leetcode/solution/3100-3199/3152.Special Array II

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3152.Special%20Array%20II/README_EN.md	1523	Weekly Contest 398 Q2	Array Binary Search Prefix Sum

3152. Special Array II

中文文档

Description

An array is considered special if every pair of its adjacent elements contains two numbers with different parity.

You are given an array of integer nums and a 2D integer matrix queries, where for queries[i] = [fromi, toi] your task is to check that subarray nums[fromi..toi] is special or not.

Return an array of booleans answer such that answer[i] is true if nums[fromi..toi] is special.

 

Example 1:

Input: nums = [3,4,1,2,6], queries = 0,4

Output: [false]

Explanation:

The subarray is [3,4,1,2,6]. 2 and 6 are both even.

Example 2:

Input: nums = [4,3,1,6], queries = 0,2],[2,3

Output: [false,true]

Explanation:

1. The subarray is [4,3,1]. 3 and 1 are both odd. So the answer to this query is false.
2. The subarray is [1,6]. There is only one pair: (1,6) and it contains numbers with different parity. So the answer to this query is true.

 

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 105
• 1 <= queries.length <= 105
• queries[i].length == 2
• 0 <= queries[i][0] <= queries[i][1] <= nums.length - 1

Solutions

Solution 1: Record the Leftmost Special Array Position for Each Position

We can define an array d to record the leftmost special array position for each position, initially d[i] = i. Then we traverse the array nums from left to right. If nums[i] and nums[i - 1] have different parities, then d[i] = d[i - 1].

Finally, we traverse each query and check whether d[to] <= from holds.

The time complexity is O(n) and the space complexity is O(n), where n is the length of the array.

Python3

class Solution:
    def isArraySpecial(self, nums: List[int], queries: List[List[int]]) -> List[bool]:
        n = len(nums)
        d = list(range(n))
        for i in range(1, n):
            if nums[i] % 2 != nums[i - 1] % 2:
                d[i] = d[i - 1]
        return [d[t] <= f for f, t in queries]

Java

class Solution {
    public boolean[] isArraySpecial(int[] nums, int[][] queries) {
        int n = nums.length;
        int[] d = new int[n];
        for (int i = 1; i < n; ++i) {
            if (nums[i] % 2 != nums[i - 1] % 2) {
                d[i] = d[i - 1];
            } else {
                d[i] = i;
            }
        }
        int m = queries.length;
        boolean[] ans = new boolean[m];
        for (int i = 0; i < m; ++i) {
            ans[i] = d[queries[i][1]] <= queries[i][0];
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<bool> isArraySpecial(vector<int>& nums, vector<vector<int>>& queries) {
        int n = nums.size();
        vector<int> d(n);
        iota(d.begin(), d.end(), 0);
        for (int i = 1; i < n; ++i) {
            if (nums[i] % 2 != nums[i - 1] % 2) {
                d[i] = d[i - 1];
            }
        }
        vector<bool> ans;
        for (auto& q : queries) {
            ans.push_back(d[q[1]] <= q[0]);
        }
        return ans;
    }
};

Go

func isArraySpecial(nums []int, queries [][]int) (ans []bool) {
	n := len(nums)
	d := make([]int, n)
	for i := range d {
		d[i] = i
	}
	for i := 1; i < len(nums); i++ {
		if nums[i]%2 != nums[i-1]%2 {
			d[i] = d[i-1]
		}
	}
	for _, q := range queries {
		ans = append(ans, d[q[1]] <= q[0])
	}
	return
}

TypeScript

function isArraySpecial(nums: number[], queries: number[][]): boolean[] {
    const n = nums.length;
    const d: number[] = Array.from({ length: n }, (_, i) => i);
    for (let i = 1; i < n; ++i) {
        if (nums[i] % 2 !== nums[i - 1] % 2) {
            d[i] = d[i - 1];
        }
    }
    return queries.map(([from, to]) => d[to] <= from);
}