Maximum Number Of Ways To Partition An Array - Leetcode Solution

Problem Statement: You are given an array nums. You can partition the array into at most k non-empty adjacent subarrays. After such a partitioning, each subarray has its own sum, greater than or equal to x. Your task is to find the maximum number of ways to partition the array.

Example: Input: nums = [2,2,2,2,5,5,5,8], k = 4, x = 4 Output: 3 Explanation: The optimal way to partition the array is into [2,2,2,2], [5,5,5], [8] and the minimum subarray sum is 4. There are 3 ways to partition the array.

Approach:

This problem can be solved using binary search and sliding window techniques. Here's the overall approach to solve this problem:

1. Initialize l=1 and r=n.
2. while l <= r: a. For mid = (l + r) / 2, check if it is possible to partition the array in at most k non-empty adjacent subarrays such that each subarray has its own sum, greater than or equal to x. To do this, use a sliding window approach - start with the first element and move the window until the sum of the current subarray is less than x. At this point, you have found the first subarray. Repeat the same process with the remaining elements to find the remaining subarrays. If the number of subarrays found is less than or equal to k, we can try to find a larger mid. b. If the number of subarrays found is greater than k, we need to try a smaller mid and move to the left half of the array by setting r = mid - 1.
3. Return l.

Solution:

Here's the Python code for the solution:

class Solution: def maxNumOfSubarrays(self, nums: List[int], k: int, x: int) -> int: n = len(nums) l, r = 1, n while l <= r: mid = (l + r) // 2 cnt, currSum = 0, 0 for i in range(n): if nums[i] > x: cnt = n + 1 break currSum += nums[i] if currSum > mid: currSum = nums[i] cnt += 1 cnt += 1 if cnt <= k: l = mid + 1 else: r = mid - 1 return r

Time Complexity: The time complexity of this solution is O(N*logN), where N is the length of the input array. The binary search iterates logN times, and each iteration takes O(N) time to calculate the sum of subarrays using sliding windows.

Space Complexity: The space complexity of this solution is O(1), as we are not using any extra data structures.

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