LeetCode #1787 — HARD

Make the XOR of All Segments Equal to Zero

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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 [0,0,0,0,0].

Example 2:

Input: nums = [3,4,5,2,1,7,3,4,7], k = 3
Output: 3
Explanation: Modify the array from [3,4,5,2,1,7,3,4,7] to [3,4,7,3,4,7,3,4,7].

Example 3:

Input: nums = [1,2,4,1,2,5,1,2,6], k = 3
Output: 3
Explanation: Modify the array from [1,2,4,1,2,5,1,2,6] to [1,2,3,1,2,3,1,2,3].

Constraints:

1 <= k <= nums.length <= 2000
0 <= nums[i] < 2¹⁰

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Dynamic Programming · Bit Manipulation

Example 1

[1,2,0,3,0]
1

Example 2

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

Example 3

[1,2,4,1,2,5,1,2,6]
3

Core Insight

What unlocks the optimal approach

Let's note that for the XOR of all segments with size K to be equal to zeros, nums[i] has to be equal to nums[i+k]
Basically, we need to make the first K elements have XOR = 0 and then modify them.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1787: Make the XOR of All Segments Equal to Zero
class Solution {
    public int minChanges(int[] nums, int k) {
        int n = 1 << 10;
        Map<Integer, Integer>[] cnt = new Map[k];
        Arrays.setAll(cnt, i -> new HashMap<>());
        int[] size = new int[k];
        for (int i = 0; i < nums.length; ++i) {
            int j = i % k;
            cnt[j].merge(nums[i], 1, Integer::sum);
            size[j]++;
        }
        int[] f = new int[n];
        final int inf = 1 << 30;
        Arrays.fill(f, inf);
        f[0] = 0;
        for (int i = 0; i < k; ++i) {
            int[] g = new int[n];
            Arrays.fill(g, min(f) + size[i]);
            for (int j = 0; j < n; ++j) {
                for (var e : cnt[i].entrySet()) {
                    int v = e.getKey(), c = e.getValue();
                    g[j] = Math.min(g[j], f[j ^ v] + size[i] - c);
                }
            }
            f = g;
        }
        return f[0];
    }

    private int min(int[] arr) {
        int mi = arr[0];
        for (int v : arr) {
            mi = Math.min(mi, v);
        }
        return mi;
    }
}

// Accepted solution for LeetCode #1787: Make the XOR of All Segments Equal to Zero
func minChanges(nums []int, k int) int {
	n := 1 << 10
	cnt := make([]map[int]int, k)
	for i := range cnt {
		cnt[i] = map[int]int{}
	}
	size := make([]int, k)
	for i, v := range nums {
		cnt[i%k][v]++
		size[i%k]++
	}
	f := make([]int, n)
	for i := 1; i < n; i++ {
		f[i] = 0x3f3f3f3f
	}
	for i, sz := range size {
		g := make([]int, n)
		x := slices.Min(f) + sz
		for i := range g {
			g[i] = x
		}
		for j := 0; j < n; j++ {
			for v, c := range cnt[i] {
				g[j] = min(g[j], f[j^v]+sz-c)
			}
		}
		f = g
	}
	return f[0]
}

# Accepted solution for LeetCode #1787: Make the XOR of All Segments Equal to Zero
class Solution:
    def minChanges(self, nums: List[int], k: int) -> int:
        n = 1 << 10
        cnt = [Counter() for _ in range(k)]
        size = [0] * k
        for i, v in enumerate(nums):
            cnt[i % k][v] += 1
            size[i % k] += 1
        f = [inf] * n
        f[0] = 0
        for i in range(k):
            g = [min(f) + size[i]] * n
            for j in range(n):
                for v, c in cnt[i].items():
                    g[j] = min(g[j], f[j ^ v] + size[i] - c)
            f = g
        return f[0]

// Accepted solution for LeetCode #1787: Make the XOR of All Segments Equal to Zero
/**
 * [1787] Make the XOR of All Segments Equal to Zero
 *
 * You are given an array nums and an integer k. The <font face="monospace">XOR</font> 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 [<u>1</u>,<u>2</u>,0,<u>3</u>,0] to from [<u>0</u>,<u>0</u>,0,<u>0</u>,0].
 *
 * Example 2:
 *
 * Input: nums = [3,4,5,2,1,7,3,4,7], k = 3
 * Output: 3
 * Explanation: Modify the array from [3,4,<u>5</u>,<u>2</u>,<u>1</u>,7,3,4,7] to [3,4,<u>7</u>,<u>3</u>,<u>4</u>,7,3,4,7].
 *
 * Example 3:
 *
 * Input: nums = [1,2,4,1,2,5,1,2,6], k = 3
 * Output: 3
 * Explanation: Modify the array from [1,2,<u>4,</u>1,2,<u>5</u>,1,2,<u>6</u>] to [1,2,<u>3</u>,1,2,<u>3</u>,1,2,<u>3</u>].
 *  
 * Constraints:
 *
 * 	1 <= k <= nums.length <= 2000
 * 	0 <= nums[i] < 2^10
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/make-the-xor-of-all-segments-equal-to-zero/
// discuss: https://leetcode.com/problems/make-the-xor-of-all-segments-equal-to-zero/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn min_changes(nums: Vec<i32>, k: i32) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_1787_example_1() {
        let nums = vec![1, 2, 0, 3, 0];
        let k = 1;

        let result = 3;

        assert_eq!(Solution::min_changes(nums, k), result);
    }

    #[test]
    #[ignore]
    fn test_1787_example_2() {
        let nums = vec![3, 4, 5, 2, 1, 7, 3, 4, 7];
        let k = 3;

        let result = 3;

        assert_eq!(Solution::min_changes(nums, k), result);
    }

    #[test]
    #[ignore]
    fn test_1787_example_3() {
        let nums = vec![1, 2, 4, 1, 2, 5, 1, 2, 6];
        let k = 3;

        let result = 3;

        assert_eq!(Solution::min_changes(nums, k), result);
    }
}

// Accepted solution for LeetCode #1787: Make the XOR of All Segments Equal to Zero
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1787: Make the XOR of All Segments Equal to Zero
// class Solution {
//     public int minChanges(int[] nums, int k) {
//         int n = 1 << 10;
//         Map<Integer, Integer>[] cnt = new Map[k];
//         Arrays.setAll(cnt, i -> new HashMap<>());
//         int[] size = new int[k];
//         for (int i = 0; i < nums.length; ++i) {
//             int j = i % k;
//             cnt[j].merge(nums[i], 1, Integer::sum);
//             size[j]++;
//         }
//         int[] f = new int[n];
//         final int inf = 1 << 30;
//         Arrays.fill(f, inf);
//         f[0] = 0;
//         for (int i = 0; i < k; ++i) {
//             int[] g = new int[n];
//             Arrays.fill(g, min(f) + size[i]);
//             for (int j = 0; j < n; ++j) {
//                 for (var e : cnt[i].entrySet()) {
//                     int v = e.getKey(), c = e.getValue();
//                     g[j] = Math.min(g[j], f[j ^ v] + size[i] - c);
//                 }
//             }
//             f = g;
//         }
//         return f[0];
//     }
// 
//     private int min(int[] arr) {
//         int mi = arr[0];
//         for (int v : arr) {
//             mi = Math.min(mi, v);
//         }
//         return mi;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.