Package g2001_2100.s2025_maximum_number_of_ways_to_partition_an_array

Class Solution

java.lang.Object
g2001_2100.s2025_maximum_number_of_ways_to_partition_an_array.Solution
public class Solution extends Object
2025 - Maximum Number of Ways to Partition an Array\. Hard You are given a **0-indexed** integer array `nums` of length `n`. The number of ways to **partition** `nums` is the number of `pivot` indices that satisfy both conditions: * `1 <= pivot < n` * `nums[0] + nums[1] + ... + nums[pivot - 1] == nums[pivot] + nums[pivot + 1] + ... + nums[n - 1]` You are also given an integer `k`. You can choose to change the value of **one** element of `nums` to `k`, or to leave the array **unchanged**. Return _the **maximum** possible number of ways to **partition**_ `nums` _to satisfy both conditions after changing **at most** one element_. **Example 1:** **Input:** nums = [2,-1,2], k = 3 **Output:** 1 **Explanation:** One optimal approach is to change nums[0] to k. The array becomes [**3** ,-1,2]. There is one way to partition the array: - For pivot = 2, we have the partition [3,-1 | 2]: 3 + -1 == 2. **Example 2:** **Input:** nums = [0,0,0], k = 1 **Output:** 2 **Explanation:** The optimal approach is to leave the array unchanged. There are two ways to partition the array: - For pivot = 1, we have the partition [0 | 0,0]: 0 == 0 + 0. - For pivot = 2, we have the partition [0,0 | 0]: 0 + 0 == 0. **Example 3:** **Input:** nums = [22,4,-25,-20,-15,15,-16,7,19,-10,0,-13,-14], k = -33 **Output:** 4 **Explanation:** One optimal approach is to change nums[2] to k. The array becomes [22,4, **\-33** ,-20,-15,15,-16,7,19,-10,0,-13,-14]. There are four ways to partition the array. **Constraints:** * `n == nums.length` * 2 <= n <= 105 * -105 <= k, nums[i] <= 105

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • waysToPartition

      public int waysToPartition(int[] nums, int k)