Steps to make Array non-decreasing

/*
https://leetcode.com/problems/steps-to-make-array-non-decreasing/discuss/2085967/Python-Explanation-with-pictures-stack

Logic:
arr[i] will be removed the nearest greater element in left.
We will maintain a decreasing stack of pair where first element is element of array
and second element will be the steps required to remove that element.
An element will be only removed if all the element between current element and left greater
element will be removed.so the steps required to remove current element is maximum steps
required to remove all the smaller elements between current and left greater element+1.


*/


class Solution {
public:
    int totalSteps(vector<int>& arr) {
        stack<pair<int,int>> st;
        st.push({arr[0],0});
        int ans=0;
        for(int i=1;i<arr.size();i++){
            int cnt=0;
            while(!st.empty() && arr[i]>=st.top().first){
                cnt=max(st.top().second,cnt);
                st.pop();
            }
            if(st.empty()) cnt=0;   //This means that no greater element in left,so no one will remove it.
            else cnt+=1;
            ans=max(cnt,ans);
            st.push({arr[i],cnt});
        }
        return ans;
    }
};