README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1200-1299/1247.Minimum%20Swaps%20to%20Make%20Strings%20Equal/README_EN.md	1597	Weekly Contest 161 Q1	Greedy Math String

1247. Minimum Swaps to Make Strings Equal

中文文档

Description

You are given two strings s1 and s2 of equal length consisting of letters "x" and "y" only. Your task is to make these two strings equal to each other. You can swap any two characters that belong to different strings, which means: swap s1[i] and s2[j].

Return the minimum number of swaps required to make s1 and s2 equal, or return -1 if it is impossible to do so.

 

Example 1:

Input: s1 = "xx", s2 = "yy"
Output: 1
Explanation: Swap s1[0] and s2[1], s1 = "yx", s2 = "yx".

Example 2:

Input: s1 = "xy", s2 = "yx"
Output: 2
Explanation: Swap s1[0] and s2[0], s1 = "yy", s2 = "xx".
Swap s1[0] and s2[1], s1 = "xy", s2 = "xy".
Note that you cannot swap s1[0] and s1[1] to make s1 equal to "yx", cause we can only swap chars in different strings.

Example 3:

Input: s1 = "xx", s2 = "xy"
Output: -1

 

Constraints:

• 1 <= s1.length, s2.length <= 1000
• s1.length == s2.length
• s1, s2 only contain 'x' or 'y'.

Solutions

Solution 1: Greedy

According to the problem description, both strings $s_1$ and $s_2$ contain only the characters $x$ and $y$, and they have the same length. Therefore, we can match the characters in $s_1$ and $s_2$ one by one, i.e., $s_1[i]$ and $s_2[i]$.

If $s_1[i] = s_2[i]$, no swap is needed, and we can skip to the next character. If $s_1[i] \neq s_2[i]$, a swap is needed. We count the combinations of $s_1[i]$ and $s_2[i]$: if $s_1[i] = x$ and $s_2[i] = y$, we denote it as $xy$; if $s_1[i] = y$ and $s_2[i] = x$, we denote it as $yx$.

If $xy + yx$ is odd, it is impossible to complete the swaps, and we return $-1$. If $xy + yx$ is even, the number of swaps needed is $\left \lfloor \frac{xy}{2} \right \rfloor + \left \lfloor \frac{yx}{2} \right \rfloor + xy \bmod{2} + yx \bmod{2}$.

The time complexity is $O(n)$, where $n$ is the length of the strings $s_1$ and $s_2$. The space complexity is $O(1)$.

Python3

class Solution:
    def minimumSwap(self, s1: str, s2: str) -> int:
        xy = yx = 0
        for a, b in zip(s1, s2):
            xy += a < b
            yx += a > b
        if (xy + yx) % 2:
            return -1
        return xy // 2 + yx // 2 + xy % 2 + yx % 2

Java

class Solution {
    public int minimumSwap(String s1, String s2) {
        int xy = 0, yx = 0;
        for (int i = 0; i < s1.length(); ++i) {
            char a = s1.charAt(i), b = s2.charAt(i);
            if (a < b) {
                ++xy;
            }
            if (a > b) {
                ++yx;
            }
        }
        if ((xy + yx) % 2 == 1) {
            return -1;
        }
        return xy / 2 + yx / 2 + xy % 2 + yx % 2;
    }
}

C++

class Solution {
public:
    int minimumSwap(string s1, string s2) {
        int xy = 0, yx = 0;
        for (int i = 0; i < s1.size(); ++i) {
            char a = s1[i], b = s2[i];
            xy += a < b;
            yx += a > b;
        }
        if ((xy + yx) % 2) {
            return -1;
        }
        return xy / 2 + yx / 2 + xy % 2 + yx % 2;
    }
};

Go

func minimumSwap(s1 string, s2 string) int {
	xy, yx := 0, 0
	for i := range s1 {
		if s1[i] < s2[i] {
			xy++
		}
		if s1[i] > s2[i] {
			yx++
		}
	}
	if (xy+yx)%2 == 1 {
		return -1
	}
	return xy/2 + yx/2 + xy%2 + yx%2
}

TypeScript

function minimumSwap(s1: string, s2: string): number {
    let xy = 0,
        yx = 0;

    for (let i = 0; i < s1.length; ++i) {
        const a = s1[i],
            b = s2[i];
        xy += a < b ? 1 : 0;
        yx += a > b ? 1 : 0;
    }

    if ((xy + yx) % 2 !== 0) {
        return -1;
    }

    return Math.floor(xy / 2) + Math.floor(yx / 2) + (xy % 2) + (yx % 2);
}

JavaScript

var minimumSwap = function (s1, s2) {
    let xy = 0,
        yx = 0;
    for (let i = 0; i < s1.length; ++i) {
        const a = s1[i],
            b = s2[i];
        if (a < b) {
            ++xy;
        }
        if (a > b) {
            ++yx;
        }
    }
    if ((xy + yx) % 2 === 1) {
        return -1;
    }
    return Math.floor(xy / 2) + Math.floor(yx / 2) + (xy % 2) + (yx % 2);
};