What is the time complexity of Make the XOR of All Segments Equal to Zero?

The optimal solution has a time complexity of O(k * MAXXOR * MAXXOR).

What is the space complexity of Make the XOR of All Segments Equal to Zero?

The space complexity is O(k * MAXXOR).

What data structures are used to solve Make the XOR of All Segments Equal to Zero?

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

What are the key insights for solving Make the XOR of All Segments Equal to Zero?

Insight 1: The key insight is to recognize that if the XOR of every segment of size `k` is zero, then `nums[i] ^ nums[i+1] ^ ... ^ nums[i+k-1] == 0` for all `i`. This implies `nums[i] == nums[i+k]` for all valid `i`. In other words, the array becomes periodic with period `k`. Insight 2: We can use dynamic programming to determine the minimum changes for each subproblem. The subproblems are defined by each index `i` from 0 to `k-1`. For each index `i`, we need to find the optimal value to which we should change the element to maintain the periodicity.

Which companies ask LeetCode 1787. Make the XOR of All Segments Equal to Zero in interviews?

This problem is frequently asked at Media.net.

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Make the XOR of All Segments Equal to Zero - LeetCode 1787 Solution

Q: How to solve Make the XOR of All Segments Equal to Zero?

The provided solution employs dynamic programming. The algorithm breaks the problem into `k` subproblems where the `i`-th subproblem (0 <= i < k) focuses on minimizing changes in the elements at indices i, i+k, i+2k, and so on. For each subproblem, it iterates through all possible XOR values (0 to 2^10 - 1). It maintains a `dp` array where `dp[j]` represents the minimum number of changes needed to achieve a XOR value of `j` considering previous `i` values (modulo k). For each possible value `v` at index `i`, the solution calculates the `cost` of changing the elements at `i` (modulo `k`) to `v`. This cost is added to the previously calculated minimum value (`dp[prev_xor]`). The solution then updates `dp[j]` with the minimum cost to achieve `j` XOR value. Finally, `dp[0]` gives the minimum number of changes required.

Make the XOR of All Segments Equal to Zero - Complete Solution Guide

Make the XOR of All Segments Equal to Zero is LeetCode problem 1787, 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 an array nums and an integer k . The XOR of a segment [left, right] where left <= right is the XOR of all the elements with indices between left and right , inclusive: nums[left] XOR nums[left+1] XOR ... XOR nums[right] . Return the minimum number of elements to change in the array such that the XOR of all segments of size k is equal to zero. Example 1: Input: nums = [1,2,0,3,0], k = 1 Output: 3 Explanation: Modify the array from [ 1 , 2 ,0, 3 ,0] to from [ 0 , 0 ,0

Detailed Explanation

The problem requires minimizing the number of changes in an array `nums` such that the XOR of all segments of size `k` is equal to zero. A segment is a contiguous subarray of length `k`. The XOR of a segment is the bitwise XOR of all its elements. We need to find the fewest modifications to `nums` to satisfy this XOR constraint. The constraints are: 1 <= k <= nums.length <= 2000 and 0 <= nums[i] < 2^10.

Solution Approach

The provided solution employs dynamic programming. The algorithm breaks the problem into `k` subproblems where the `i`-th subproblem (0 <= i < k) focuses on minimizing changes in the elements at indices i, i+k, i+2k, and so on. For each subproblem, it iterates through all possible XOR values (0 to 2^10 - 1). It maintains a `dp` array where `dp[j]` represents the minimum number of changes needed to achieve a XOR value of `j` considering previous `i` values (modulo k). For each possible value `v` at index `i`, the solution calculates the `cost` of changing the elements at `i` (modulo `k`) to `v`. This cost is added to the previously calculated minimum value (`dp[prev_xor]`). The solution then updates `dp[j]` with the minimum cost to achieve `j` XOR value. Finally, `dp[0]` gives the minimum number of changes required.

Step-by-Step Algorithm

Step 1: Initialize `counts`: Create `k` counters to store the frequency of each number at indices `i % k`.
Step 2: Calculate `sizes`: Record the number of elements belonging to each index `i % k`.
Step 3: Initialize `dp`: Create a DP array `dp` of size `MAXXOR` (2^10), where `dp[j]` stores the minimum number of changes to achieve a XOR value of `j` for the first column (i.e., i % k == 0). This is initialized by calculating the number of changes if every entry at index `i % k == 0` are changed to a single value.
Step 4: Iterate through subproblems: For each index `i` from 1 to `k-1`, update the `dp` array. For each possible XOR value `j`, calculate the cost of making the XOR of `i % k` equal to `j`. This involves XORing each value `v` with `j` (`prev_xor = j ^ v`) and updating the DP array with the minimum cost.
Step 5: Optimize with `min_prev_dp`: To optimize, the solution maintains `min_prev_dp`, representing the minimum value in the previous `dp` array. This is because, without any specific value being picked for `i % k == currentColumn`, the best would be to take the precomputed minimum and changing the total amount of cells in that currentColumn. `dp[j]` is initialized to `min_prev_dp + sizes[i]`.
Step 6: Calculate cost: The `cost` to change a value to `j ^ prev_xor`, is `dp[prev_xor] + sizes[i] - count` where count is how many of elements were actually equal to `j ^ prev_xor` at position `i % k`
Step 7: Return `dp[0]`: The final answer is stored in `dp[0]`, which represents the minimum number of changes to make the XOR of all segments of size `k` equal to zero.

Key Insights

Insight 1: The key insight is to recognize that if the XOR of every segment of size `k` is zero, then `nums[i] ^ nums[i+1] ^ ... ^ nums[i+k-1] == 0` for all `i`. This implies `nums[i] == nums[i+k]` for all valid `i`. In other words, the array becomes periodic with period `k`.
Insight 2: We can use dynamic programming to determine the minimum changes for each subproblem. The subproblems are defined by each index `i` from 0 to `k-1`. For each index `i`, we need to find the optimal value to which we should change the element to maintain the periodicity.
Insight 3: Optimizing the dynamic programming involves considering the trade-off between changing all elements in one group (at index i%k for all i) to a specific value, or using pre-calculated minimum values from previous subproblems.

Complexity Analysis

Time Complexity: O(k * MAXXOR * MAXXOR)

Space Complexity: O(k * MAXXOR)

Topics

This problem involves: Array, Dynamic Programming, Bit Manipulation.

Companies

Asked at: Media.net.

Make the XOR of All Segments Equal to Zero | LeetCode 1787

Hard

Array Dynamic Programming Bit Manipulation

Time: O(k * MAXXOR * MAXXOR)Space: O(k * MAXXOR)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public int minChanges(int[] nums, int k) {
        int MAXXOR = 1 << 10;
        
        List<Map<Integer, Integer>> counts = new ArrayList<>();
        for (int i = 0; i < k; i++) {
            counts.add(new HashMap<>());
        }
        
        int[] sizes = new int[k];
        
        for (int i = 0; i < nums.length; i++) {
            int num = nums[i];
            int index = i % k;
            
            Map<Integer, Integer> countMap = counts.get(index);
            countMap.put(num, countMap.getOrDefault(num, 0) + 1);
            sizes[index]++;
        }
        
        int[] dp = new int[MAXXOR];
        for (int j = 0; j < MAXXOR; j++) {
            dp[j] = sizes[0] - counts.get(0).getOrDefault(j, 0);
        }
        
        for (int i = 1; i < k; i++) {
            int min_prev_dp = Integer.MAX_VALUE;
            for (int val : dp) {
                min_prev_dp = Math.min(min_prev_dp, val);
            }
            
            int[] new_dp = new int[MAXXOR];
            for (int j = 0; j < MAXXOR; j++) {
                new_dp[j] = min_prev_dp + sizes[i];
            }
            
            Map<Integer, Integer> countMap = counts.get(i);
            for (Map.Entry<Integer, Integer> entry : countMap.entrySet()) {
                int v = entry.getKey();
                int count = entry.getValue();
                
                for (int j = 0; j < MAXXOR; j++) {
                    int prev_xor = j ^ v;
                    int cost = dp[prev_xor] + sizes[i] - count;
                    if (cost < new_dp[j]) {
                        new_dp[j] = cost;
                    }
                }
            }
            
            dp = new_dp;
        }
        
        return dp[0];
    }
}

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Problem Info

Description

Problem 1787: Make the XOR of All Segments Equal to Zero - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Bit Manipulation

Companies

Media.net

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1787

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 `counts`: Create `k` counters to store the frequency of each number at indices `i % k`.
2
Step 2: Calculate `sizes`: Record the number of elements belonging to each index `i % k`.
3
Step 3: Initialize `dp`: Create a DP array `dp` of size `MAXXOR` (2^10), where `dp[j]` stores the minimum number of changes to achieve a XOR value of `j` for the first column (i.e., i % k == 0). This is initialized by calculating the number of changes if every entry at index `i % k == 0` are changed to a single value.
4
Step 4: Iterate through subproblems: For each index `i` from 1 to `k-1`, update the `dp` array. For each possible XOR value `j`, calculate the cost of making the XOR of `i % k` equal to `j`. This involves XORing each value `v` with `j` (`prev_xor = j ^ v`) and updating the DP array with the minimum cost.
5
Step 5: Optimize with `min_prev_dp`: To optimize, the solution maintains `min_prev_dp`, representing the minimum value in the previous `dp` array. This is because, without any specific value being picked for `i % k == currentColumn`, the best would be to take the precomputed minimum and changing the total amount of cells in that currentColumn. `dp[j]` is initialized to `min_prev_dp + sizes[i]`.
6
Step 6: Calculate cost: The `cost` to change a value to `j ^ prev_xor`, is `dp[prev_xor] + sizes[i] - count` where count is how many of elements were actually equal to `j ^ prev_xor` at position `i % k`
7
Step 7: Return `dp[0]`: The final answer is stored in `dp[0]`, which represents the minimum number of changes to make the XOR of all segments of size `k` equal to zero.

💡

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(k * MAXXOR * MAXXOR)

💾

Space Complexity

O(k * MAXXOR)

📝

Why?

The time complexity stems from the three nested loops. The outer loop iterates `k` times. The middle loop iterates through all possible XOR values (up to MAXXOR = 2^10). The inner loop iterates over each value that appeared in `counts[i]`. In the worst case, the number of unique values in each group `counts[i]` can reach `MAXXOR`, making it O(k * MAXXOR * MAXXOR). The space complexity comes from storing the `counts` array and `dp` array, both with a size proportional to `k * MAXXOR` and `MAXXOR` respectively.

📖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 (Not feasible)

Trying all possible combinations of changes would be extremely inefficient because of the large number of combinations. This would involve trying every possible value for each element in `nums`, and checking if the XOR of all segments of size `k` is zero.

Time: O(MAXXOR^n), where n is the length of `nums` and MAXXOR is 2^10.Space: O(1)

Trade-offs: Not feasible due to exponential time complexity.

✨

Key Insights

1Insight 1: The key insight is to recognize that if the XOR of every segment of size `k` is zero, then `nums[i] ^ nums[i+1] ^ ... ^ nums[i+k-1] == 0` for all `i`. This implies `nums[i] == nums[i+k]` for all valid `i`. In other words, the array becomes periodic with period `k`.
2Insight 2: We can use dynamic programming to determine the minimum changes for each subproblem. The subproblems are defined by each index `i` from 0 to `k-1`. For each index `i`, we need to find the optimal value to which we should change the element to maintain the periodicity.
3Insight 3: Optimizing the dynamic programming involves considering the trade-off between changing all elements in one group (at index i%k for all i) to a specific value, or using pre-calculated minimum values from previous subproblems.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `k = 1`. In this case, all elements must be zeroed out. The solution correctly handles this because in first loop, it'll simply reduce the array to array of all zeros.
Edge case 2: Empty input array (though the constraints state `nums.length >= 1`). An empty array would require no changes, and the solution would likely need a check for this trivial input if allowed by the constraints.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not recognizing the periodicity constraint. Failing to identify that `nums[i] == nums[i+k]` is crucial for an efficient solution.

✗Common mistake 2: Incorrectly calculating the cost of changing elements. The cost should be the number of elements that are *not* equal to the chosen XOR value.

✗Common mistake 3: Incorrectly implementing the dynamic programming transitions. The transitions must consider the previous XOR value and the cost of changing the current element to maintain the periodic pattern.

✅

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. Finding patterns in data segments to compress them effectively. The XOR operation is frequently used in compression algorithms.

🏦Real-world scenario 2: Error detection and correction. Using XOR to create parity bits that can detect and correct errors in data transmission or storage. The segments of data can be considered as the array in this problem.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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: November 15, 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

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

@TechLeadJun 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.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Make the XOR of All Segments Equal to Zero"?

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

🤖 Need help with this problem?