What is the time complexity of Minimize OR of Remaining Elements Using Operations?

The optimal solution has a time complexity of O(n * log(max(nums))).

What data structures are used to solve Minimize OR of Remaining Elements Using Operations?

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

What are the key insights for solving Minimize OR of Remaining Elements Using Operations?

Insight 1: Binary Search is not directly applicable. We're not searching a range of possible OR values. Instead, we must iteratively determine the bits of the minimized OR value. Insight 2: The key is to determine the bits of the answer from the most significant bit to the least significant bit. We try to set each bit to 0 if possible, and if it's not possible, we keep the bit as 1.

Which companies ask LeetCode 3022. Minimize OR of Remaining Elements Using Operations in interviews?

This problem is frequently asked at Aon.

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Minimize OR of Remaining Elements Using Operations - LeetCode 3022 Solution

Q: How to solve Minimize OR of Remaining Elements Using Operations?

The solution employs a greedy approach combined with bit manipulation. It iterates through the bits of the numbers from the most significant bit (MSB) to the least significant bit (LSB), attempting to set each bit of the result to 0. For each bit position, it checks if setting that bit to 0 in the result is possible while still satisfying the constraint of performing at most `k` operations. If setting the bit to 0 is not feasible, it means that bit must be set to 1 in the final result. This process gradually builds up the minimum possible OR value.

Minimize OR of Remaining Elements Using Operations - Complete Solution Guide

Minimize OR of Remaining Elements Using Operations is LeetCode problem 3022, a Hard 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 and an integer k . In one operation, you can pick any index i of nums such that 0 <= i < nums.length - 1 and replace nums[i] and nums[i + 1] with a single occurrence of nums[i] & nums[i + 1] , where & represents the bitwise AND operator. Return the minimum possible value of the bitwise OR of the remaining elements of nums after applying at most k operations . Example 1: Input: nums = [3,5,3,2,7], k = 2 Output: 3 Explanation: Let's do the following ope

Detailed Explanation

The problem asks us to minimize the bitwise OR of the remaining elements in an array `nums` after applying at most `k` operations. Each operation allows us to replace two adjacent elements in `nums` with their bitwise AND. The goal is to find the smallest possible bitwise OR of the resulting array after performing up to `k` such operations.

Solution Approach

The solution employs a greedy approach combined with bit manipulation. It iterates through the bits of the numbers from the most significant bit (MSB) to the least significant bit (LSB), attempting to set each bit of the result to 0. For each bit position, it checks if setting that bit to 0 in the result is possible while still satisfying the constraint of performing at most `k` operations. If setting the bit to 0 is not feasible, it means that bit must be set to 1 in the final result. This process gradually builds up the minimum possible OR value.

Step-by-Step Algorithm

Step 1: Initialize the result `ans` to 0.
Step 2: Iterate from the most significant bit (29) to the least significant bit (0).
Step 3: In each iteration, calculate a `potential_ans` by setting the current bit `b` to 1 in `ans` (`ans | (1 << b)`).
Step 4: Create a `test_target` value. This value represents what the final bitwise OR must be smaller than or equal to (a submask of). It is calculated as `potential_ans - 1`, ensuring all bits higher than 'b' match 'ans', bit 'b' is 0 and all lower bits are 1 (we want to know if we can find a way that results in the 'b' bit being 0 in the minimized OR)
Step 5: Call the `check` function with `test_target`, `nums`, and `k` to see if we can reach a solution whose OR is a submask of `test_target` with at most `k` operations.
Step 6: If `check(test_target)` returns `false`, it means setting the `b`-th bit to 0 is not possible without exceeding `k` operations or violating the bitwise OR constraints. Therefore, we update `ans` to include the `b`-th bit (`ans = potential_ans`).
Step 7: After iterating through all bits, the `ans` variable holds the minimum possible bitwise OR value.

Key Insights

Insight 1: Binary Search is not directly applicable. We're not searching a range of possible OR values. Instead, we must iteratively determine the bits of the minimized OR value.
Insight 2: The key is to determine the bits of the answer from the most significant bit to the least significant bit. We try to set each bit to 0 if possible, and if it's not possible, we keep the bit as 1.
Insight 3: The `check` function efficiently verifies if a target OR value (or a submask of it) can be achieved with at most `k` operations by greedily forming groups where the AND of elements is a submask of the target.

Complexity Analysis

Time Complexity: O(n * log(max(nums)))

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy, Bit Manipulation.

Companies

Asked at: Aon.

Minimize OR of Remaining Elements Using Operations | LeetCode 3022

Hard

Array Greedy Bit Manipulation

Time: O(n * log(max(nums)))Space: O(1)

Solution Code

Available in:

class Solution {
    public int minOrAfterOperations(int[] nums, int k) {
        int n = nums.length;

        // Checks if it's possible to achieve a final OR value that is a submask of target_or_mask.
        // This is possible if we can partition nums into at least n-k groups,
        // where the bitwise AND of each group is a submask of target_or_mask.
        // We greedily form as many groups as possible by making each group as short as possible.
        
        
        class Helper {
            public boolean check(int target_or_mask, int[] nums, int k) {
                int groups = 0;
                int current_and = -1; // Represents all bits set to 1

                for (int num : nums) {
                    current_and &= num;
                    if ((current_and | target_or_mask) == target_or_mask) {
                        groups += 1;
                        current_and = -1;
                    }
                }

                return groups >= n - k;
            }
        }

        Helper helper = new Helper();

        int ans = 0;
        // We determine the bits of the minimum OR value from most significant to least significant.
        for (int b = 29; b >= 0; --b) {
            // We try to make the b-th bit of the result 0.
            // `ans` holds the bits greater than b that are determined to be 1.
            // `potential_ans` is the number with bits > b from `ans` and bit `b` set to 1.
            int potential_ans = ans | (1 << b);

            // `test_target` is a mask where bits > b match `ans`, bit `b` is 0,
            // and bits < b are 1 (wildcards).
            int test_target = potential_ans - 1;

            if (!helper.check(test_target, nums, k)) {
                // If check(test_target) is false, it's impossible to have a final OR value
                // that is a submask of `test_target`. This means the b-th bit must be 1
                // in the minimal result.
                ans = potential_ans;
            }
        }

        return ans;
    }
}

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Problem Info

Description

Problem 3022: Minimize OR of Remaining Elements Using Operations - Detailed explanation and solution approaches.

Topics

Array Greedy Bit Manipulation

Companies

Aon

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3022

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

7 steps

1
Step 1: Initialize the result `ans` to 0.
2
Step 2: Iterate from the most significant bit (29) to the least significant bit (0).
3
Step 3: In each iteration, calculate a `potential_ans` by setting the current bit `b` to 1 in `ans` (`ans | (1 << b)`).
4
Step 4: Create a `test_target` value. This value represents what the final bitwise OR must be smaller than or equal to (a submask of). It is calculated as `potential_ans - 1`, ensuring all bits higher than 'b' match 'ans', bit 'b' is 0 and all lower bits are 1 (we want to know if we can find a way that results in the 'b' bit being 0 in the minimized OR)
5
Step 5: Call the `check` function with `test_target`, `nums`, and `k` to see if we can reach a solution whose OR is a submask of `test_target` with at most `k` operations.
6
Step 6: If `check(test_target)` returns `false`, it means setting the `b`-th bit to 0 is not possible without exceeding `k` operations or violating the bitwise OR constraints. Therefore, we update `ans` to include the `b`-th bit (`ans = potential_ans`).
7
Step 7: After iterating through all bits, the `ans` variable holds the minimum possible bitwise OR value.

💡

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 * log(max(nums)))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The outer loop iterates 30 times (from bit 29 to 0), which contributes a factor of O(log(max(nums))). The `check` function iterates through the entire `nums` array once, contributing O(n). The bitwise operations inside `check` are constant time. Therefore, the overall time complexity is O(n * log(max(nums))). The algorithm uses only a constant amount of extra space for variables, hence the space complexity is O(1).

📖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

Dynamic Programming

A dynamic programming approach could be used to determine the minimum OR value. DP[i][j][mask] could store whether it is possible to obtain an array of length 'i' using 'j' operations whose OR value is equal to mask. However, this approach is likely to be less efficient due to the large state space.

Time: O(n^2 * 2^30)Space: O(n^2 * 2^30)

Trade-offs: The DP approach has significantly higher time and space complexity, making it impractical. The greedy approach is much more efficient for this problem.

✨

Key Insights

1Insight 1: Binary Search is not directly applicable. We're not searching a range of possible OR values. Instead, we must iteratively determine the bits of the minimized OR value.
2Insight 2: The key is to determine the bits of the answer from the most significant bit to the least significant bit. We try to set each bit to 0 if possible, and if it's not possible, we keep the bit as 1.
3Insight 3: The `check` function efficiently verifies if a target OR value (or a submask of it) can be achieved with at most `k` operations by greedily forming groups where the AND of elements is a submask of the target.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: `k = 0`. In this case, no operations can be performed, so the minimum OR is just the OR of all elements in the original array. The code correctly handles this because the check function will require groups equal to n.
Edge case 2: `k = nums.length - 1`. In this case, we can reduce the array to a single element, which will be the AND of all elements in the array. The binary search approach implicitly handles this as it attempts to make bits zero if possible.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the bits from most significant to least significant. Processing the bits in the wrong order will not guarantee the minimum OR value.

✗Common mistake 2: Incorrect implementation of the `check` function. Failing to correctly determine the number of groups that can be formed under the constraint of `k` operations will lead to an incorrect result. For example, failing to reinitialize current_and within the `check` function.

✅

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: Network routing, where the 'OR' operation represents combining network capabilities. Minimizing the OR of the routes corresponds to finding routes with the least combination of capabilities necessary to meet a certain criteria.

🏦Real-world scenario 2: Data compression, where bitwise AND operations are used to compress data. The problem mirrors a scenario where you want to minimize the information loss (represented by the OR of remaining elements) while applying a certain number of compression operations (bitwise AND).

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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 21, 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

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

@InterviewProJul 13

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

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimize OR of Remaining Elements Using Operations"?

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

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