Count subarrays with given xor K

Intuition

We will check the XOR of every possible subarray and count how many of them are equal to k. To do this, use three nested loops. The first two loops (let's call them i and j) will iterate over every possible starting and ending index of a subarray. In each iteration, the subarray range will be from index i to index j. Using another loop, calculate the XOR of the elements in the subarray [i...j]. Count the number of subarrays where the XOR is equal to k.

Note: Select every possible subarray using two nested loops, and for each subarray, calculate the XOR of all its elements using another loop.

Approach

First, run a loop (let's call it i) that selects every possible starting index of the subarray. The starting indices can range from index 0 to n−1 (where n is the size of the array). Inside this loop, run another loop (let's call it j) that signifies the ending index of the subarray. For each subarray starting from index i, the ending index can range from i to n−1.
For each subarray from index i to j (i.e., arr[i...j]), run another loop to calculate the XOR of all the elements in that subarray.
If the XOR equals k, increase the count by 1.

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    // Function to count the number of subarrays with XOR k
    int subarraysWithXorK(vector<int>& nums, int k) {
        int n = nums.size(); 
        int cnt = 0;

        // Step 1: Generate subarrays
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                int xorr = 0;
                /* Step 2: Calculate XOR of 
                   all elements in the subarray */
                for (int K = i; K <= j; K++) {
                    xorr = xorr ^ nums[K];
                }
                // Step 3: Check XOR and count
                if (xorr == k) cnt++;
            }
        }
        return cnt;
    }
};

int main() {
    vector<int> a = {4, 2, 2, 6, 4}; 
    int k = 6; 

    // Create an instance of the Solution class
    Solution solution;

    // Function call to get the result
    int ans = solution.subarraysWithXorK(a, k);
  
    cout << "The number of subarrays with XOR k is: " << ans << "\n";
    return 0;
}

import java.util.*;

class Solution {
    // Function to count the number of subarrays with XOR k
    public int subarraysWithXorK(int[] nums, int k) {
        int n = nums.length; 
        int cnt = 0;

        // Step 1: Generate subarrays
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                int xorr = 0;
                /* Step 2: Calculate XOR of 
                   all elements in the subarray */
                for (int K = i; K <= j; K++) {
                    xorr = xorr ^ nums[K];
                }
                // Step 3: Check XOR and count
                if (xorr == k) cnt++;
            }
        }
        return cnt;
    }

    public static void main(String[] args) {
        int[] a = {4, 2, 2, 6, 4}; 
        int k = 6; 

        // Create an instance of the Solution class
        Solution solution = new Solution();

        // Function call to get the result
        int ans = solution.subarraysWithXorK(a, k);
   
        System.out.println("The number of subarrays with XOR k is: " + ans);
    }
}

class Solution:
    # Function to count the number of subarrays with XOR k
    def subarraysWithXorK(self, nums, k):
        n = len(nums)
        cnt = 0

        # Step 1: Generate subarrays
        for i in range(n):
            for j in range(i, n):
                xorr = 0
                # Step 2: Calculate XOR of all elements in the subarray
                for K in range(i, j + 1):
                    xorr ^= nums[K]
                # Step 3: Check XOR and count
                if xorr == k:
                    cnt += 1
        return cnt

if __name__ == "__main__":
    a = [4, 2, 2, 6, 4]
    k = 6

    # Create an instance of the Solution class
    solution = Solution()

    # Function call to get the result
    ans = solution.subarraysWithXorK(a, k)
 
    print("The number of subarrays with XOR k is:", ans)
class Solution {
    // Function to count the number of subarrays with XOR k
    subarraysWithXorK(nums, k) {
        const n = nums.length;
        let cnt = 0;

        // Step 1: Generate subarrays
        for (let i = 0; i < n; i++) {
            for (let j = i; j < n; j++) {
                let xorr = 0;
                /* Step 2: Calculate XOR of 
                   all elements in the subarray */
                for (let K = i; K <= j; K++) {
                    xorr ^= nums[K];
                }
                // Step 3: Check XOR and count
                if (xorr === k) cnt++;
            }
        }
        return cnt;
    }
}

const a = [4, 2, 2, 6, 4];
const k = 6;

// Create an instance of the Solution class
const solution = new Solution();

// Function call to get the result
const ans = solution.subarraysWithXorK(a, k);

console.log("The number of subarrays with XOR k is:", ans);

Complexity Analysis

Time Complexity: O(N³), where N is the size of the array. This is because we are using three nested loops, each running approximately N times.

Space Complexity: O(1) since we are not using any additional space.

Intuition

If we carefully observe, we can notice that to get the XOR of the current subarray, we just need to XOR the current element (i.e., arr[j]) with the XOR of the previous subarray (i.e., arr[i...j-1]). Assume the previous subarray is arr[i...j-1] and the current subarray is arr[i...j]. The XOR of arr[i...j] can be calculated as (XOR of arr[i...j-1]) ^ arr[j]. This approach allows us to remove the third loop, and calculate the XOR while moving the j pointer.

Approach

First, run a loop (let's call it i) that selects every possible starting index of the subarray. The starting indices can range from index 0 to n−1 (where n is the size of the array). Inside this loop, run another loop (let's call it j) that signifies the ending index of the subarray. For each subarray starting from index i, the ending index can range from i to n−1.
For each subarray from index i to j, calculate the XOR of all the elements in that subarray by simply doing XOR operation on the previous XOR and the element at j index.
If the XOR equals k, increase the count by 1.

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    // Function to count the number of subarrays with XOR k
    int subarraysWithXorK(vector<int>& nums, int k) {
        int n = nums.size(); 
        int cnt = 0;

        // Step 1: Generate subarrays
        for (int i = 0; i < n; i++) {
            int xorr = 0;
            for (int j = i; j < n; j++) {
                /* Step 2: Calculate XOR of
                   all elements in the subarray */
                xorr = xorr ^ nums[j];

                // Step 3: Check XOR and count
                if (xorr == k) cnt++;
            }
        }
        return cnt;
    }
};

int main() {
    vector<int> a = {4, 2, 2, 6, 4}; 
    int k = 6; 

    // Create an instance of the Solution class
    Solution solution;
    // Function call to get the result
    int ans = solution.subarraysWithXorK(a, k);

    
    cout << "The number of subarrays with XOR k is: " << ans << "\n";
    return 0;
}

import java.util.*;

class Solution {
    // Function to count the number of subarrays with XOR k
    public int subarraysWithXorK(int[] nums, int k) {
        int n = nums.length; 
        int cnt = 0;

        // Step 1: Generate subarrays
        for (int i = 0; i < n; i++) {
            int xorr = 0;
            for (int j = i; j < n; j++) {
                /* Step 2: Calculate XOR of
                   all elements in the subarray */
                xorr = xorr ^ nums[j];

                // Step 3: Check XOR and count
                if (xorr == k) cnt++;
            }
        }
        return cnt;
    }

    public static void main(String[] args) {
        int[] a = {4, 2, 2, 6, 4}; 
        int k = 6; 

        // Create an instance of the Solution class
        Solution solution = new Solution();
        // Function call to get the result
        int ans = solution.subarraysWithXorK(a, k);
      
        System.out.println("The number of subarrays with XOR k is: " + ans);
    }
}

class Solution:
    # Function to count the number of subarrays with XOR k
    def subarraysWithXorK(self, nums, k):
        n = len(nums)
        cnt = 0

        # Step 1: Generate subarrays
        for i in range(n):
            xorr = 0
            for j in range(i, n):
                # Step 2: Calculate XOR of all elements in the subarray
                xorr ^= nums[j]

                # Step 3: Check XOR and count
                if xorr == k:
                    cnt += 1
        return cnt

if __name__ == "__main__":
    a = [4, 2, 2, 6, 4]
    k = 6

    # Create an instance of the Solution class
    solution = Solution()

    # Function call to get the result
    ans = solution.subarraysWithXorK(a, k)
    
    print("The number of subarrays with XOR k is:", ans)

class Solution {
    // Function to count the number of subarrays with XOR k
    subarraysWithXorK(nums, k) {
        const n = nums.length;
        let cnt = 0;

        // Step 1: Generate subarrays
        for (let i = 0; i < n; i++) {
            let xorr = 0;
            for (let j = i; j < n; j++) {
                /* Step 2: Calculate XOR of
                   all elements in the subarray */
                xorr ^= nums[j];

                // Step 3: Check XOR and count
                if (xorr === k) cnt++;
            }
        }
        return cnt;
    }
}

const a = [4, 2, 2, 6, 4];
const k = 6;

// Create an instance of the Solution class
const solution = new Solution();

// Function call to get the result
const ans = solution.subarraysWithXorK(a, k);

console.log("The number of subarrays with XOR k is:", ans);

Complexity Analysis

Time Complexity: O(N²), where N is the size of the array. Since we are using two nested loops, each running for N times, the time complexity will be approximately O(N²).

Space Complexity: O(1) as we are not using any additional space.

Intuition

In this approach, tryutilize the concept of prefix XOR to solve this problem. The prefix XOR of a subarray ending at index 𝑖 is simply the XOR of all the elements from the start of the array up to index 𝑖.

Observation:

Assume the prefix XOR of a subarray ending at index i is xr. To find subarrays with XOR equal to k, we note that if a subarray has XOR k, then the prefix XOR of the remaining part will be xr XOR k (where XOR denotes the XOR operation). So, for a subarray ending at index i with prefix XOR xr, if we remove the part with prefix XOR xr XOR k, the remaining part will have XOR k. The below image will clarify this concept.

Instead of directly searching for subarrays with XOR k, we use a map to count subarrays with prefix XOR xr XOR k. The map stores each prefix XOR and its count. At each index i, we check the map for xr XOR k and add the count to our result. We repeat this for all indices from 0 to n-1. This approach efficiently finds the number of subarrays with XOR k using the prefix XOR concept and a map.

Approach

The steps are as follows:

First start by declaring a map to store the prefix XORs and their counts, and set the value of 0 as 1 in the map. This setup is crucial because it helps us handle cases where the prefix XOR itself equals the target value k.
Run a loop from index 0 to n-1 (where n is the size of the array). For each index i, XOR the current element arr[i] with the existing prefix XOR. Then calculate the required prefix XOR xr^ k to check its occurrence.
Now, add the occurrence of the prefix XOR xr ^ k from the map to our answer. This gives the count of subarrays that have the desired XOR value k. Finally, store the current prefix XOR xr in the map, increasing its occurrence by 1 to keep track of its count for future subarrays.

Question: Why set the value of 0 beforehand?

To understand this, consider the array [3, 3, 1, 1, 1] with k = 3. At index 0, the total prefix XOR is 3, and k is also 3. So, the prefix XOR xr XOR k will be 3 XOR 3 = 0. If the value 0 is not previously set in the map, we will add 0 to our answer, incorrectly indicating that no subarray with XOR 3 has been found. However, index 0 itself is a subarray with XOR k = 3. Setting the value of 0 as 1 in the map beforehand avoids this issue.

Dry Run

Please refer to the video for complete dry-run.

#include <bits/stdc++.h>
using namespace std;

class Solution
{
public:
    int subarraysWithXorK(vector<int>& nums, int k)
    {
        int n = nums.size(); 
        int xr = 0;
        map<int, int> mpp; 
        // setting the value of 0.
        mpp[xr]++; 
        int cnt = 0;

        for (int i = 0; i < n; i++) {
            // prefix XOR till index i:
            xr = xr ^ nums[i];

            // By formula: x = xr^k:
            int x = xr ^ k;

            // add the occurrence of xr^k to the count:
            cnt += mpp[x];

            // Insert the prefix xor till index i into the map:
            mpp[xr]++;
        }
        return cnt;
    }
};

int main()
{
    vector<int> a = {4, 2, 2, 6, 4};
    int k = 6;

    // Create an instance of the Solution class
    Solution solution;

    // Function call to get the result
    int ans = solution.subarraysWithXorK(a, k);

    cout << "The number of subarrays with XOR k is: " << ans << "\n";
    return 0;
}

import java.util.*;

class Solution {
    public int subarraysWithXorK(int[] nums, int k) {
        int n = nums.length;
        int xr = 0;
        Map<Integer, Integer> mpp = new HashMap<>();
        // setting the value of 0.
        mpp.put(xr, mpp.getOrDefault(xr, 0) + 1);
        int cnt = 0;

        for (int i = 0; i < n; i++) {
            // prefix XOR till index i:
            xr = xr ^ nums[i];

            // By formula: x = xr ^ k:
            int x = xr ^ k;

            // add the occurrence of xr ^ k to the count:
            cnt += mpp.getOrDefault(x, 0);

            // Insert the prefix xor till index i into the map:
            mpp.put(xr, mpp.getOrDefault(xr, 0) + 1);
        }
        return cnt;
    }

    public static void main(String[] args) {
        int[] a = {4, 2, 2, 6, 4};
        int k = 6;

        // Create an instance of the Solution class
        Solution solution = new Solution();

        // Function call to get the result
        int ans = solution.subarraysWithXorK(a, k);

        System.out.println("The number of subarrays with XOR k is: " + ans);
    }
}

class Solution:
    def subarraysWithXorK(self, nums, k):
        n = len(nums)
        xr = 0
        mpp = {}
        # setting the value of 0.
        mpp[xr] = mpp.get(xr, 0) + 1
        cnt = 0

        for i in range(n):
            # prefix XOR till index i:
            xr = xr ^ nums[i]

            # By formula: x = xr ^ k:
            x = xr ^ k

            # add the occurrence of xr ^ k to the count:
            cnt += mpp.get(x, 0)

            # Insert the prefix xor till index i into the map:
            mpp[xr] = mpp.get(xr, 0) + 1

        return cnt

if __name__ == "__main__":
    a = [4, 2, 2, 6, 4]
    k = 6

    # Create an instance of the Solution class
    solution = Solution()

    # Function call to get the result
    ans = solution.subarraysWithXorK(a, k)

    print("The number of subarrays with XOR k is:", ans)

class Solution {
    subarraysWithXorK(nums, k) {
        const n = nums.length;
        let xr = 0;
        const mpp = new Map();
        // setting the value of 0.
        mpp.set(xr, (mpp.get(xr) || 0) + 1);
        let cnt = 0;

        for (let i = 0; i < n; i++) {
            // prefix XOR till index i:
            xr = xr ^ nums[i];

            // By formula: x = xr ^ k:
            const x = xr ^ k;

            // add the occurrence of xr ^ k to the count:
            cnt += mpp.get(x) || 0;

            // Insert the prefix xor till index i into the map:
            mpp.set(xr, (mpp.get(xr) || 0) + 1);
        }
        return cnt;
    }
}

const a = [4, 2, 2, 6, 4];
const k = 6;

// Create an instance of the Solution class
const solution = new Solution();

// Function call to get the result
const ans = solution.subarraysWithXorK(a, k);

console.log("The number of subarrays with XOR k is:", ans);

Complexity Analysis

Time Complexity: O(N) or O(NxlogN), where N is the size of the array. If we use an unordered_map in C++, the time complexity is O(N). However, with a map data structure, the time complexity is O(NxlogN). In the worst case for an unordered_map, the searching time can increase to O(N), making the overall time complexity O(N²).

Space Complexity: O(N), as we are using a map data structure.

Problem List

Count subarrays with given xor K

Examples:

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