Question

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

Example 3:

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

Note:

  1. 3 <= A.length <= 50000
  2. -10000 <= A[i] <= 10000

Difficulty:Easy

Category:

Solution

Time complexity: O(n), Space complexity: O(1)

class Solution {
 public:
  bool canThreePartsEqualSum(vector<int>& A) {
    int sum = 0;
    for (int& a : A) sum += a;
    if (sum % 3 != 0) return false;

    int target = sum / 3;
    int preSum = 0, ind1 = -1, ind2 = -1;
    for (int i = 0; i < A.size(); ++i) {
      preSum += A[i];
      if (preSum == target && ind1 == -1) ind1 = 1;
      else if (preSum == 2 * target && ind1 == 1) {
        ind2 = 1;
        break;
      }
    }

    if (ind1 == 1 && ind2 == 1) return true;
    return false;
  }
};
By guozetang            Updated: 2020-09-19 13:02:30

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