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3152. Special Array II

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.

  • 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;
    }
}

  • 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;
    }
};

  • 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]


  • 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
}

  • 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);
}

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