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Formatted question description: https://leetcode.ca/all/1013.html

1013. Partition Array Into Three Parts With Equal Sum (Easy)

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

Related Topics:
Array

Solution 1.

  • class Solution {
    public boolean canThreePartsEqualSum(int[] A) {
        int sum = 0;
        for (int num : A)
            sum += num;
        if (sum % 3 != 0)
            return false;
        int partSum = sum / 3;
        int count = 0;
        int curSum = 0;
        int length = A.length;
        for (int i = 0; i < length; i++) {
            curSum += A[i];
            if (curSum == partSum * (count + 1))
                count++;
        }
        return count >= 3;
    }
}

############

class Solution {
    public boolean canThreePartsEqualSum(int[] arr) {
        int s = 0;
        for (int v : arr) {
            s += v;
        }
        if (s % 3 != 0) {
            return false;
        }
        int i = 0, j = arr.length - 1;
        int a = 0, b = 0;
        while (i < arr.length) {
            a += arr[i];
            if (a == s / 3) {
                break;
            }
            ++i;
        }
        while (j >= 0) {
            b += arr[j];
            if (b == s / 3) {
                break;
            }
            --j;
        }
        return i < j - 1;
    }
}

  • // OJ: https://leetcode.com/problems/partition-array-into-three-parts-with-equal-sum/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    bool canThreePartsEqualSum(vector<int>& A) {
        int total = accumulate(begin(A), end(A), 0), cnt = 0, sum = 0;
        if (total % 3) return false;
        total /= 3;
        for (int i = 0; i < A.size() - 1; ++i) {
            sum += A[i];
            if (sum == total) {
                sum = 0;
                ++cnt;
            }
            if (cnt == 2) return true;
        }
        return false;
    }
};

  • class Solution:
    def canThreePartsEqualSum(self, arr: List[int]) -> bool:
        s = sum(arr)
        if s % 3 != 0:
            return False
        i, j = 0, len(arr) - 1
        a = b = 0
        while i < len(arr):
            a += arr[i]
            if a == s // 3:
                break
            i += 1
        while ~j:
            b += arr[j]
            if b == s // 3:
                break
            j -= 1
        return i < j - 1

############

class Solution(object):
    def numPairsDivisibleBy60(self, time):
        """
        :type time: List[int]
        :rtype: int
        """
        count = collections.Counter(time)
        key = list(count.keys())
        N = len(key)
        res = 0
        for i, t in enumerate(key):
            for j in range(i, N):
                if (t + key[j]) % 60 == 0:
                    if i == j:
                        res += count[t] * (count[t] - 1) / 2
                    else:
                        res += count[t] * count[key[j]]
        return res

  • func canThreePartsEqualSum(arr []int) bool {
	s := 0
	for _, v := range arr {
		s += v
	}
	if s%3 != 0 {
		return false
	}
	i, j := 0, len(arr)-1
	a, b := 0, 0
	for i < len(arr) {
		a += arr[i]
		if a == s/3 {
			break
		}
		i++
	}
	for j >= 0 {
		b += arr[j]
		if b == s/3 {
			break
		}
		j--
	}
	return i < j-1
}

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