<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>

class Solution {
public:
    vector<int> prevPermOpt1(vector<int>& arr) {
        set<int> st, pos[10002];
        
        for (int i = arr.size() - 1; i >= 0; i--) {
            
            set<int>::iterator it = st.upper_bound(-arr[i]);
            if (it == st.end()) {
                st.insert(-arr[i]);
                pos[arr[i]].insert(i);
                continue;
            }
            
            int val = -*it;
            swap(arr[i], arr[*pos[val].begin()]);
            return arr;
        }
        return arr;
    }
};
// -9 -9 -9 -1 -1 -1 -1 
// [1,9,4,6,7]
// [1,7,4,6,9]

// 4 3 2 1 3 3 , 4 3 2 1 3 3
// 3 4 2 1 3 3 , 3 3 2 1 3 4
// swap(A[i], max(A[i + 1, i + 2, ..., i + k])) such that A[i] > A[i + k] and k is min

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