What is the time complexity of Steps to Make Array Non-decreasing?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Steps to Make Array Non-decreasing?

This problem uses Array, Linked List, Stack, Monotonic Stack. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Steps to Make Array Non-decreasing?

Insight 1: The core idea is to identify and remove elements that violate the non-decreasing property at each step. The number of steps is determined by the element that requires the maximum number of removals before it is no longer violating the condition. Insight 2: A monotonic stack is the most suitable data structure to efficiently track potential violating elements and determine the removal steps required for each element.

Which companies ask LeetCode 2289. Steps to Make Array Non-decreasing in interviews?

This problem is frequently asked at Morgan Stanley.

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Steps to Make Array Non-decreasing - LeetCode 2289 Solution

Steps to Make Array Non-decreasing - Complete Solution Guide

Steps to Make Array Non-decreasing is LeetCode problem 2289, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem requires us to determine the number of steps needed to transform a given array `nums` into a non-decreasing array. In each step, we remove all elements `nums[i]` where `nums[i-1] > nums[i]`. We repeat this process until the array becomes non-decreasing, and the output is the number of steps taken.

Solution Approach

The provided solution uses a monotonic stack and dynamic programming to efficiently calculate the number of steps. The stack stores indices of elements that could potentially be greater than the next element (violating the non-decreasing property). The `dp` array stores the number of steps each element has to wait before becoming part of a non-decreasing sequence. The algorithm iterates through the input array, maintaining the stack's monotonicity. When an element is encountered that is greater than or equal to the top of the stack, the elements that are smaller than it are popped, updating the current element's `dp` value based on the popped elements' `dp` values. Finally, the maximum `dp` value is the answer.

Step-by-Step Algorithm

Step 1: Initialize an empty stack to store indices and a `dp` array of the same size as the input `nums` array, filled with zeros. Also, initialize `ans = 0` to store the final result.
Step 2: Iterate through the `nums` array from left to right.
Step 3: For each element `nums[i]`, initialize `cur_steps = 0`.
Step 4: While the stack is not empty and `nums[stack[-1]] <= nums[i]`, pop the top element from the stack and update `cur_steps` as `max(cur_steps, dp[popped_index])`.
Step 5: If the stack is not empty after popping elements, it means there's an element `nums[stack[-1]] > nums[i]` to the left. Therefore set `dp[i] = cur_steps + 1`.
Step 6: Push the current index `i` onto the stack.
Step 7: Update the `ans` with the maximum `dp` value seen so far: `ans = max(ans, dp[i])`.
Step 8: After iterating through all elements, return `ans`.

Key Insights

Insight 1: The core idea is to identify and remove elements that violate the non-decreasing property at each step. The number of steps is determined by the element that requires the maximum number of removals before it is no longer violating the condition.
Insight 2: A monotonic stack is the most suitable data structure to efficiently track potential violating elements and determine the removal steps required for each element.
Insight 3: Dynamic programming (DP) helps in storing the number of steps required to remove an element, avoiding redundant calculations.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Linked List, Stack, Monotonic Stack.

Companies

Asked at: Morgan Stanley.

Steps to Make Array Non-decreasing | LeetCode 2289

Medium

Array Linked List Stack Monotonic Stack

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public int totalSteps(int[] nums) {
        int n = nums.length;
        int[] dp = new int[n];
        Stack<Integer> stack = new Stack<>();
        int ans = 0;
        for (int i = 0; i < n; i++) {
            int cur_steps = 0;
            while (!stack.isEmpty() && nums[stack.peek()] <= nums[i]) {
                cur_steps = Math.max(cur_steps, dp[stack.pop()]);
            }

            if (!stack.isEmpty()) {
                dp[i] = cur_steps + 1;
            }

            stack.push(i);
            ans = Math.max(ans, dp[i]);
        }

        return ans;
    }
}

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Problem Info

Description

Problem 2289: Steps to Make Array Non-decreasing - Detailed explanation and solution approaches.

Topics

Array Linked List Stack Monotonic Stack

Companies

Morgan Stanley

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2289

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize an empty stack to store indices and a `dp` array of the same size as the input `nums` array, filled with zeros. Also, initialize `ans = 0` to store the final result.
2
Step 2: Iterate through the `nums` array from left to right.
3
Step 3: For each element `nums[i]`, initialize `cur_steps = 0`.
4
Step 4: While the stack is not empty and `nums[stack[-1]] <= nums[i]`, pop the top element from the stack and update `cur_steps` as `max(cur_steps, dp[popped_index])`.
5
Step 5: If the stack is not empty after popping elements, it means there's an element `nums[stack[-1]] > nums[i]` to the left. Therefore set `dp[i] = cur_steps + 1`.
6
Step 6: Push the current index `i` onto the stack.
7
Step 7: Update the `ans` with the maximum `dp` value seen so far: `ans = max(ans, dp[i])`.
8
Step 8: After iterating through all elements, return `ans`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because each element is visited and processed a constant number of times (pushed and popped from the stack at most once). The space complexity is O(n) due to the stack and the dp array, both needing to store information for each element in the input array.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Simulate the steps by repeatedly iterating through the array and removing elements that violate the non-decreasing condition. Repeat this process until the array becomes non-decreasing.

Time: O(n^2) in the worst case (e.g., reverse sorted array), where n is the length of the `nums` array.Space: O(1) or O(n) depending on if you modify the array in place or create a copy.

Trade-offs: The brute force approach is simpler to understand but much less efficient than the stack-based solution. Use the brute force approach for very small input sizes only, as it may cause Time Limit Exceeded errors for larger inputs.

✨

Key Insights

1Insight 1: The core idea is to identify and remove elements that violate the non-decreasing property at each step. The number of steps is determined by the element that requires the maximum number of removals before it is no longer violating the condition.
2Insight 2: A monotonic stack is the most suitable data structure to efficiently track potential violating elements and determine the removal steps required for each element.
3Insight 3: Dynamic programming (DP) helps in storing the number of steps required to remove an element, avoiding redundant calculations.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Linked List

Elements connected via pointers. Like a chain where each link knows the next.

Edge Cases

Edge case 1: Input array is already non-decreasing (e.g., `[1, 2, 3, 4, 5]`). In this case, the stack will always be increasing, and no elements will be removed, resulting in `ans = 0` which is the correct answer.
Edge case 2: Input array is strictly decreasing (e.g., `[5, 4, 3, 2, 1]`). In this case, elements will be continuously popped from the stack. The algorithm correctly computes the number of steps.
Edge case 3: Input contains duplicate elements (e.g. `[5, 3, 4, 4, 7, 3, 6, 11, 8, 5, 11]`). The stack will pop elements as long as they are less than or equal to the current element, hence duplicates are handled correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a stack and trying to update the array in place while iterating. This makes it difficult to track the correct number of steps because the indices change with each removal.

✗Common mistake 2: Incorrectly updating the `dp` values. The `cur_steps` variable must track the maximum number of steps of the popped elements to determine the correct number of steps required for the current element.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Stock price analysis. Identifying trends in stock prices by removing elements that violate a generally increasing trend. The steps would represent the number of corrections needed to establish an upward trend.

🏦Real-world scenario 2: Data filtering or smoothing. Imagine sensor data where you need to remove outliers to obtain a more stable trend. The algorithm can be used to filter out data points that create decreasing sequences, and the steps count how many smoothing iterations are necessary.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 8, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How does the fast/slow pointer technique work?

Fast pointer moves 2 steps while slow moves 1 step. When fast reaches the end, slow is at middle. If there's a cycle, they will meet. This technique is used for: finding middle node, detecting cycles, finding cycle start point, and checking palindromes.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Steps to Make Array Non-decreasing"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether linked list techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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