1538. Guess the Majority in a Hidden Array
Problem Description
In this problem, we're given an array of integers called nums
, where every integer is either 0 or 1. Direct access to this array is not permitted; instead, we have access to an API ArrayReader
that provides two functions. The query(a, b, c, d)
function can be used to query the array with four different indices (with the condition 0 <= a < b < c < d < ArrayReader.length()
) and returns the distribution of the four selected elements as follows:
- Returns
4
if all four elements are 0's or all are 1's. - Returns
2
if three elements are the same (either three 0's and one 1, or three 1's and one 0). - Returns
0
if there's an even split (two 0's and two 1's).
The length()
function returns the size of the array.
Our goal is to find the index of any element with the most frequent value, which could be either 0 or 1. In case of a tie, where 0 and 1 are equally frequent, we return -1.
Intuition
To solve this problem, we start with the understanding that the query()
function can give us an insight into the majority element by comparing the distribution of 0s and 1s among different subsets of four elements. Since we're allowed to call query()
2 * n
times where n
is the array length, we have enough queries at our disposal to deduce the majority element's index.
The solution includes querying the first four elements to establish the initial distribution (let's call this the baseline distribution). Afterwards, we loop through the rest of the elements, using query()
with the first three indices fixed and the fourth one changing to include each new index from the rest of the array. If the distribution matches the baseline, the new element shares the majority value with the first three; if it doesn't match, it's different.
The first element not included in the initial query (index 4
) is specially treated by checking its distribution with the next 3 consecutive elements. This helps us in confirming whether the initially queried elements were a majority or a split.
After looping through all elements, we compare the count of elements that matched the baseline distribution (a
) with those that didn't (b
). The index that occurs more is the majority. If both counts are equal, we have a tie, and we return -1
.
The intuition is based on iteratively narrowing down the possible indices for the majority element by comparing the distribution of subsets of the array using the query()
function.
Learn more about Math patterns.
Solution Approach
The implementation of the solution approach involves using a simple iteration to leverage the ArrayReader
API effectively. Here's a breakdown of each part of the code and the algorithms, data structures, or patterns used:
-
Initialization: The
n
variable is initialized with the size of the array usingreader.length()
. Two counters,a
andb
, are initialized to keep track of the counts of indices that match or do not match the initial baseline distribution. Variablek
is initialized to store an index where the value is different from the baseline if such an index is found. -
Initial Query: We perform the first
query(0, 1, 2, 3)
to establish a baseline distribution stored in variablex
. This tells us the distribution pattern of the first four elements. -
Iterative Comparison: A
for
loop is used to iterate through the elements starting from index4
to the last indexn-1
. In each iteration, we query a subset that includes the first three fixed indices (0, 1, 2
) and the current indexi
.- If the return value of this query matches
x
, we increment countera
; - If the return value doesn't match
x
, we increment counterb
and updatek
with the current indexi
.
- If the return value of this query matches
-
Special Cases: After the loop, we need to verify the initial three elements (which are fixed in the previous
query()
calls) to ensure their distribution. This requires three additional queries, each omitting one of the first three indices alongside the fourth index4
. This helps in confirming whether the initially queried indices were a majority.- The results of these queries are compared against a different baseline distribution stored in
y
which is obtained byquery(0, 1, 2, 4)
. - If the result of these comparisons is equal to
y
, incrementa
since the omitted element matches the initial query (suggesting it's a majority element); otherwise, incrementb
and updatek
respectively.
- The results of these queries are compared against a different baseline distribution stored in
-
Determine Majority Index: Finally, we determine whether there is a majority by comparing the counts
a
andb
.- If
a
equalsb
, we conclude there is a tie and return-1
; - If
a
is greater, we return3
which was part of the original subset and, therefore, in the majority. - If
b
is greater, the indexk
is returned, indicating thatk
is part of the frequent value.
- If
Data structures used in this solution are primarily basic integer variables for counting. The pattern used is iterative comparisons with early stopping as soon as the majority element is confirmed. The algorithm doesn't require complex data structures as it makes effective use of the provided API to solve the problem.
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Start EvaluatorExample Walkthrough
Let's assume we have a nums
array of size 8, and we perform the following steps:
-
Initialization:
- The array size
n
is 8 (determined usingreader.length()
). - We initialize two counters,
a
andb
, to zero. Countera
will track indices matching the baseline distribution, whileb
tracks the others. - The
k
variable is initialized but not yet set.
- The array size
-
Initial Query:
- We run
query(0, 1, 2, 3)
and let's say it returns4
. This suggests all elements are the same (all 0s or all 1s), so our baseline distributionx
is4
.
- We run
-
Iterative Comparison:
- We iterate with a loop for indices
4
to7
(the rest of the array). - We perform
query(0, 1, 2, 4)
. Assume it returns4
, identical to our baseline, so we incrementa
. - Next,
query(0, 1, 2, 5)
might return2
, unlike our baseline. We incrementb
and setk
to5
. - We continue with
query(0, 1, 2, 6)
, returning4
, so we incrementa
. - Lastly,
query(0, 1, 2, 7)
returns4
, incrementinga
again.
- We iterate with a loop for indices
-
Special Cases:
- We then create a separate baseline
y
withquery(0, 1, 2, 4)
(we have already done this and know it's4
). - We run
query(1, 2, 3, 4)
for special case handling, and if it returns4
, we incrementa
—let's assume it does; - Next,
query(0, 2, 3, 4)
and if it returns4
, we incrementa
—let's say it does; - Finally,
query(0, 1, 3, 4)
and if it returns4
, we incrementa
—let's assume it also does.
- We then create a separate baseline
At the end of this process, let's assume a
is 6
, and b
is 1
. We now decide the majority element:
- Determine Majority Index:
- Since
a
is greater thanb
, we return index3
because it belongs to the subset that matched our initial baseline the most.
- Since
This example confirmed that the majority values were present in the initial subset, and indexes that matched this distribution were in the majority. If b
had been greater, we would return k
, which in this walkthrough was 5
. If a
and b
were equal, indicating an even split between 0s and 1s, we would return -1
.
Solution Implementation
1class Solution:
2 def guessMajority(self, reader: "ArrayReader") -> int:
3 # Get the length of the array
4 n = reader.length()
5
6 # Initialize the counter for the two possible majorities and an index tracker
7 majority_count, minority_count = 1, 0
8 minority_index = 0
9
10 # Perform the initial query to establish a baseline comparison
11 baseline_query = reader.query(0, 1, 2, 3)
12
13 # Iterate through the rest of the array starting from index 4
14 for i in range(4, n):
15 # Query with the new index and compare to the baseline
16 if reader.query(0, 1, 2, i) == baseline_query:
17 majority_count += 1
18 else:
19 minority_count += 1
20 minority_index = i # Store the index of the first occurrence in minority
21
22 # Perform additional queries to determine the first three elements' majority/minority status
23 # These queries involve the element at index 4 to compare with the baseline value
24 initial_queries_comparison = reader.query(0, 1, 2, 4)
25 for i in range(3):
26 if reader.query(i, (i + 1) % 3, (i + 2) % 3, 4) == initial_queries_comparison:
27 majority_count += 1
28 else:
29 minority_count += 1
30 minority_index = i # Update the index of the first occurrence in minority
31
32 # If the counts are the same, we return -1, as there is no majority
33 if majority_count == minority_count:
34 return -1
35
36 # The element at index 3 has a different status compared to the first three elements due to the setup
37 # Hence, if the majority count is greater, the index of the minority is returned; otherwise, 3 is returned
38 return minority_index if majority_count > minority_count else 3
39
1class Solution {
2 public int guessMajority(ArrayReader reader) {
3 // Get the length of the array
4 int arrayLength = reader.length();
5
6 // Initial query to determine the pattern to compare against
7 int initialQueryResult = reader.query(0, 1, 2, 3);
8
9 // 'sameMajorityCount' is the count of indices following the initial pattern
10 // 'differentMajorityCount' is the count of indices that differ from the initial pattern
11 int sameMajorityCount = 1, differentMajorityCount = 0;
12
13 // 'differentIndex' keeps track of the index where a different pattern was first noticed
14 int differentIndex = 0;
15
16 // Check rest of the elements with the initial ones except the last three
17 for (int i = 4; i < arrayLength; ++i) {
18 if (reader.query(0, 1, 2, i) == initialQueryResult) {
19 sameMajorityCount++; // Same pattern detected, increment count
20 } else {
21 differentMajorityCount++; // Different pattern detected, increment count and update the index
22 differentIndex = i;
23 }
24 }
25
26 // Perform queries swapping out one of the first three elements with the fourth one
27 // This checks the pattern of the last three indices
28 int subsequentQueryResult = reader.query(0, 1, 2, 4);
29 for (int i = 0; i < 3; i++) {
30 int queryResult = reader.query(i == 0 ? 1 : 0, i == 1 ? 2 : 1, i == 2 ? 3 : 2, 4);
31 if (queryResult == subsequentQueryResult) {
32 sameMajorityCount++; // Increment count if the same as the subsequent pattern
33 } else {
34 differentMajorityCount++; // Increment different count and update index with the swapped element
35 differentIndex = i;
36 }
37 }
38
39 // If counts are equal, no unique majority index exists
40 if (sameMajorityCount == differentMajorityCount) {
41 return -1;
42 }
43
44 // Return the index which has the majority pattern (most occurrences)
45 // If the counts of the same pattern is greater, return the constant 3 as per initial result
46 // If counts of the different pattern is greater, return the index where the difference was first noted
47 return sameMajorityCount > differentMajorityCount ? 3 : differentIndex;
48 }
49}
50
1class Solution {
2public:
3 int guessMajority(ArrayReader& reader) {
4 // Get the array length
5 int arrayLength = reader.length();
6
7 // Query the first four elements to get a baseline comparison
8 int firstQueryResult = reader.query(0, 1, 2, 3);
9
10 // Initialize the count for the majority and minority elements
11 int majorityCount = 1; // We start with 1 since we know the first four elements are the baseline
12 int minorityCount = 0;
13
14 // Keep track of the latest minority element index
15 int minorityIndex = 0;
16
17 // Compare the rest of the array elements starting from the fifth element
18 for (int i = 4; i < arrayLength; ++i) {
19 // If the next element belongs to the majority, increase the majority count
20 if (reader.query(0, 1, 2, i) == firstQueryResult) {
21 ++majorityCount;
22 } else {
23 // Otherwise, increase the minority count and update the minorityIndex
24 ++minorityCount;
25 minorityIndex = i;
26 }
27 }
28
29 // Perform another query with 0, 1, 2 and a known different element (4 in this case)
30 int secondQueryResult = reader.query(0, 1, 2, 4);
31
32 // Check the remaining three elements against the new query to determine if they belong to the majority or minority
33 if (reader.query(1, 2, 3, 4) == secondQueryResult) {
34 ++majorityCount;
35 } else {
36 ++minorityCount;
37 minorityIndex = 0; // Element 0 is different
38 }
39 if (reader.query(0, 2, 3, 4) == secondQueryResult) {
40 ++majorityCount;
41 } else {
42 ++minorityCount;
43 minorityIndex = 1; // Element 1 is different
44 }
45 if (reader.query(0, 1, 3, 4) == secondQueryResult) {
46 ++majorityCount;
47 } else {
48 ++minorityCount;
49 minorityIndex = 2; // Element 2 is different
50 }
51
52 // If the majority and minority counts are the same, the result is ambiguous
53 if (majorityCount == minorityCount) {
54 return -1;
55 }
56
57 // If the majority count is greater, the answer is the initial subset index (3)
58 // Otherwise, return the index of the minority element
59 return majorityCount > minorityCount ? 3 : minorityIndex;
60 }
61};
62
1function guessMajority(reader: ArrayReader): number {
2 // Retrieve the length of the array from the reader
3 const arrayLength = reader.length();
4
5 // Perform the initial query on the first four elements
6 const firstQueryResult = reader.query(0, 1, 2, 3);
7
8 // Initialize counter for the majority and minority elements, as well as the index of the first different element
9 let majorityCount = 1;
10 let minorityCount = 0;
11 let firstDifferentIndex = 0;
12
13 // Iterate over the rest of the elements, starting from the 5th element
14 for (let i = 4; i < arrayLength; ++i) {
15 // Compare each new element with the initial three elements to determine its type
16 if (reader.query(0, 1, 2, i) === firstQueryResult) {
17 ++majorityCount;
18 } else {
19 ++minorityCount;
20 firstDifferentIndex = i;
21 }
22 }
23
24 // Perform additional queries by excluding one element from the first four each time
25 const secondQueryBase = reader.query(0, 1, 2, 4);
26 if (reader.query(1, 2, 3, 4) === secondQueryBase) {
27 ++majorityCount;
28 } else {
29 ++minorityCount;
30 firstDifferentIndex = 0;
31 }
32 if (reader.query(0, 2, 3, 4) === secondQueryBase) {
33 ++majorityCount;
34 } else {
35 ++minorityCount;
36 firstDifferentIndex = 1;
37 }
38 if (reader.query(0, 1, 3, 4) === secondQueryBase) {
39 ++majorityCount;
40 } else {
41 ++minorityCount;
42 firstDifferentIndex = 2;
43 }
44
45 // If there is an equal number of majority and minority elements, there is no majority
46 if (majorityCount === minorityCount) {
47 return -1;
48 }
49
50 // The majority is determined by the element with more occurrences
51 // If the majority is in the first four elements, we return 3, which points to the index 3 in the initial query
52 // Otherwise, we return the index of the first element that was different from the initial majority
53 return majorityCount > minorityCount ? 3 : firstDifferentIndex;
54}
55
Time and Space Complexity
Time Complexity
The time complexity of the provided guessMajority
function primarily depends on the number of calls made to the reader.query
method within a loop and outside of it.
- There is a fixed number of calls to
reader.query
made outside of the loop for the indices0, 1, 2, 3
, and four additional calls to compare with index4
. - The loop iterates from index
4
ton - 1
(wheren
is the length of the array), making one call toreader.query
per iteration. - Hence, the number of
query
calls is4 (initial) + 4 (comparisons with index 4) + (n - 4)
for the loop.
Thus, the total number of query
calls is 8 + (n - 4) = n + 4
. Given that each query
call is assumed to have a constant time complexity, the time complexity of the guessMajority
function is O(n)
.
Space Complexity
The space complexity of the code is determined by the variables used in the guessMajority
function.
- Constant extra space is used for the variables
a
,b
,k
,x
, andy
, independent of the input sizen
. - No additional data structures that grow with the input size are used.
Therefore, the space complexity of the guessMajority
function is O(1)
, which means it requires constant space.
Learn more about how to find time and space complexity quickly using problem constraints.
Is the following code DFS or BFS?
1void search(Node root) { 2 if (!root) return; 3 visit(root); 4 root.visited = true; 5 for (Node node in root.adjacent) { 6 if (!node.visited) { 7 search(node); 8 } 9 } 10}
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