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548. Split Array with Equal Sum πŸ”’

Description

Given an integer array nums of length n, return true if there is a triplet (i, j, k) which satisfies the following conditions:

  • 0 < i, i + 1 < j, j + 1 < k < n - 1
  • The sum of subarrays (0, i - 1), (i + 1, j - 1), (j + 1, k - 1) and (k + 1, n - 1) is equal.

A subarray (l, r) represents a slice of the original array starting from the element indexed l to the element indexed r.

 

Example 1:

Input: nums = [1,2,1,2,1,2,1]
Output: true
Explanation:
i = 1, j = 3, k = 5. 
sum(0, i - 1) = sum(0, 0) = 1
sum(i + 1, j - 1) = sum(2, 2) = 1
sum(j + 1, k - 1) = sum(4, 4) = 1
sum(k + 1, n - 1) = sum(6, 6) = 1

Example 2:

Input: nums = [1,2,1,2,1,2,1,2]
Output: false

 

Constraints:

  • n == nums.length
  • 1 <= n <= 2000
  • -106 <= nums[i] <= 106

Solutions

Solution 1

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class Solution:
    def splitArray(self, nums: List[int]) -> bool:
        n = len(nums)
        s = [0] * (n + 1)
        for i, v in enumerate(nums):
            s[i + 1] = s[i] + v
        for j in range(3, n - 3):
            seen = set()
            for i in range(1, j - 1):
                if s[i] == s[j] - s[i + 1]:
                    seen.add(s[i])
            for k in range(j + 2, n - 1):
                if s[n] - s[k + 1] == s[k] - s[j + 1] and s[n] - s[k + 1] in seen:
                    return True
        return False

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class Solution {
    public boolean splitArray(int[] nums) {
        int n = nums.length;
        int[] s = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        for (int j = 3; j < n - 3; ++j) {
            Set<Integer> seen = new HashSet<>();
            for (int i = 1; i < j - 1; ++i) {
                if (s[i] == s[j] - s[i + 1]) {
                    seen.add(s[i]);
                }
            }
            for (int k = j + 2; k < n - 1; ++k) {
                if (s[n] - s[k + 1] == s[k] - s[j + 1] && seen.contains(s[n] - s[k + 1])) {
                    return true;
                }
            }
        }
        return false;
    }
}

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class Solution {
public:
    bool splitArray(vector<int>& nums) {
        int n = nums.size();
        vector<int> s(n + 1);
        for (int i = 0; i < n; ++i) s[i + 1] = s[i] + nums[i];
        for (int j = 3; j < n - 3; ++j) {
            unordered_set<int> seen;
            for (int i = 1; i < j - 1; ++i)
                if (s[i] == s[j] - s[i + 1])
                    seen.insert(s[i]);
            for (int k = j + 2; k < n - 1; ++k)
                if (s[n] - s[k + 1] == s[k] - s[j + 1] && seen.count(s[n] - s[k + 1]))
                    return true;
        }
        return false;
    }
};

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func splitArray(nums []int) bool {
    n := len(nums)
    s := make([]int, n+1)
    for i, v := range nums {
        s[i+1] = s[i] + v
    }
    for j := 3; j < n-3; j++ {
        seen := map[int]bool{}
        for i := 1; i < j-1; i++ {
            if s[i] == s[j]-s[i+1] {
                seen[s[i]] = true
            }
        }
        for k := j + 2; k < n-1; k++ {
            if s[n]-s[k+1] == s[k]-s[j+1] && seen[s[n]-s[k+1]] {
                return true
            }
        }
    }
    return false
}

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