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Copy file name to clipboardexpand all lines: solution/1200-1299/1247.Minimum Swaps to Make Strings Equal/README_EN.md
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### Solution 1: Greedy
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According to the problem description, both strings $s1$ and $s2$ only contain characters $x$ and $y$, and have the same length. Therefore, we can pair the characters in $s1$ and $s2$, i.e., $s1[i]$ and $s2[i]$.
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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 $s_2[i]$.
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If $s1[i] = s2[i]$, no swap is needed, and we can skip it. If $s1[i] \neq s2[i]$, a swap is needed. We count the combination of $s1[i]$ and $s2[i]$, i.e., the situation where $s1[i] = x$ and $s2[i] = y$, denoted as $xy$, and the situation where $s1[i] = y$ and $s2[i] = x$, denoted as $yx$.
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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 $yx$.
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If $xy + yx$ is odd, the swap cannot be completed, and we return $-1$. If $xy + yx$ is even, the number of swaps needed is $\left \lfloor \frac{x}{2} \right \rfloor$ + $\left \lfloor \frac{y}{2} \right \rfloor$ + $xy \bmod{2}$ + $yx \bmod{2}$.
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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 $\left \lfloor \frac{xy}{2} \right \rfloor + \left \lfloor \frac{yx}{2} \right \rfloor + xy \bmod{2} + yx \bmod{2}$.
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The time complexity is $O(n)$, where $n$ is the length of the strings $s1$ and $s2$. The space complexity is $O(1)$.
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The time complexity is $O(n)$, where $n$ is the length of the strings $s_1$ and $s_2$. The space complexity is $O(1)$.
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