The "Guess the Majority in a Hidden Array" problem on Leetcode presents you with a hidden array of length n, where each element is either 0 or 1. You do not have direct access to the array, but you are given an interface with a query function. This function takes in any four distinct indices and returns the majority value among those four elements (i.e., whether most of them are 0 or 1).
Your task is to find any index of a majority element in the hidden array (an index where the value equals the majority value in the whole array). If there is no majority, you may return -1. The constraints guarantee that there is at most one majority value.
query function to gather information.At first glance, the problem seems tricky because you cannot directly access the array elements. A brute-force approach would be to try every possible combination of indices, but this would require a large number of queries and is inefficient.
Instead, the key insight is that, since the query function returns the majority among four indices, you can compare the results of different queries to deduce relationships between the values at different positions. By fixing a set of reference indices, you can classify other indices as being "the same as" or "different from" the reference, based on the query outcomes.
This approach is similar to the classic "majority element" problem, but with the challenge of indirect access and limited information per query. The goal is to systematically build up enough information to confidently identify a majority index.
To solve this problem efficiently, we use the following step-by-step strategy:
i from 4 to n-1, replace one of the reference indices with i in the query.i is likely the same as the replaced reference index.i is likely different from the replaced reference.-1 (no majority).
This strategy is efficient because each query gives you information about one new element, and you only need O(n) queries.
Let's walk through an example where the hidden array is [1, 1, 0, 1, 0, 1].
This process ensures that we correctly identify a majority index using the minimum number of queries.
Brute-force Approach:
O(n^4).O(n) queries (one for each index, after the initial four).O(1) (only a few variables to keep track of counts and indices).This makes the optimized approach very efficient and scalable for large arrays.
The "Guess the Majority in a Hidden Array" problem can be solved efficiently by cleverly using the query function to compare other elements to a fixed set of reference indices. By tracking which indices are "the same as" or "different from" the reference group, you can confidently identify the majority index with only O(n) queries and constant extra space. The elegance of this solution lies in its systematic classification and minimal query usage, making it both practical and optimal.
# This is the interface that allows for interaction with the hidden array.
# You should not implement it, or speculate about its implementation.
# class MajorityChecker:
# def query(self, i: int, j: int, k: int, l: int) -> int:
class Solution:
def guessMajority(self, checker: 'MajorityChecker') -> int:
n = checker.length # Provided by the problem, or you may need to pass it in
same = 1
diff = 0
same_idx = 0
diff_idx = -1
# First, get the result of the first four indices as reference
ref = checker.query(0, 1, 2, 3)
for i in range(4, n):
res = checker.query(1, 2, 3, i)
if res == ref:
same += 1
same_idx = i
else:
diff += 1
diff_idx = i
# Now, check the first three indices by replacing each with index 4
for idx in range(3):
indices = [j for j in range(4) if j != idx] + [4]
res = checker.query(*indices)
if res == ref:
same += 1
same_idx = idx
else:
diff += 1
diff_idx = idx
if same > diff:
return same_idx
elif diff > same:
return diff_idx
else:
return -1
// This is the interface that allows for interaction with the hidden array.
// You should not implement it, or speculate about its implementation.
// class MajorityChecker {
// public:
// int query(int i, int j, int k, int l);
// int length;
// };
class Solution {
public:
int guessMajority(MajorityChecker* checker) {
int n = checker->length;
int same = 1, diff = 0;
int same_idx = 0, diff_idx = -1;
int ref = checker->query(0, 1, 2, 3);
for (int i = 4; i < n; ++i) {
int res = checker->query(1, 2, 3, i);
if (res == ref) {
++same;
same_idx = i;
} else {
++diff;
diff_idx = i;
}
}
for (int idx = 0; idx < 3; ++idx) {
vector indices;
for (int j = 0; j < 4; ++j)
if (j != idx) indices.push_back(j);
indices.push_back(4);
int res = checker->query(indices[0], indices[1], indices[2], indices[3]);
if (res == ref) {
++same;
same_idx = idx;
} else {
++diff;
diff_idx = idx;
}
}
if (same > diff) return same_idx;
else if (diff > same) return diff_idx;
else return -1;
}
};
// This is the interface that allows for interaction with the hidden array.
// You should not implement it, or speculate about its implementation.
// class MajorityChecker {
// public int query(int i, int j, int k, int l) {}
// public int length;
// }
class Solution {
public int guessMajority(MajorityChecker checker) {
int n = checker.length;
int same = 1, diff = 0;
int sameIdx = 0, diffIdx = -1;
int ref = checker.query(0, 1, 2, 3);
for (int i = 4; i < n; i++) {
int res = checker.query(1, 2, 3, i);
if (res == ref) {
same++;
sameIdx = i;
} else {
diff++;
diffIdx = i;
}
}
for (int idx = 0; idx < 3; idx++) {
int[] indices = new int[4];
int pos = 0;
for (int j = 0; j < 4; j++) {
if (j != idx) {
indices[pos++] = j;
}
}
indices[3] = 4;
int res = checker.query(indices[0], indices[1], indices[2], indices[3]);
if (res == ref) {
same++;
sameIdx = idx;
} else {
diff++;
diffIdx = idx;
}
}
if (same > diff) return sameIdx;
else if (diff > same) return diffIdx;
else return -1;
}
}
/**
* // This is the interface that allows for interaction with the hidden array.
* // You should not implement it, or speculate about its implementation.
* // class MajorityChecker {
* // query(i: number, j: number, k: number, l: number): number {}
* // length: number;
* // }
*/
var guessMajority = function(checker) {
const n = checker.length;
let same = 1, diff = 0;
let sameIdx = 0, diffIdx = -1;
const ref = checker.query(0, 1, 2, 3);
for (let i = 4; i < n; i++) {
const res = checker.query(1, 2, 3, i);
if (res === ref) {
same++;
sameIdx = i;
} else {
diff++;
diffIdx = i;
}
}
for (let idx = 0; idx < 3; idx++) {
const indices = [];
for (let j = 0; j < 4; j++) {
if (j !== idx) indices.push(j);
}
indices.push(4);
const res = checker.query(indices[0], indices[1], indices[2], indices[3]);
if (res === ref) {
same++;
sameIdx = idx;
} else {
diff++;
diffIdx = idx;
}
}
if (same > diff) return sameIdx;
else if (diff > same) return diffIdx;
else return -1;
};