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<p>We can put the elements in the array $nums$ into a min heap one by one, and each timetake out two elements $a$ and $b$ from the min heap, then put $b$ and $a$ into the answer array in turn, until the min heap is empty.</p>
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<p>Time complexity is $O(n \times \log n)$, and space complexity is $O(n)$. Where $n$ is the length of the array $nums$.</p>
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<p>We can put the elements of the array $\textit{nums}$ into a min heap one by one. Each time, we take out two elements $a$ and $b$ from the min heap, and then sequentially put $b$ and $a$ into the answer array until the min heap is empty.</p>
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<p>The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.</p>
<p>We can sort the array $nums$, and then swap the positions of every two adjacent elements in sequence to get the answer array.</p>
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<p>The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Where $n$ is the length of the array $nums$.</p>
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<p>We can sort the array $\textit{nums}$, and then iterate through the array, swapping adjacent elements each time until the iteration is complete, and return the swapped array.</p>
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<p>The time complexity is $O(n \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $\textit{nums}$.</p>
<p>We can enumerate any two horizontal fences $a$ and $b$ in $hFences$, calculate the distance $d$ between $a$ and $b$, and record it in the hash table $hs$. Then, we enumerate any two vertical fences $c$ and $d$ in $vFences$, calculate the distance $d$ between $c$ and $d$, and record it in the hash table $vs$. Finally, we traverse the hash table $hs$. If a distance $d$ in $hs$ also exists in the hash table $vs$, it means that there exists a square field with a side length of $d$ and an area of $d^2$. We just need to take the largest $d$ and calculate $d^2 \bmod 10^9 + 7$.</p>
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<p>The time complexity is $O(h^2 + v^2)$, and the space complexity is $O(h^2 + v^2)$. Where $h$ and $v$ are the lengths of $hFences$ and $vFences$ respectively.</p>
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<p>We can enumerate any two horizontal fences $a$ and $b$ in $\textit{hFences}$, calculate the distance $d$ between $a$ and $b$, and record it in the hash table $hs$. Then, we enumerate any two vertical fences $c$ and $d$ in $\textit{vFences}$, calculate the distance $d$ between $c$ and $d$, and record it in the hash table $vs$. Finally, we traverse the hash table $hs$. If a certain distance $d$ in $hs$ also exists in the hash table $vs$, it indicates that there exists a square field with a side length of $d$, and the area is $d^2$. We just need to take the largest $d$ and calculate $d^2 \bmod 10^9 + 7$.</p>
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<p>The time complexity is $O(h^2 + v^2)$, and the space complexity is $O(h^2 + v^2)$. Here, $h$ and $v$ are the lengths of $\textit{hFences}$ and $\textit{vFences}$, respectively.</p>
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