1013. Divide the array into three parts equal to and

Address ：

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

subject ：

Give you an array of integers arr, Only can it be divided into three and equal Non empty Return only when partial  true, Otherwise return to false.

Formally , If you can find the index  i + 1 < j  And meet  (arr[0] + arr[1] + ... + arr[i] == arr[i + 1] + arr[i + 2] + ... + arr[j - 1] == arr[j] + arr[j + 1] + ... + arr[arr.length - 1])  You can divide the array into three equal parts .

Example 1：

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

Example 2：

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

Example 3：

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

Tips ：

3 <= arr.length <= 5 * 104 -104 <= arr[i] <= 104

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/partition-array-into-three-parts-with-equal-sum
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Ideas ：

Since it is divided into three groups and , Then first of all, ask and , Every part and is its 1/3

If the sum cannot be 3 to be divisible by , direct false

Use double pointers to separate from the array [0], Array [len-1] Close to the middle , Find the left value and , Right value sum

Intermediate and = The sum of the - L value and - Right value sum

The pit here is ：

1. Can't find L value and = Right value sum , Just jump out of the loop , At this time, the left value has not reached 1/3, Nature needs to continue to accumulate

2. Can't find L value and = Right value sum , And equal to 1/3 , Jump out of the loop and find the middle sum is also equal , There is a possibility that

        The sum of the = 0, L value and = Right value sum = 0, such as [1,-1, -1, 1]

    So it also depends on whether the number of the middle number is 0, Only when it is not, can we continue to judge

There are many calculations in this way , And consider more boundaries , The test case has a high probability of error

Another idea is relatively simple , Reference method 2

Method 1 、 Double pointer summation

bool canThreePartsEqualSum(int* arr, int arrSize){
    int goal = 0;
	int lsum = 0, msum = 0, rsum = 0;
	int sums = 0;
    int cnt = arrSize;
	int i,j;
	
	for(i=0; i<arrSize; i++)
		sums += arr[i];
	
    if(sums % 3 != 0)
        return false;

    goal = sums/3;

    i = 0;
    j = arrSize - 1;
    lsum += arr[i];
    rsum += arr[arrSize-1];
    cnt-=2;

    while( cnt > 0)
    {
        if(lsum != goal)
        {
            i++;
            lsum += arr[i];
            cnt--;
        }
        
        if(rsum != goal)
        {
            j--;
            rsum += arr[j];
            cnt--;
        }

        if(lsum == goal && lsum == rsum)
            break;
        
    }
    
    msum = sums - lsum - rsum;
    
    if(cnt > 0 && lsum == msum && rsum == msum)
		return true;
    else
    	return false;    
}

Method 2 、 Find the number of multiples

We know 1/3 And , Then at least there is 3 Such a sum , There may also be more , For example, He Wei 0 perhaps 3n A combination like this

So we only need two steps to complete

1. Statistical sum

2. Ask again 1/3 And , Number of Statistics ++, If the number of statistics is less than 3 , That's it false

bool canThreePartsEqualSum(int* arr, int arrSize){
    int sum = 0;
    int tmp = 0;
    int goal = 0;
    int cnt = 0;
    int i;

    for(i=0; i<arrSize; i++)
        sum += arr[i];
    
    if(sum % 3 != 0)
        return false;
    
    goal = sum / 3;

    for(i=0; i<arrSize; i++)
    {
        tmp += arr[i];
        if(tmp == goal)
        {
            cnt++;
            tmp = 0;
        }
    }

    if(cnt < 3)
        return false;
    
    return true;
}

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