<p>Given an array of integers <code>temperatures</code> represents the daily temperatures, return <em>an array</em> <code>answer</code> <em>such that</em> <code>answer[i]</code> <em>is the number of days you have to wait after the</em> <code>i<sup>th</sup></code> <em>day to get a warmer temperature</em>. If there is no future day for which this is possible, keep <code>answer[i] == 0</code> instead.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> temperatures = [73,74,75,71,69,72,76,73]
<strong>Output:</strong> [1,1,4,2,1,1,0,0]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> temperatures = [30,40,50,60]
<strong>Output:</strong> [1,1,1,0]
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> temperatures = [30,60,90]
<strong>Output:</strong> [1,1,0]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;=&nbsp;temperatures.length &lt;= 10<sup>5</sup></code></li>
	<li><code>30 &lt;=&nbsp;temperatures[i] &lt;= 100</code></li>
</ul>

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {
        unordered_map<int, int> history;
        int i, l=temperatures.size();
        vector<int> result(l,0);
        for(i=temperatures.size()-1;i>=0;--i){
            int temp = temperatures[i];
            int minm = INT_MAX;
            for(int j = temp+1;j<101;++j){
                if(history[j] && (history[j]-i) > 0){
                    minm = min(minm, history[j]-i);
                }
            }
            if(minm==INT_MAX){
                result[i] = 0;
            }
            else {
                result[i] = minm;
            }
            history[temp] = i;
        }
        return result;
    }
};
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