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

Difficulty: 🔴 Hard
Topics: Array, Hash Table, Dynamic Programming, Bit Manipulation, Counting

Solution

Problem Understanding

The problem is asking us to modify an array nums such that the XOR of all contiguous subarrays (segments) of length k is equal to zero, while minimizing the number of changes. Each element of nums is a non-negative integer less than $2^{10} = 1024$. The input array has a length up to 2000, and k can range from 1 to the length of the array.

In simpler terms, for every consecutive sequence of k elements, the XOR of that sequence must be zero. The task is not to directly compute these XORs, but to figure out the minimum number of elements that need to be altered in the original array to achieve this property.

Key observations include that XOR is associative and commutative, which allows breaking the problem into independent positions modulo k. Specifically, elements that are k apart affect different segments independently in terms of XOR contributions. This hints at a dynamic programming approach where we consider "groups" of positions modulo k rather than individual segments.

Important edge cases include k = 1 (every element must be zero), k = n (the XOR of the whole array must be zero), and arrays where all elements are already zero or uniform. These can help verify correctness and ensure the algorithm handles minimal or maximal change scenarios efficiently.

Approaches

The brute-force approach would consider every possible modification of the array and check whether all segments of length k XOR to zero. This requires exploring all combinations of changes, which is exponential in n and clearly infeasible for the maximum constraints.

The optimal approach leverages dynamic programming and hash maps. By viewing the array in k interleaved groups based on their modulo k indices, we can track the minimum changes needed to achieve each possible XOR value for the segments. For each group, we record the frequency of numbers and use dynamic programming to propagate the minimum changes required to maintain the XOR property for all previous groups. The insight is that each group only affects the XOR sum for positions of the same modulo, which allows the problem to be split into k independent subproblems.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(2^n)	O(n)	Explore all possible modifications; infeasible
Optimal	O(n * 2^10)	O(2^10)	DP over XOR values with hash maps, feasible because nums[i] < 1024

Algorithm Walkthrough

Group the array indices by modulo k, forming k independent groups. For example, indices 0, k, 2k, ... form group 0, indices 1, k+1, 2k+1, ... form group 1, and so on.
Initialize a dynamic programming table dp where dp[x] represents the minimum number of changes needed to achieve XOR x with the groups processed so far. Initially, dp[0] = 0 and all other dp[x] are set to infinity.
For each group, count the frequency of each value appearing in the group. This allows us to know how many elements already have a particular value and thus do not need to be changed.
Create a temporary DP table for the current group. For each possible XOR value xor_val from the previous DP, and for each number num in the current group, update the temporary DP by computing the new XOR (xor_val ^ num) and adding the required changes (len(group) - frequency[num]).
After processing the current group, update the main DP table with the temporary DP table.
After all groups are processed, the answer is dp[0], the minimum changes needed to make the XOR of all segments zero.

Why it works: The algorithm works because XOR is associative and each position modulo k only affects one "slot" in every segment. By independently solving for each modulo group and propagating the XOR values using dynamic programming, we ensure the overall XOR across segments becomes zero with the fewest changes.

Python Solution

from typing import List
from collections import Counter

class Solution:
    def minChanges(self, nums: List[int], k: int) -> int:
        n = len(nums)
        dp = {0: 0}
        
        for i in range(k):
            count = Counter()
            length = 0
            for j in range(i, n, k):
                count[nums[j]] += 1
                length += 1
            
            temp_dp = {}
            min_dp = min(dp.values())
            
            for xor_val in range(1024):
                temp_dp[xor_val] = min_dp + length
            
            for xor_val, changes in dp.items():
                for num, freq in count.items():
                    new_xor = xor_val ^ num
                    temp_dp[new_xor] = min(temp_dp[new_xor], changes + length - freq)
            
            dp = temp_dp
        
        return dp[0]

This implementation first groups indices modulo k and counts the frequency of each value. The DP dictionary stores the minimum changes required to reach each XOR value. For each group, the algorithm updates the DP using the current group's frequencies, ensuring the minimum changes propagate efficiently. Finally, dp[0] holds the answer.

Go Solution

package main

func minChanges(nums []int, k int) int {
    n := len(nums)
    dp := make(map[int]int)
    dp[0] = 0

    for i := 0; i < k; i++ {
        count := make(map[int]int)
        length := 0
        for j := i; j < n; j += k {
            count[nums[j]]++
            length++
        }

        tempDP := make(map[int]int)
        minDP := int(1e9)
        for _, v := range dp {
            if v < minDP {
                minDP = v
            }
        }

        for xorVal := 0; xorVal < 1024; xorVal++ {
            tempDP[xorVal] = minDP + length
        }

        for xorVal, changes := range dp {
            for num, freq := range count {
                newXor := xorVal ^ num
                if tempDP[newXor] > changes+length-freq {
                    tempDP[newXor] = changes + length - freq
                }
            }
        }

        dp = tempDP
    }

    return dp[0]
}

The Go implementation mirrors Python's logic. The main differences are explicit map initializations and handling large numbers with int(1e9) as infinity. Iteration over maps ensures all keys are considered when updating DP.

Worked Examples

Example 1: nums = [1,2,0,3,0], k = 1

Group: [1], [2], [0], [3], [0]

DP evolves as:

Initial dp: {0:0}

Processing each group:

1 -> changes=1, xor 1 -> temp_dp[1]=1, temp_dp[0]=1

2 -> temp_dp[2]=2, temp_dp[0]=2

... final dp[0]=3

Example 2: nums = [3,4,5,2,1,7,3,4,7], k = 3

Groups: [3,2,3], [4,1,4], [5,7,7]

DP updates for each group considering frequency counts. Final dp[0]=3.

Example 3: nums = [1,2,4,1,2,5,1,2,6], k = 3

Groups: [1,1,1], [2,2,2], [4,5,6]

Final dp[0]=3 after computing minimum changes using frequencies.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n * 2^10)	Each of n elements is processed in k groups, updating DP for up to 1024 XOR values
Space	O(2^10)	DP table stores 1024 possible XOR values

The complexity is feasible because 1024 is small enough to handle in memory and iterations, and the grouping by modulo k avoids repeated recomputation across segments.

Test Cases

# Provided examples
assert Solution().minChanges([1,2,0,3,0], 1) == 3  # k=1, every element must become 0
assert Solution().minChanges([3,4,5,2,1,7,3,4,7], 3) == 3  # grouped patterns
assert Solution().minChanges([1,2,4,1,2,5,1,2,6], 3) == 3  # adjust last group

# Boundary cases
assert Solution().minChanges([0], 1) == 0  # single element already zero
assert Solution().minChanges([1], 1) == 1  # single element needs change
assert Solution().minChanges([1,1,1,1], 4) == 1  # whole array XOR must be 0

# Stress test
assert Solution().minChanges([i%1024 for i in range(2000)], 1000) >= 0  # large n, large k

Test	Why
k=1,