deque<int> maxi(k); // 2 dequeue bne hai maxi or mini nam ki

deque<int> mini(k);

// addition of first k size window

for(int i = 0; i < k; i++) {

while(!maxi.empty() && arr[maxi.back()] <= arr[i])

maxi.pop_back();

while(!mini.empty() && arr[mini.back()] >= arr[i])

mini.pop_back();

maxi.push_back(i);

mini.push_back(i);

}

int ans = 0;

ans += arr[max.front()] + arr[mini.front()];

// remaining windows ko process karlo

for(int i = k; i < n; i++) {

// next window

// removal

while(!maxi.empty() && i - maxi.front() >=k) {

maxi.pop_front();

}

while(!mini.empty() && i - mini.front() >=k) {

mini.pop_front();

}

//addition

while(!maxi.empty() && arr[maxi.back()] <= arr[i])

maxi.pop_back();

while(!mini.empty() && arr[mini.back()] >= arr[i])

mini.pop_back();

maxi.push_back(i);

mini.push_back(i);

ans += arr[maxi.front()] + arr[mini.front()];

}

return ans;

int arr[7] = {2, 5, -1, 7, -3, -1, -2};

int k = 4;

cout << solve(arr, 7, k) << endl;

return 0;