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# Steps to Make Array Non-decreasing LeetCode Solution

## Problem – 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`

### Steps to Make Array Non-decreasing LeetCode Solution in Python

``````class Solution:
def totalSteps(self, nums: List[int]) -> int:
ans = 0
nums.reverse()
lst = [[nums[0], 0]]
for i in range(1, len(nums)):
cnt = 0
while lst and lst[-1][0] < nums[i]:
cnt = max(cnt + 1, lst[-1][1])
lst.pop()
lst.append([nums[i], cnt])
ans = max(ans, cnt)
return ans``````

### Steps to Make Array Non-decreasing LeetCode Solution in C++

``````class Solution {
public:
int totalSteps(vector<int>& nums)
{
int n=nums.size(),i,ans=0,cnt,prev;
stack <pair<int,int>> st;
st.push({nums[n-1],0});
for(i=n-2;i>=0;i--)
{
cnt=0;
while(!st.empty() && nums[i]>st.top().first)
{
cnt=max(cnt+1,st.top().second);
st.pop();
}
ans=max(ans,cnt);
st.push({nums[i],cnt});
}
return ans;
}
};``````

### Steps to Make Array Non-decreasing LeetCode Solution in Java

``````   public int totalSteps(int[] nums) {
int ans = 0;
Stack<int[]> stack = new Stack<int[]>();
for(int i = nums.length-1; i >= 0; i--) {
if(stack.isEmpty() || stack.peek()[0] >= nums[i]) {
stack.push(new int[]{nums[i], 0});
}else{
int count = 0;
while(!stack.isEmpty() && stack.peek()[0] < nums[i]) {
count++;
int[] item = stack.pop();
if(count < item[1]) count += (item[1] - count);
}
stack.push(new int[]{nums[i], count});
ans = Math.max(ans, count);
}
}

return ans;
}``````
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