What is the time complexity of Find if Array Can Be Sorted?

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

What data structures are used to solve Find if Array Can Be Sorted?

This problem uses Array, Bit Manipulation, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Find if Array Can Be Sorted?

Insight 1: The key is to recognize that numbers with the same number of set bits form independent groups that can be internally sorted, but cannot be intermixed with numbers from other groups. Insight 2: The problem can be solved by iterating through the array and checking if the minimum value within each group of numbers with the same number of set bits is greater than or equal to the maximum value of the previous group.

Which companies ask LeetCode 3011. Find if Array Can Be Sorted in interviews?

This problem is frequently asked at Edelweiss Group.

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Find if Array Can Be Sorted - LeetCode 3011 Solution

Find if Array Can Be Sorted - Complete Solution Guide

Find if Array Can Be Sorted is LeetCode problem 3011, 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 array of positive integers nums . In one operation , you can swap any two adjacent elements if they have the same number of set bits . You are allowed to do this operation any number of times ( including zero ). Return true if you can sort the array in ascending order, else return false . Example 1: Input: nums = [8,4,2,30,15] Output: true Explanation: Let's look at the binary representation of every element. The numbers 2, 4, and 8 have one set bit each with binary rep

Detailed Explanation

The problem asks whether a given array of positive integers can be sorted into ascending order by repeatedly swapping adjacent elements that have the same number of set bits (i.e., the same number of 1s in their binary representation). The goal is to determine if a series of such swaps can result in a sorted array.

Solution Approach

The provided solution approach works by iterating through the input array and identifying contiguous groups of numbers with the same number of set bits. For each group, it finds the minimum and maximum values. It then checks if the minimum value of the current group is less than the maximum value of the previous group. If it is, it means that even with swaps within the groups with equal number of set bits, it's impossible to sort the array into ascending order because an element from the current group would need to be moved before the previous group to achieve sorting. The algorithm maintains the maximum value seen in the previous group and compares it with the minimum value of the current group of numbers with the same bit count.

Step-by-Step Algorithm

Step 1: Initialize `prev_max` to 0. This variable will store the maximum value of the previous group of numbers having the same number of set bits.
Step 2: Iterate through the `nums` array using a `while` loop controlled by the index `i`.
Step 3: Inside the outer loop, determine the number of set bits for the current element `nums[i]` using the appropriate bit counting function (e.g., `bit_count` in C, `Integer.bitCount` in Java, `__builtin_popcount` in C++, `bit_count()` method in Python)
Step 4: Initialize `curr_min` and `curr_max` to the current element `nums[i]`. These variables will store the minimum and maximum values within the current group.
Step 5: Use an inner `while` loop (controlled by the index `j`) to traverse the contiguous elements that have the same number of set bits as `nums[i]`.
Step 6: Within the inner loop, update `curr_min` and `curr_max` by comparing them with each element `nums[j]`.
Step 7: After the inner loop finishes, check if `curr_min` is less than `prev_max`. If it is, it means that the array cannot be sorted using the allowed swaps, so return `false`.
Step 8: Update `prev_max` to `curr_max` as we finish processing current group.
Step 9: Increment the outer loop index `i` to `j`, effectively moving to the next group of numbers.
Step 10: If the outer loop completes without returning `false`, it means that the array can be sorted using the allowed swaps, so return `true`.

Key Insights

Insight 1: The key is to recognize that numbers with the same number of set bits form independent groups that can be internally sorted, but cannot be intermixed with numbers from other groups.
Insight 2: The problem can be solved by iterating through the array and checking if the minimum value within each group of numbers with the same number of set bits is greater than or equal to the maximum value of the previous group.
Insight 3: No explicit sorting is required; the solution only needs to verify if sorting via adjacent swaps based on the set bits constraint is possible.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Bit Manipulation, Sorting.

Companies

Asked at: Edelweiss Group.

Find if Array Can Be Sorted | LeetCode 3011

Medium

Array Bit Manipulation Sorting

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean canSortArray(int[] nums) {
        int n = nums.length;
        int i = 0;
        int prev_max = 0;
        while (i < n) {
            int j = i;
            int bits = Integer.bitCount(nums[i]);
            
            int curr_min = nums[i];
            int curr_max = nums[i];
            
            j += 1;
            while (j < n && Integer.bitCount(nums[j]) == bits) {
                curr_min = Math.min(curr_min, nums[j]);
                curr_max = Math.max(curr_max, nums[j]);
                j += 1;
            }

            if (curr_min < prev_max) {
                return false;
            }

            prev_max = curr_max;
            i = j;
            
        }
        return true;
    }
}

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Problem Info

Description

Problem 3011: Find if Array Can Be Sorted - Detailed explanation and solution approaches.

Topics

Array Bit Manipulation Sorting

Companies

Edelweiss Group

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3011

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

10 steps

1
Step 1: Initialize `prev_max` to 0. This variable will store the maximum value of the previous group of numbers having the same number of set bits.
2
Step 2: Iterate through the `nums` array using a `while` loop controlled by the index `i`.
3
Step 3: Inside the outer loop, determine the number of set bits for the current element `nums[i]` using the appropriate bit counting function (e.g., `bit_count` in C, `Integer.bitCount` in Java, `__builtin_popcount` in C++, `bit_count()` method in Python)
4
Step 4: Initialize `curr_min` and `curr_max` to the current element `nums[i]`. These variables will store the minimum and maximum values within the current group.
5
Step 5: Use an inner `while` loop (controlled by the index `j`) to traverse the contiguous elements that have the same number of set bits as `nums[i]`.
6
Step 6: Within the inner loop, update `curr_min` and `curr_max` by comparing them with each element `nums[j]`.
7
Step 7: After the inner loop finishes, check if `curr_min` is less than `prev_max`. If it is, it means that the array cannot be sorted using the allowed swaps, so return `false`.
8
Step 8: Update `prev_max` to `curr_max` as we finish processing current group.
9
Step 9: Increment the outer loop index `i` to `j`, effectively moving to the next group of numbers.
10
Step 10: If the outer loop completes without returning `false`, it means that the array can be sorted using the allowed swaps, so return `true`.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because the outer loop iterates through the array once, and the inner loop processes a contiguous group of elements with the same number of set bits. In the worst case, the inner loop might iterate over the entire array, but overall, each element of the array is visited at most a constant number of times. The space complexity is O(1) because the algorithm uses a fixed number of variables regardless of the input array size.

📖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

Explicit Sorting with Custom Comparator

An alternative approach would be to explicitly sort the array using a custom comparator that first compares the number of set bits and then compares the numbers themselves if the number of set bits is the same. This would involve implementing a sorting algorithm (e.g., merge sort, quicksort) or using a built-in sorting function with the custom comparator.

Time: O(n log n), where n is the length of the array. Sorting algorithms typically have a time complexity of O(n log n).Space: O(n) or O(log n) depending on the sorting algorithm used. Merge sort has O(n) space complexity, while quicksort has O(log n) on average.

Trade-offs: This approach is less efficient than the described solution due to the higher time complexity of sorting. However, it might be easier to understand for some programmers, especially those familiar with sorting algorithms. Additionally, the sorting approach provides the actual sorted array based on the swap conditions, while current solution only checks if sorting is possible.

✨

Key Insights

1Insight 1: The key is to recognize that numbers with the same number of set bits form independent groups that can be internally sorted, but cannot be intermixed with numbers from other groups.
2Insight 2: The problem can be solved by iterating through the array and checking if the minimum value within each group of numbers with the same number of set bits is greater than or equal to the maximum value of the previous group.
3Insight 3: No explicit sorting is required; the solution only needs to verify if sorting via adjacent swaps based on the set bits constraint is possible.

📚

Key Terms Explained

For Students

Array

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty array. The problem states that 1 <= nums.length <= 100, so an empty array will not be an input. If it were, the algorithm would either not execute (depending on the language's array access behavior), or would need a check for an empty array at the beginning.
Edge case 2: Already sorted array. If the input array is already sorted, the algorithm will correctly return `true` because `curr_min` will always be greater than or equal to `prev_max` in each group.
Edge case 3: Array with all elements having the same number of set bits. The algorithm will correctly process such an array, checking if it is sorted. The result will depend on the actual ordering of elements.
Edge case 4: Array with repeated elements. The algorithm will correctly handle duplicate elements as long as they can be grouped and 'sorted' with the adjacent swap rule. If numbers with same set bit counts are out of order, then the algorithm correctly return `false`

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the number of set bits. It's important to use the correct bit counting function for the programming language being used (e.g., `bit_count` in Python, `Integer.bitCount` in Java, `__builtin_popcount` in C++). An incorrect calculation will lead to incorrect grouping and sorting decisions.

✗Common mistake 2: Failing to account for the interaction between adjacent groups. The key insight is to realize that the minimum value of a group must be greater than or equal to the maximum value of the previous group. Forgetting this constraint will lead to incorrect results.

✗Common mistake 3: Trying to explicitly sort the array using bubble sort with the swap condition. While conceptually correct, bubble sort can lead to TLE since the problem does not guarantee that elements with the same set bit count can be adjacent in all scenarios.

✅

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: Data compression. In some data compression algorithms, data elements with a similar number of set bits might be grouped together to improve compression efficiency. This problem's concept could be applied to analyze if such grouping is possible while maintaining a certain order or structure.

🏦Real-world scenario 2: Task scheduling. Imagine scheduling tasks where certain tasks can only be reordered if they have similar resource requirements (represented by the number of set bits). This algorithm could be used to determine if a given task list can be sorted based on resource constraints by only swapping similar-resource tasks.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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 10, 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

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

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 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 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 "Find if Array Can Be Sorted"?

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 bit manipulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?