<p>Given a positive integer <code>n</code>, find <em>the smallest integer which has exactly the same digits existing in the integer</em> <code>n</code> <em>and is greater in value than</em> <code>n</code>. If no such positive integer exists, return <code>-1</code>.</p>

<p><strong>Note</strong> that the returned integer should fit in <strong>32-bit integer</strong>, if there is a valid answer but it does not fit in <strong>32-bit integer</strong>, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> n = 12
<strong>Output:</strong> 21
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> n = 21
<strong>Output:</strong> -1
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution {
public:
    int nextGreaterElement(int n) {
        vector<int> nums;
        while(n>0){
            nums.push_back(n%10);
            n = n/10;
        }
        for(int i=1;i<nums.size();i++){
            if(nums[i]<nums[i-1]){
                for(int j=0;j<i;j++){
                    if(nums[j]>nums[i]){
                        swap(nums[i],nums[j]);
                        break;
                    }
                }
                reverse(nums.begin(),nums.begin()+i);
                long ans = 0;
                for(int i=nums.size()-1;i>=0;i--){
                    ans = ans*10 + nums[i];
                }
                if(ans>INT_MAX) return -1;
                return ans;
            }
        }
        return -1;
    }
};
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