2289. Steps to Make Array Non-decreasing

Description

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

Solutions

• class Solution {
public int totalSteps(int[] nums) {
Deque<Integer> stk = new ArrayDeque<>();
int ans = 0;
int n = nums.length;
int[] dp = new int[n];
for (int i = n - 1; i >= 0; --i) {
while (!stk.isEmpty() && nums[i] > nums[stk.peek()]) {
dp[i] = Math.max(dp[i] + 1, dp[stk.pop()]);
ans = Math.max(ans, dp[i]);
}
stk.push(i);
}
return ans;
}
}

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

• class Solution:
def totalSteps(self, nums: List[int]) -> int:
stk = []
ans, n = 0, len(nums)
dp = [0] * n
for i in range(n - 1, -1, -1):
while stk and nums[i] > nums[stk[-1]]:
dp[i] = max(dp[i] + 1, dp[stk.pop()])
stk.append(i)
return max(dp)


• func totalSteps(nums []int) int {
stk := []int{}
ans, n := 0, len(nums)
dp := make([]int, n)
for i := n - 1; i >= 0; i-- {
for len(stk) > 0 && nums[i] > nums[stk[len(stk)-1]] {
dp[i] = max(dp[i]+1, dp[stk[len(stk)-1]])
stk = stk[:len(stk)-1]
ans = max(ans, dp[i])
}
stk = append(stk, i)
}
return ans
}

• function totalSteps(nums: number[]): number {
let ans = 0;
let stack = [];
for (let num of nums) {
let max = 0;
while (stack.length && stack[0][0] <= num) {
max = Math.max(stack[0][1], max);
stack.shift();
}
if (stack.length) max++;
ans = Math.max(max, ans);
stack.unshift([num, max]);
}
return ans;
}