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:
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: 0
Example 3:
Input: nums = [1,0,1,0,1,0,1,0] Output: -1
Constraints:
5 <= nums.length <= 10^50 <= nums[i] <= 1struct Solution;
struct ArrayReader {
nums: Vec<i32>,
}
impl ArrayReader {
fn new(nums: Vec<i32>) -> Self {
ArrayReader { nums }
}
fn query(&self, a: i32, b: i32, c: i32, d: i32) -> i32 {
let a = a as usize;
let b = b as usize;
let c = c as usize;
let d = d as usize;
let mut diff: i32 = 0;
for &i in &[a, b, c, d] {
if self.nums[i] == 1 {
diff += 1;
} else {
diff -= 1;
}
}
diff.abs()
}
fn length(&self) -> i32 {
self.nums.len() as i32
}
}
impl Solution {
pub fn get_majority(reader: &ArrayReader) -> i32 {
let n = reader.length() as usize;
let mut zero = 1;
let mut one = 0;
for i in 1..n {
let mut other = vec![];
for j in 1..n {
if j != i {
other.push(j);
}
if other.len() == 3 {
break;
}
}
let mut q1 = vec![0];
let mut q2 = vec![i as i32];
for j in other {
q1.push(j as i32);
q2.push(j as i32);
}
q2.sort_unstable();
if reader.query(q1[0], q1[1], q1[2], q1[3]) != reader.query(q2[0], q2[1], q2[2], q2[3])
{
one += 1;
} else {
zero += 1;
}
if one > n / 2 || zero > n / 2 {
return i as i32;
}
}
-1
}
}
#[test]
fn test() {
let nums = vec![0, 0, 1, 0, 1, 1, 1, 1];
let reader = ArrayReader::new(nums);
let res = 7;
assert_eq!(Solution::get_majority(&reader), res);
}