Maximum Partition Factor

Maximum Partition Factor

What is this problem about?

The Maximum Partition Factor coding problem gives you an array of integers nums. You need to partition the array into one or more contiguous subarrays. For each subarray, you calculate a "partition factor", which is the count of distinct elements in that subarray. The goal is to maximize the minimum partition factor among all the partitioned subarrays.

Why is this asked in interviews?

Gameskraft uses this problem to test a candidate's understanding of array partitioning, subarray properties, and binary search on the answer. It requires finding a monotonic property for the minimum partition factor, which allows for an efficient verification function.

Algorithmic pattern used

Binary Search on the Answer + Greedy Check:

1. Binary Search on min_partition_factor: The maximum possible value for the minimum partition factor (X) is monotonic. If it's possible to partition the array such that all subarrays have a partition factor of at least X, then it's also possible to do so for any X' < X. This enables binary search for the maximum X in the range [1, N] (where N is the number of elements in nums).
2. check(X) function: Given a target minimum partition factor X, can we partition nums into contiguous subarrays such that each subarray has at least X distinct elements?
   • Initialize current_start = 0.
   • Initialize current_distinct_count = 0.
   • Initialize current_freq_map (hash map) for the current subarray.
   • Iterate i from current_start to N-1:
     • Add nums[i] to current_freq_map. If nums[i] is new to this subarray, current_distinct_count++.
     • If current_distinct_count >= X: We have found a valid subarray.
       • Increment the count of valid partitions.
       • Reset current_freq_map, current_distinct_count = 0.
       • Set current_start = i + 1.
   • After iterating through the entire array, if we successfully formed partitions for the whole array (i.e., the last subarray also met the X distinct elements criteria), return true. Otherwise, false. This greedy approach for check(X) works: If we can form a valid subarray [current_start, i] with X distinct elements, it is always optimal to make this subarray as short as possible (current_start to i) to leave more elements for subsequent subarrays. This greedy choice makes the check function O(N).

Time Complexity: O(N * log N) (binary search log N iterations, each check(X) is O(N)).

Example explanation

nums = [1, 2, 1, 3], N = 4. Binary search X in [1, 4].

Try check(X=2): current_start = 0. current_freq_map = {}. current_distinct_count = 0.

• i=0, nums[0]=1: current_freq_map={1:1}, current_distinct_count=1.
• i=1, nums[1]=2: current_freq_map={1:1, 2:1}, current_distinct_count=2.
  • current_distinct_count (2) >= X (2). Yes. Form partition [1,2].
  • Reset: current_freq_map = {}, current_distinct_count = 0, current_start = 2.
• i=2, nums[2]=1: current_freq_map={1:1}, current_distinct_count=1.
• i=3, nums[3]=3: current_freq_map={1:1, 3:1}, current_distinct_count=2.
  • current_distinct_count (2) >= X (2). Yes. Form partition [1,3].
  • Reset: current_freq_map = {}, current_distinct_count = 0, current_start = 4. We reached end of array. All elements partitioned. check(2) returns true. ans = 2, low = 3.

Try check(X=3): current_start = 0. current_freq_map = {}. current_distinct_count = 0.

• i=0, nums[0]=1: current_distinct_count=1.
• i=1, nums[1]=2: current_distinct_count=2.
• i=2, nums[2]=1: current_distinct_count=2.
• i=3, nums[3]=3: current_distinct_count=3.
  • current_distinct_count (3) >= X (3). Yes. Form partition [1,2,1,3].
  • Reset: current_start = 4. We reached end of array. All elements partitioned. check(3) returns true. ans = 3, low = 4.

Try check(X=4): current_start = 0. current_freq_map = {}. current_distinct_count = 0.

• ... after iterating through entire array, current_distinct_count will be 3 for [1,2,1,3].
  • current_distinct_count (3) >= X (4). No. check(4) returns false. high = 3.

Loop terminates (low=4, high=3). ans = 3.

Common mistakes candidates make

• Brute-force approach: Trying all partitions (exponential complexity).
• Flawed check function: Not using a greedy approach for partitioning, or incorrectly counting distinct elements within a sliding window.
• Incorrect binary search range: Ensure low and high cover the full possible range of min_partition_factor.

Interview preparation tip

For the Array Union Find Breadth-First Search Graph Depth-First Search Binary Search (This problem heavily emphasizes Binary Search and Array with Greedy) interview pattern, problems asking to maximize a "minimum" quantity (or minimize a "maximum" quantity) are often solved with binary search on the answer. The check function typically uses a greedy strategy or a specific data structure.