<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) {
        int n=arr.size();
        int maxi=-1;
        int swapped=0;

        for(int i=n-2;i>=0;i--){
            if(arr[i]>arr[i+1]){
               maxi=i;
               break;
            }
        }

        if(maxi==-1)return arr;
        int index=-1;
        for(int i=maxi+1;i<n;i++){
            if(arr[i]<arr[maxi]){
               if(index==-1)
               index=i;
               else if(arr[index]<arr[i]){
                     index=i;
               }
            }
        }
         
         swap(arr[index],arr[maxi]);
        return arr;
    }
};
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