LeetCode 1538. Guess the Majority in a Hidden Array

文章目录

- 1. 题目
- 2. 解题

1. 题目

We have an integer array nums, where all the integers in nums are 0 or 1. You will not be given direct access to the array, instead, you will have an API ArrayReader which have the following functions:

int query(int a, int b, int c, int d): where 0 <= a < b < c < d < ArrayReader.length().
The function returns the distribution of the value of the 4 elements and returns:
4 : if the values of the 4 elements are the same (0 or 1).
2 : if three elements have a value equal to 0 and one element has value equal to 1 or vice versa.
0 : if two element have a value equal to 0 and two elements have a value equal to 1.
int length(): Returns the size of the array.
You are allowed to call query() 2 * n times at most where n is equal to ArrayReader.length().

Return any index of the most frequent value in nums, in case of tie, return -1.

Follow up: What is the minimum number of calls needed to find the majority element?

Example 1:
Input: nums = [0,0,1,0,1,1,1,1]
Output: 5
Explanation: The following calls to the API
reader.length() 
// returns 8 because there are 8 elements in the hidden array.
reader.query(0,1,2,3) 
// returns 2 this is a query that compares the elements nums[0], nums[1], nums[2], nums[3]
// Three elements have a value equal to 0 and one element has value equal to 1 or viceversa.
reader.query(4,5,6,7) 
// returns 4 because nums[4], nums[5], nums[6], nums[7] have the same value.
we can infer that the most frequent value is found in the last 4 elements.
Index 2, 4, 6, 7 is also a correct answer.Example 2:
Input: nums = [0,0,1,1,0]
Output: 0Example 3:
Input: nums = [1,0,1,0,1,0,1,0]
Output: -1Constraints:
5 <= nums.length <= 10^5
0 <= nums[i] <= 1

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/guess-the-majority-in-a-hidden-array
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2. 解题

参考题解

/*** // This is the ArrayReader's API interface.* // You should not implement it, or speculate about its implementation* class ArrayReader {*   public:*     // Compares 4 different elements in the array*     // return 4 if the values of the 4 elements are the same (0 or 1).*     // return 2 if three elements have a value equal to 0 and one element has value equal to 1 or vice versa.*     // return 0 : if two element have a value equal to 0 and two elements have a value equal to 1.*     int query(int a, int b, int c, int d);**     // Returns the length of the array*     int length();* };*/class Solution {
public:int guessMajority(ArrayReader &reader) {int n = reader.length();int start = reader.query(0,1,2,3);int g1 = 1, g2 = 0, idx1 = 0, idx2 = -1;//假设idx = 0 的数是第一类，其个数为 g1 = 1for(int i = 4; i < n; ++i){if(reader.query(1,2,3,i)==start)//0， i 是否是一类g1++;elseg2++, idx2 = i;}//0 与 4,5...n-1 是否是一类//还要确定1,2,3int q = reader.query(0,2,3,4);int p = reader.query(1,2,3,4);if(q == p)//0和1是一类g1++;elseg2++, idx2 = 1;q = reader.query(0,1,3,4);// p = reader.query(1,2,3,4);if(q == p)//0和2是否是一类g1++;elseg2++,idx2 = 2;q = reader.query(0,1,2,4);// p = reader.query(1,2,3,4);if(q == p)//0和3是否是一类g1++;elseg2++,idx2 = 3;if(g1 == g2) return -1;if(g1 > g2) return idx1;return idx2;}
};