<p>Given an integer array <code>nums</code> and two integers <code>lower</code> and <code>upper</code>, return <em>the number of range sums that lie in</em> <code>[lower, upper]</code> <em>inclusive</em>.</p>

<p>Range sum <code>S(i, j)</code> is defined as the sum of the elements in <code>nums</code> between indices <code>i</code> and <code>j</code> inclusive, where <code>i &lt;= j</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [-2,5,-1], lower = -2, upper = 2
<strong>Output:</strong> 3
<strong>Explanation:</strong> The three ranges are: [0,0], [2,2], and [0,2] and their respective sums are: -2, -1, 2.
</pre>

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

<pre>
<strong>Input:</strong> nums = [0], lower = 0, upper = 0
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>-10<sup>5</sup> &lt;= lower &lt;= upper &lt;= 10<sup>5</sup></code></li>
	<li>The answer is <strong>guaranteed</strong> to fit in a <strong>32-bit</strong> integer.</li>
</ul>

#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including tree_order_statistics_node_update
#include <ext/pb_ds/detail/standard_policies.hpp>
#include <functional>
using namespace __gnu_pbds;
#define ll long long

typedef tree<ll, null_type, less_equal<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
class Solution {
public:
    int countRangeSum(vector<int>& nums, int l, int u) {
        ordered_set st;
        int n=nums.size();
        vector<ll> psum(nums.begin(),nums.end());
        for(int i=1;i<n;i++){
            psum[i]+=psum[i-1];
        }
        ll ans=0;
        for(int i=n-1;i>=0;i--){
            st.insert(psum[i]);
            ll lo=l;
            ll hi=u;
            if(i-1>=0){
                lo+=psum[i-1];
                hi+=psum[i-1];
            }
            ll s=st.order_of_key(lo);
            ll en=st.order_of_key(hi+1);
            ans+=(en-s);
        }
        return ans;
    }
};
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