Question

2811. Check if it is Possible to Split Array

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Description

You are given an array nums of length n and an integer m. You need to determine if it is possible to split the array into n non-empty arrays by performing a series of steps.

In each step, you can select an existing array (which may be the result of previous steps) with a length of at least two and split it into two subarrays, if, for each resulting subarray, at least one of the following holds:

• The length of the subarray is one, or
• The sum of elements of the subarray is greater than or equal to m.

Return true_ if you can split the given array into _n_ arrays, otherwise return_ false.

Note: A subarray is _a contiguous non-empty sequence of elements within an array_.

 

Example 1:

Input: nums = [2, 2, 1], m = 4
Output: true
Explanation: We can split the array into [2, 2] and [1] in the first step. Then, in the second step, we can split [2, 2] into [2] and [2]. As a result, the answer is true.

Example 2:

Input: nums = [2, 1, 3], m = 5 
Output: false
Explanation: We can try splitting the array in two different ways: the first way is to have [2, 1] and [3], and the second way is to have [2] and [1, 3]. However, both of these ways are not valid. So, the answer is false.

Example 3:

Input: nums = [2, 3, 3, 2, 3], m = 6
Output: true
Explanation: We can split the array into [2, 3, 3, 2] and [3] in the first step. Then, in the second step, we can split [2, 3, 3, 2] into [2, 3, 3] and [2]. Then, in the third step, we can split [2, 3, 3] into [2] and [3, 3]. And in the last step we can split [3, 3] into [3] and [3]. As a result, the answer is true.

 

Constraints:

• 1 <= n == nums.length <= 100
• 1 <= nums[i] <= 100
• 1 <= m <= 200

Solutions

Solution 1: Memoization Search

First, we preprocess to get the prefix sum array $s$, where $s[i]$ represents the sum of the first $i$ elements of the array $nums$.

Next, we design a function $dfs(i, j)$, which represents whether there is a way to split the index range $[i, j]$ of the array $nums$ that meets the conditions. If it exists, return true, otherwise return false.

The calculation process of the function $dfs(i, j)$ is as follows:

If $i = j$, then there is only one element, no need to split, return true;

Otherwise, we enumerate the split point $k$, where $k \in [i, j]$, if the following conditions are met, then it can be split into two subarrays $nums[i,.. k]$ and $nums[k + 1,.. j]$:

• The subarray $nums[i,..k]$ has only one element, or the sum of the elements of the subarray $nums[i,..k]$ is greater than or equal to $m$;
• The subarray $nums[k + 1,..j]$ has only one element, or the sum of the elements of the subarray $nums[k + 1,..j]$ is greater than or equal to $m$;
• Both $dfs(i, k)$ and $dfs(k + 1, j)$ are true.

To avoid repeated calculations, we use the method of memoization search, and use a two-dimensional array $f$ to record all the return values of $dfs(i, j)$, where $f[i][j]$ represents the return value of $dfs(i, j)$.

The time complexity is $O(n^3)$, and the space complexity is $O(n^2)$, where $n$ is the length of the array $nums$.

class Solution:
  def canSplitArray(self, nums: List[int], m: int) -> bool:
    @cache
    def dfs(i: int, j: int) -> bool:
      if i == j:
        return True
      for k in range(i, j):
        a = k == i or s[k + 1] - s[i] >= m
        b = k == j - 1 or s[j + 1] - s[k + 1] >= m
        if a and b and dfs(i, k) and dfs(k + 1, j):
          return True
      return False

    s = list(accumulate(nums, initial=0))
    return dfs(0, len(nums) - 1)

class Solution {
  private Boolean[][] f;
  private int[] s;
  private int m;

  public boolean canSplitArray(List<Integer> nums, int m) {
    int n = nums.size();
    f = new Boolean[n][n];
    s = new int[n + 1];
    for (int i = 1; i <= n; ++i) {
      s[i] = s[i - 1] + nums.get(i - 1);
    }
    this.m = m;
    return dfs(0, n - 1);
  }

  private boolean dfs(int i, int j) {
    if (i == j) {
      return true;
    }
    if (f[i][j] != null) {
      return f[i][j];
    }
    for (int k = i; k < j; ++k) {
      boolean a = k == i || s[k + 1] - s[i] >= m;
      boolean b = k == j - 1 || s[j + 1] - s[k + 1] >= m;
      if (a && b && dfs(i, k) && dfs(k + 1, j)) {
        return f[i][j] = true;
      }
    }
    return f[i][j] = false;
  }
}

class Solution {
public:
  bool canSplitArray(vector<int>& nums, int m) {
    int n = nums.size();
    vector<int> s(n + 1);
    for (int i = 1; i <= n; ++i) {
      s[i] = s[i - 1] + nums[i - 1];
    }
    int f[n][n];
    memset(f, -1, sizeof f);
    function<bool(int, int)> dfs = [&](int i, int j) {
      if (i == j) {
        return true;
      }
      if (f[i][j] != -1) {
        return f[i][j] == 1;
      }
      for (int k = i; k < j; ++k) {
        bool a = k == i || s[k + 1] - s[i] >= m;
        bool b = k == j - 1 || s[j + 1] - s[k + 1] >= m;
        if (a && b && dfs(i, k) && dfs(k + 1, j)) {
          f[i][j] = 1;
          return true;
        }
      }
      f[i][j] = 0;
      return false;
    };
    return dfs(0, n - 1);
  }
};

func canSplitArray(nums []int, m int) bool {
  n := len(nums)
  f := make([][]int, n)
  s := make([]int, n+1)
  for i, x := range nums {
    s[i+1] = s[i] + x
  }
  for i := range f {
    f[i] = make([]int, n)
  }
  var dfs func(i, j int) bool
  dfs = func(i, j int) bool {
    if i == j {
      return true
    }
    if f[i][j] != 0 {
      return f[i][j] == 1
    }
    for k := i; k < j; k++ {
      a := k == i || s[k+1]-s[i] >= m
      b := k == j-1 || s[j+1]-s[k+1] >= m
      if a && b && dfs(i, k) && dfs(k+1, j) {
        f[i][j] = 1
        return true
      }
    }
    f[i][j] = -1
    return false
  }
  return dfs(0, n-1)
}

function canSplitArray(nums: number[], m: number): boolean {
  const n = nums.length;
  const s: number[] = new Array(n + 1).fill(0);
  for (let i = 1; i <= n; ++i) {
    s[i] = s[i - 1] + nums[i - 1];
  }
  const f: number[][] = Array(n)
    .fill(0)
    .map(() => Array(n).fill(-1));
  const dfs = (i: number, j: number): boolean => {
    if (i === j) {
      return true;
    }
    if (f[i][j] !== -1) {
      return f[i][j] === 1;
    }
    for (let k = i; k < j; ++k) {
      const a = k === i || s[k + 1] - s[i] >= m;
      const b = k === j - 1 || s[j + 1] - s[k + 1] >= m;
      if (a && b && dfs(i, k) && dfs(k + 1, j)) {
        f[i][j] = 1;
        return true;
      }
    }
    f[i][j] = 0;
    return false;
  };
  return dfs(0, n - 1);
}

impl Solution {
  pub fn can_split_array(nums: Vec<i32>, m: i32) -> bool {
    let n = nums.len();
    if n <= 2 {
      return true;
    }
    for i in 1..n {
      if nums[i - 1] + nums[i] >= m {
        return true;
      }
    }
    false
  }
}

Solution 2: Quick Thinking

No matter how you operate, there will always be a length == 2 subarray left in the end. Since there are no negative numbers in the elements, as the split operation proceeds, the length and sum of the subarray will gradually decrease. The sum of other length > 2 subarrays must be larger than the sum of this subarray. Therefore, we only need to consider whether there is a length == 2 subarray with a sum greater than or equal to m.

???? Note that when nums.length <= 2, no operation is needed.

The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

function canSplitArray(nums: number[], m: number): boolean {
  const n = nums.length;
  if (n <= 2) {
    return true;
  }
  for (let i = 1; i < n; i++) {
    if (nums[i - 1] + nums[i] >= m) {
      return true;
    }
  }
  return false;
}