Make The Xor Of All Segments Equal To Zero - Leetcode Solution

Problem Description

Given an array nums consisting of n integers. Find the minimum number of operations required to make all the XOR of all segments of nums equal to 0.

A XOR of a segment l, r (l-indexed and r-indexed inclusive) of an array is the XOR of all elements between and including l and r (nums[l] XOR nums[l+1] XOR ... XOR nums[r]).

Solution

We can approach this problem by trying to find a pattern in the XOR of all segments of the input array. Once we find the pattern, we can easily determine the number of operations required to make all XOR of all segments equal to zero.

Let's assume that the XOR of all segments of the array is already 0, as then no operations are required. Now suppose we apply any operation on this array: If we add an element x to any position i in the array, then all segments that contained this position i will now have x added to them. This will change the XOR of these segments from 0 to x. This results in all other segments also being changed by applying x to them. Therefore, we can conclude that one operation changes the XOR of all segments to a fixed value x, which means that we need to perform n operations to get all XOR of all segments to 0.

But, we do not need to actually perform n operations. Instead, we need to find a pattern in the XOR of all segments, and perform the minimum number of operations that would change the XOR of all segments to 0.

To find the pattern in the XOR of all segments, let's take an example. Suppose we have an array nums = [1, 2, 3, 4, 5, 6]. We can visualize all the segments of the array as a matrix, where each element m[i][j] of the matrix contains the XOR of the segment i, j of the array.

1   3   0   4   1   7
    2   1   5   6   3
        3   7   4   5
            4   0   1
                5   6
                    6

We can see that the diagonal elements of the matrix contain only one element, which means that their value is the element itself. For example, m[0][0] = 1, m[1][1] = 2, and so on.

Similarly, the elements above the diagonal contain the XOR of the current element and the previous element in the row. For example, m[0][1] is the XOR of 1 and 2, which is 3.

Similarly, the elements below the diagonal contain the XOR of the current element and the previous element in the column. For example, m[1][0] is the XOR of 1 and 2, which is 3.

Since the matrix is symmetric, we can focus only on the upper triangular part of the matrix. From the above observations, we can conclude that the XOR of all segments can be expressed as:

m[0][0] XOR
m[1][1] XOR
m[2][2] XOR
.
.
.
m[n-1][n-1] XOR

2 * (m[0][1] XOR m[0][2] XOR ... XOR m[0][n-1]) XOR
2 * (m[1][2] XOR m[1][3] XOR ... XOR m[1][n-1]) XOR
.
.
.
2 * (m[n-3][n-2] XOR m[n-3][n-1]) XOR
2 * (m[n-2][n-1])

We can simplify this expression by using XOR properties A XOR A = 0 and A XOR 0 = A.

m[0][0] XOR
m[1][1] XOR
m[2][2] XOR
.
.
.
m[n-1][n-1] XOR

2 * (m[0][1] XOR m[1][2] XOR ... XOR m[n-2][n-1])

Now we only need to focus on finding the XOR of all elements in the upper triangular part of the matrix. To do this efficiently, we can use dynamic programming.

Let dp[i][j] be the XOR of all elements in the upper triangular part of the matrix, such that the last element is m[i][j]. Then, we can express dp[i][j] as follows.

dp[i][j] = m[i][j] ^
           dp[i-1][j] ^ dp[i][j-1] ^
           dp[i-1][j-1]

The first term m[i][j] is the current element. The second term dp[i-1][j] ^ dp[i][j-1] is the XOR of all elements in the upper triangle part of the matrix, excluding the current element. This is the XOR of all elements on the row above m[i][j] and the column to the left of m[i][j]. The third term dp[i-1][j-1] is the XOR of all elements in the upper triangle part of the matrix, excluding the current element and the elements in the row above and column to the left of it. This is the value of dp[i-1][j-1] for the element m[i][j].

We can fill the dp table using the above formula and find the XOR of all elements in the upper triangle part of the matrix as dp[n-1][n-1].

Finally, we can calculate the minimum number of operations required to make all XOR of all segments equal to 0 as n - (dp[n-1][n-1]), since we need to perform n - (dp[n-1][n-1]) operations to change the XOR of all segments to 0.

Time Complexity

• Calculating the matrix and dp table takes O(n^2) time.
• Finding the XOR of all elements in the upper triangle part of the matrix takes O(1) time.
• Therefore, the overall time complexity of the solution is O(n^2).

Space Complexity

• We need to store the matrix, which takes O(n^2) space.
• We also need to store the dp table, which also takes O(n^2) space.
• Therefore, the overall space complexity of the solution is O(n^2).

Pseudo Code

function minOperations(nums: List[int]) -> int:
    n = len(nums)

    # Calculate the matrix
    matrix = [[0] * n for _ in range(n)]
    for i in range(n):
        for j in range(i, n):
            if i == j:
                matrix[i][j] = nums[i]
            else:
                matrix[i][j] = matrix[i][j-1] ^ nums[j]

    # Calculate the dp table
    dp = [[0] * n for _ in range(n)]
    for i in range(n):
        for j in range(i, n):
            if i == j:
                dp[i][j] = matrix[i][j]
            elif i == 0:
                dp[i][j] = dp[i][j-1] ^ matrix[i][j]
            elif j == 0:
                dp[i][j] = dp[i-1][j] ^ matrix[i][j]
            else:
                dp[i][j] = matrix[i][j] ^ dp[i-1][j] ^ dp[i][j-1] ^ dp[i-1][j-1]

    # Find the XOR of all elements in the upper triangular part of the matrix
    xor_all = dp[n-1][n-1]

    # Calculate the minimum number of operations required to make all XOR of all segments equal to 0
    return n - xor_all

1