Problem

There is are famous problem for given array of numbers and number $$$K$$$ find minimums for all segments of length $$$K$$$.

More general variant: for given array of size $$$n$$$ and $$$q$$$ queries $$$[l_i,r_i]$$$ need to find minimums on segments at that $$$l_i \leq l_{i+1}$$$ and $$$r_i \leq r_{i+1}$$$. Solution must be $$$O(n+q)$$$.

To solve this problem we need sliding window method, where numbers from segments is window. Needed data structure should be able to push_back (add number to the end of window), pop_front (remove first element) and get_min — current minimum in window.

I was surprised to find that there are two types of such data structure. First is very popular and can be easily found in web. But type that I use whole life I can't find. I'll describe both of them.

Solution #1

The idea is to store sequence of increasing minimums. First minimum is minimum of whole window, second is minimum of elements after first minimum, and so on. For example, for window [2,3,1,5,4,8,7,6,9] sequence is [1,4,6,9].

When element is adding to the right of window (incR, push_back), the sequence changes as follows: all higher elements are gone from sequence, new element is goes to the end of sequence.

Couple of samples for [1,4,6,9]:

Adding 5: [1,4,6,9] -> [1,4] -> [1,4,5]

Adding 100: [1,4,6,9] -> [1,4,6,9,100]

Adding 0: [1,4,6,9] -> [] -> [0]

In removing left element from window (incL, pop_front) we need to know if first element of sequence is first element of window. Impossible to know this only by values, so we need to store sequence of pairs (value, index) except of only values. Previous example turns to [(1,2), (4,4), (6,7), (9,8)], where indices are numbered from zero. Bounds of window stored as pair of numbers $$$L$$$ and $$$R$$$. So, if first element of sequence is first element of window, that is its index equals to $$$L$$$, than only needed is to remove this first element from sequence.

Realization requires data structure that can efficiently perform operations of removing elements from left and right and adding new element to the right. Here we can use deque (std::deque in c++).

int inf = 2e9;
struct sliding_window{
    deque<pair<int,int>> d;
    int l, r;
    
    sliding_window(){
        l = r = 0;
    }
    
    void push_back(int x){
        while(!d.empty() && d.back().first >= x) d.pop_back();
        d.emplace_back(x,r);
        ++r;
    }
    
    void pop_front(){
        if(d.front().second == l) d.pop_front();
        ++l;
    }
    
    int get_min(){
        return d.empty() ? inf : d.front().first;
    }
};

Solution #2

Represent window as two stacks: prefix of window stored in left stack $$$L$$$ so that first element of window in top of $$$L$$$, and suffix of window is stored in right stack $$$R$$$ so that last element of window in top of $$$R$$$.

[2,3,1,5,4,8,7,6,9]
<-------|--------->

L = [5,1,3,2]
R = [4,8,7,6,9]

When element is adding to the right of window (incR, push_back), it just goes to the top of $$$R$$$.

In removing left element from window (incL, pop_front), its pops from $$$L$$$, except of case when $$$L$$$ is empty. In this case we need to move right stack to left: while $$$R$$$ is not empty top of $$$R$$$ goes to the top of $$$L$$$. Example:

[4,3,1,2]
|------->

L = []
R = [4,3,1,2]

L = [2]
R = [4,3,1]

L = [2,1]
R = [4,3]

L = [2,1,3]
R = [4]

L = [2,1,3,4]
R = []

[4,3,2,1]
<-------|

Now, after moving, we can pop top from $$$L$$$.

To get current minimum of window we need to know minimums of both stacks. Minimum of $$$R$$$ stored in $$$Rmin$$$ can be easily updated when elements are added to $$$R$$$.

In $$$L$$$ except of value need to store minimum of elements from this value to the bottom of stack. For example for stack [5,7,3,4,2,1,8,6] it is stack [5,5,3,3,2,1,1,1].

Thus, minimum in window is $$$Rmin$$$ or top of $$$L$$$.

int inf = 2e9;
struct sliding_window{
    vector<int> l, r;
    int rmin;
    
    sliding_window(){
        rmin = inf;
    }
    
    void push_back(int x){
        r.push_back(x);
        rmin = min(rmin, x);
    }
    
    void pop_front(){
        if(l.empty()){
            for(int mn = inf; !r.empty();){
                mn = min(mn, r.back());
                l.push_back(mn);
                r.pop_back();
            }
            rmin = inf;
        }
        l.pop_back();
    }
    
    int get_min(){
        int res = rmin;
        if(!l.empty()) res = min(res, l.back());
        return res;
    }
};

Notes

Personally, I think second solution is simpler for understanding, but first more easy to code.

Both solutions can be upgraded to operation push_front, but seems like its useless.