find a maximum product triplet in this array

O（N）扫一遍array就可以了: 找三个最大的， 两个最小的，然后根据第三个最大的正负来判断。

Approach 4: O(n) Time, O(1) Space

1. Scan the array and compute Maximum, second maximum and third maximum element present in the array.
2. Scan the array and compute Minimum and second minimum element present in the array.
3. Return the maximum of product of Maximum, second maximum and third maximum and product of Minimum, second minimum and Maximum element.

Note – Step 1 and Step 2 can be done in single traversal of the array.

int maxProduct(int arr[], int n)
{
    // if size is less than 3, no triplet exists
    if (n < 3)
        return -1;
 
    // Initialize Maximum, second maximum and third
    // maximum element
    int maxA = INT_MIN, maxB = INT_MIN, maxC = INT_MIN;
 
    // Initialize Minimum and second mimimum element
    int minA = INT_MAX, minB = INT_MAX;
 
    for (int i = 0; i < n; i++)
    {
        // Update Maximum, second maximum and third
        // maximum element
        if (arr[i] > maxA)
        {
            maxC = maxB;
            maxB = maxA;
            maxA = arr[i];
        }
 
        // Update second maximum and third maximum element
        else if (arr[i] > maxB)
        {
            maxC = maxB;
            maxB = arr[i];
        }
 
        // Update third maximum element
        else if (arr[i] > maxC)
            maxC = arr[i];
 
        // Update Minimum and second mimimum element
        if (arr[i] < minA)
        {
            minB = minA;
            minA = arr[i];
        }
 
        // Update second mimimum element
        else if(arr[i] < minB)
            minB = arr[i];
    }
 
    return max(minA * minB * maxA,
               maxA * maxB * maxC);
}