i = 0; i < s.length(); i++) {
            int x = s.charAt(i) - 'a';
            int y = t.charAt(i) - 'a';
            long c1 = s1[y + (y < x ? m : 0)] - s1[x];
            long c2 = s2[x + (x < y ? m : 0)] - s2[y];
            ans += Math.min(c1, c2);
        }
        return ans;
    }
}

C++

class Solution {
public:
    long long shiftDistance(string s, string t, vector<int>& nextCost, vector<int>& previousCost) {
        int m = 26;
        vector<long long> s1((m << 1) + 1);
        vector<long long> s2((m << 1) + 1);
        for (int i = 0; i < (m << 1); ++i) {
            s1[i + 1] = s1[i] + nextCost[i % m];
            s2[i + 1] = s2[i] + previousCost[(i + 1) % m];
        }

        long long ans = 0;
        for (int i = 0; i < s.size(); ++i) {
            int x = s[i] - 'a';
            int y = t[i] - 'a';
            long long c1 = s1[y + (y < x ? m : 0)] - s1[x];
            long long c2 = s2[x + (x < y ? m : 0)] - s2[y];
            ans += min(c1, c2);
        }

        return ans;
    }
};

Go

func shiftDistance(s string, t string, nextCost []int, previousCost []int) (ans int64) {
	m := 26
	s1 := make([]int64, (m<<1)+1)
	s2 := make([]int64, (m<<1)+1)
	for i := 0; i < (m << 1); i++ {
		s1[i+1] = s1[i] + int64(nextCost[i%m])
		s2[i+1] = s2[i] + int64(previousCost[(i+1)%m])
	}
	for i := 0; i < len(s); i++ {
		x := int(s[i] - 'a')
		y := int(t[i] - 'a')
		z := y
		if y < x {
			z += m
		}
		c1 := s1[z] - s1[x]
		z = x
		if x < y {
			z += m
		}
		c2 := s2[z] - s2[y]
		ans += min(c1, c2)
	}
	return
}

TypeScript

function shiftDistance(s: string, t: string, nextCost: number[], previousCost: number[]): number {
    const m = 26;
    const s1: number[] = Array((m << 1) + 1).fill(0);
    const s2: number[] = Array((m << 1) + 1).fill(0);
    for (let i = 0; i < m << 1; i++) {
        s1[i + 1] = s1[i] + nextCost[i % m];
        s2[i + 1] = s2[i] + previousCost[(i + 1) % m];
    }
    let ans = 0;
    const a = 'a'.charCodeAt(0);
    for (let i = 0; i < s.length; i++) {
        const x = s.charCodeAt(i) - a;
        const y = t.charCodeAt(i) - a;
        const c1 = s1[y + (y < x ? m : 0)] - s1[x];
        const c2 = s2[x + (x < y ? m : 0)] - s2[y];
        ans += Math.min(c1, c2);
    }
    return ans;
}