3653. XOR After Range Multiplication Queries I

/**
 * Problem: LeetCode 3653 - XOR After Range Multiplication Queries I
 * Difficulty: Medium
 * Language: Java
 * Author: CodeByTushu
 */
class Solution {
    public int xorAfterQueries(int[] nums, int[][] queries) {
        long MOD = 1_000_000_007L;

        for (int[] query : queries) {
            int l = query[0];
            int r = query[1];
            int k = query[2];
            int v = query[3];

            for (int idx = l; idx <= r; idx += k) {
                nums[idx] = (int) (((long) nums[idx] * v) % MOD);
            }
        }

        int xorSum = 0;
        for (int val : nums) {
            xorSum ^= val;
        }

        return xorSum;
    }
}

"""
Problem: LeetCode 3653 - XOR After Range Multiplication Queries I
Difficulty: Medium
Language: Python
Author: CodeByTushu
"""
class Solution:
    def xorAfterQueries(self, nums: list[int], queries: list[list[int]]) -> int:
        MOD = 1_000_000_007

        for l, r, k, v in queries:
            for idx in range(l, r + 1, k):
                nums[idx] = (nums[idx] * v) % MOD

        xor_sum = 0
        for val in nums:
            xor_sum ^= val

        return xor_sum

/**
 * Problem: LeetCode 3653 - XOR After Range Multiplication Queries I
 * Difficulty: Medium
 * Language: C++
 * Author: CodeByTushu
 */
#include <vector>

using namespace std;

class Solution {
public:
    int xorAfterQueries(vector<int>& nums, vector<vector<int>>& queries) {
        long long MOD = 1000000007LL;

        for (const auto& query : queries) {
            int l = query[0];
            int r = query[1];
            int k = query[2];
            int v = query[3];

            for (int idx = l; idx <= r; idx += k) {
                nums[idx] = (int) ((1LL * nums[idx] * v) % MOD);
            }
        }

        int xorSum = 0;
        for (int val : nums) {
            xorSum ^= val;
        }

        return xorSum;
    }
};

/**
 * Problem: LeetCode 3653 - XOR After Range Multiplication Queries I
 * Difficulty: Medium
 * Language: JavaScript
 * Author: CodeByTushu
 */
/**
 * @param {number[]} nums
 * @param {number[][]} queries
 * @return {number}
 */
var xorAfterQueries = function(nums, queries) {
    const MOD = 1000000007;

    for (let i = 0; i < queries.length; i++) {
        const [l, r, k, v] = queries[i];
        for (let idx = l; idx <= r; idx += k) {
            nums[idx] = (nums[idx] * v) % MOD;
        }
    }

    let xorSum = 0;
    for (let i = 0; i < nums.length; i++) {
        xorSum ^= nums[i];
    }

    return xorSum;
};