# Program to find sum of medians of all odd length sublists in C++

Suppose we have a list of numbers called nums, we have to find the sum of the medians of every odd−length sublist of the given list.

So, if the input is like nums = [2, 4, 6, 3], then the output will be 23, as the odd−length sublists are − , , , , [2, 4, 6], [4, 6, 3], so the sum of the medians is 2 + 4 + 6 + 3 + 4 + 4 = 23

To solve this, we will follow these steps −

• ret := 0

• for initialize i := 0, when i < size of nums, update (increase i by 1), do −

• define priority queue called que_max

• define another priority queue called que_min

• for initialize j := i, when j < size of nums, update (increase j by 1), do −

• insert nums[j] into que_max

• while size of que_max >= 2, do −

• insert top element of que_max into que_min

• delete top element from que_max

• while (size of que_min is not 0 and top element of que_max > top element of que_min), do −

• a := top element of que_max, delete top element from que_max

• b := top element of que_min, delete top element from que_min

• insert b into que_max

• insert a into que_min

• if i mod 2 is same as j mod 2, then −

• ret := ret + top element of que_max

• return ret

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int solve(vector<int>& nums) {
int ret = 0;
for (int i = 0; i < nums.size(); i++) {
priority_queue<int> que_max;
priority_queue<int, vector<int>, greater<int>> que_min;
for (int j = i; j < nums.size(); j++) {
que_max.push(nums[j]);
while (que_max.size() - que_min.size() >= 2) {
que_min.push(que_max.top());
que_max.pop();
}
while (que_min.size() && que_max.top() > que_min.top()) {
int a = que_max.top();
que_max.pop();
int b = que_min.top();
que_min.pop();
que_max.push(b);
que_min.push(a);
}
if (i % 2 == j % 2) {
ret += que_max.top();
}
}
}
return ret;
}
int main(){
vector<int> v = {2, 4, 6, 3};
cout << solve(v);
}

## Input

{2, 4, 6, 3}

## Output

23