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3427. Special Array Ii

Difficulty: Medium

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3427. Special Array II

Medium

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

Solution

class Solution {
    public boolean[] isArraySpecial(int[] nums, int[][] queries) {
         int n = nums.length;
        boolean[] ans = compute(nums);
        int[] prefix = new int[n];
        for (int i = 1; i < n; i++) {
            prefix[i] = prefix[i - 1] + (ans[i] ? 1 : 0);
        }


        boolean[] results = new boolean[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int u = queries[i][0];
            int v = queries[i][1];
            if (u == v) {
                results[i] = true;
            } else {
                int req = v - u;
                int current = prefix[v] - prefix[u];
                results[i] = (current == req);
            }
        }

        return results;

    }

    public static boolean[] compute(int[] nums) {
        int n = nums.length;
        boolean[] res = new boolean[n];
        for (int i = 1; i < n; i++) {
            res[i] = (nums[i] % 2) != (nums[i - 1] % 2);
        }
        return res;
    }
}

Complexity Analysis

  • Time Complexity: O(?)
  • Space Complexity: O(?)

Approach

Detailed explanation of the approach will be added here