# 2289. Steps to Make Array Non-decreasing | LeetCode Solution

## 2289. Steps to Make Array Non-decreasing | LeetCode Solution

You are given a 0-indexed integer array `nums`. In one step, remove all elements `nums[i]` where `nums[i - 1] > nums[i]` for all `0 < i < nums.length`.

Return the number of steps performed until `nums` becomes a non-decreasing array.

Example 1:

```Input: nums = [5,3,4,4,7,3,6,11,8,5,11]
Output: 3
Explanation: The following are the steps performed:
- Step 1: [5,3,4,4,7,3,6,11,8,5,11] becomes [5,4,4,7,6,11,11]
- Step 2: [5,4,4,7,6,11,11] becomes [5,4,7,11,11]
- Step 3: [5,4,7,11,11] becomes [5,7,11,11]
[5,7,11,11] is a non-decreasing array. Therefore, we return 3.
```

Example 2:

```Input: nums = [4,5,7,7,13]
Output: 0
Explanation: nums is already a non-decreasing array. Therefore, we return 0.
```

Constraints:

• `1 <= nums.length <= 105`
• `1 <= nums[i] <= 109`
class Solution {
public:
int totalSteps(vector<int>& nums) {
int ans=0;
stack<int>st;
vector<int>dp(nums.size(),0);
vector<int>v(nums.size(),-1);
for(int i=nums.size()-1; i>=0; i--){
while(!st.empty() && nums[st.top()]<nums[i]){
dp[i]=max(dp[i]+1,dp[st.top()]);
st.pop();
}
st.push(i);
ans=max(ans,dp[i]);
}

return ans;
}
};