1013.Partition Array Into Three Parts With Equal Sum¶

1013.Partition Array Into Three Parts With Equal Sum¶

Tags: Easy Array

Links: https://leetcode.com/problems/partition-array-into-three-parts-with-equal-sum/

Given an array A of integers, return true if and only if we can partition the array into three non-empty parts with equal sums.

Formally, we can partition the array if we can find indexes i+1 < j with (A[0] + A[1] + ... + A[i] == A[i+1] + A[i+2] + ... + A[j-1] == A[j] + A[j-1] + ... + A[A.length - 1])

Example 1:

Input: A = [0,2,1,-6,6,-7,9,1,2,0,1]
Output: true
Explanation: 0 + 2 + 1 = -6 + 6 - 7 + 9 + 1 = 2 + 0 + 1

Example 2:

Input: A = [0,2,1,-6,6,7,9,-1,2,0,1]
Output: false

Example 3:

Input: A = [3,3,6,5,-2,2,5,1,-9,4]
Output: true
Explanation: 3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4

Constraints:

3 <= A.length <= 50000
-10^4 <= A[i] <= 10^4

class Solution {
public:
    bool canThreePartsEqualSum(vector<int>& A) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        int n = A.size();
        vector<int> preSum(n + 1, 0);
        for (int i = 1; i <= n; ++i) {
            preSum[i] = preSum[i - 1] + A[i - 1];
        }
        int sum = preSum[n];
        if (sum % 3 != 0) return false;
        int avg = sum / 3;
        int pos = 0;
        if (!search(preSum, 1, n, avg, pos))
            return false;
        if (!search(preSum, pos + 1, n, 2 * avg, pos))
            return false;
        if (pos == n) return false;

        return true;
    }

    bool search(const vector<int> & preSum, int start, int end, int target, int & pos)
    {
        if (start >= end) return false;
        for (int i = start; i < end; ++i) {
            if (preSum[i] == target) {
                pos = i;
                return true;
            }
        }
        return false;
    }
};

首先计算出数组的前缀和，然后判断其是否整除3。然后在前缀和中寻找avg，2*avg，注意寻找2*avg一定是在找到avg的位置往后寻找。时间复杂度 $O(n)$ ，空间复杂度 $O(n)$ 。

最重要的一点是，数组里面的数据并不都是非负数，所以不能用二分查找。

之所以需要检查pos == n，是考虑1,-1,1,-1的情况。

另外注意到，其实并不一定要额外占用 $O(n)$ 的空间存储前缀和，只需一个变量存储从数组开始到当前位置的和。

class Solution {
    int pos, preSum;
public:
    bool canThreePartsEqualSum(vector<int>& A) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        int n = A.size();
        int sum = 0;
        for (auto e : A) sum += e;
        if (sum % 3 != 0) return false;
        int avg = sum / 3;

        if (!search(A, 0, n, avg))
            return false;
        if (!search(A, pos + 1, n, 2 * avg))
            return false;
        if (pos == n - 1) return false;

        return true;
    }

    bool search(const vector<int> &A, int start, int end, int target)
    {
        if (start >= end) return false;
        for (int i = start; i < end; ++i) {
            preSum += A[i];
            if (preSum == target) {
                pos = i;
                return true;
            }
        }
        return false;
    }
};

Runtime: 20 ms, faster than 100.00% of C++ online submissions for Partition Array Into Three Parts With Equal Sum.
Memory Usage: 11.9 MB, less than 100.00% of C++ online submissions for Partition Array Into Three Parts With Equal Sum.