Previous Permutation With One Swap
<p>Given an array of positive integers <code>arr</code> (not necessarily distinct), return <em>the </em><span data-keyword="lexicographically-smaller-array"><em>lexicographically</em></span><em> largest permutation that is smaller than</em> <code>arr</code>, that can be <strong>made with exactly one swap</strong>. If it cannot be done, then return the same array.</p>

<p><strong>Note</strong> that a <em>swap</em> exchanges the positions of two numbers <code>arr[i]</code> and <code>arr[j]</code></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [3,2,1]
<strong>Output:</strong> [3,1,2]
<strong>Explanation:</strong> Swapping 2 and 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,1,5]
<strong>Output:</strong> [1,1,5]
<strong>Explanation:</strong> This is already the smallest permutation.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,9,4,6,7]
<strong>Output:</strong> [1,7,4,6,9]
<strong>Explanation:</strong> Swapping 9 and 7.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= arr.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= arr[i] &lt;= 10<sup>4</sup></code></li>
</ul>
Greedy; Array
class Solution {
    public int[] prevPermOpt1(int[] arr) {
        int n = arr.length;
        for (int i = n - 1; i > 0; --i) {
            if (arr[i - 1] > arr[i]) {
                for (int j = n - 1; j > i - 1; --j) {
                    if (arr[j] < arr[i - 1] && arr[j] != arr[j - 1]) {
                        int t = arr[i - 1];
                        arr[i - 1] = arr[j];
                        arr[j] = t;
                        return arr;
                    }
                }
            }
        }
        return arr;
    }
}