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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 + nums + ... + 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.

Maximum Number of Ways to Partition an Array solution leetcode

### Maximum Number of Ways to Partition an Array solution leetcode

```Input: nums = [2,-1,2], k = 3
Output: 1
Explanation: One optimal approach is to change nums 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.
```

### Maximum Number of Ways to Partition an Array solution leetcode

```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 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.
```

### Maximum Number of Ways to Partition an Array solution leetcode

• `n == nums.length`
• `2 <= n <= 105`
• `-105 <= k, nums[i] <= 105`