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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

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

• 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;
  }
  
  

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