Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/2289.html

2289. Steps to Make Array Non-decreasing

• Difficulty: Medium.
• Related Topics: Array, Linked List, Stack, Monotonic Stack.
• Similar Questions: Remove One Element to Make the Array Strictly Increasing.

Problem

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

Solution (Java, C++, Python)

• Java
• Python
• C++
• Go
• TypeScript

• class Solution {
      public int totalSteps(int[] nums) {
          int max = 0;
          int[] pos = new int[nums.length + 1];
          int[] steps = new int[nums.length + 1];
          int top = -1;
          for (int i = 0; i <= nums.length; i++) {
              int val = i == nums.length ? Integer.MAX_VALUE : nums[i];
              while (top >= 0 && nums[pos[top]] <= val) {
                  if (top == 0) {
                      max = Math.max(max, steps[pos[top--]]);
                  } else {
                      steps[pos[--top]] = Math.max(steps[pos[top]] + 1, steps[pos[top + 1]]);
                  }
              }
              pos[++top] = i;
          }
          return max;
      }
  }
  
  ############
  
  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:
      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)
  
  ############
  
  # 2289. Steps to Make Array Non-decreasing
  # https://leetcode.com/problems/steps-to-make-array-non-decreasing
  
  from sortedcontainers import SortedList
  
  class Solution:
      def totalSteps(self, nums: List[int]) -> int:
          nums = [float('-inf')] + nums + [float('inf')]
          res = 0
          
          sl = SortedList([(i, x) for i, x in enumerate(nums)])
          p = set()
          
          for i, (a, b) in enumerate(zip(nums, nums[1:])):
              if a > b:
                  p.add((i + 1, b))
                  
          while p:
              newP = set()
              res += 1
              
              for j, b in p:
                  index = sl.bisect_left((j, b))
                  i, a = sl[index - 1]
                  k, c = sl[index + 1]
                  del sl[index]
                  
                  if a > c and (k, c) not in p:
                      newP.add((k, c))
              
              p = newP
          
          return res
  
  

• 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;
      }
  };
  

• 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
  }
  
  func max(a, b int) int {
  	if a > b {
  		return a
  	}
  	return b
  }
  

• 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;
  }
  
  

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).

All Problems

All Solutions