Saturday, June 4, 2016

Find maximum product of a triplet in array - GeeksforGeeks


Find maximum product of a triplet in array - GeeksforGeeks
Given an integer array, find a maximum product of a triplet in array.
    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.
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);
}

Approach 2: O(n) Time, O(n) Space
  1. Construct four auxiliary arrays leftMax[], rightMax[], leftMin[] and rightMin[] of same size as input array.
  2. Fill leftMax[], rightMax[], leftMin[] and rightMin[] in below manner.
    • leftMax[i] will contain maximum element on left of arr[i] excluding arr[i]. For index 0, left will contain -1.
    • leftMin[i] will contain minimum element on left of arr[i] excluding arr[i]. For index 0, left will contain -1.
    • rightMax[i] will contain maximum element on right of arr[i] excluding arr[i]. For index n-1, right will contain -1.
    • rightMin[i] will contain minimum element on right of arr[i] excluding arr[i]. For index n-1, right will contain -1.
  3. For all array indexes i except first and last index, compute maximum of arr[i]*x*y where x can be leftMax[i] or leftMin[i] and y can be rightMax[i] or rightMin[i].
  4. Return the maximum from step 3.
int maxProduct(int arr[], int n)
{
    // if size is less than 3, no triplet exists
    if (n < 3)
        return -1;
 
    // Construct four auxiliary vectors
    // of size n and initailize them by -1
    vector<int> leftMin(n, -1);
    vector<int> rightMin(n, -1);
    vector<int> leftMax(n, -1);
    vector<int> rightMax(n, -1);
 
    // will contain max product
    int max_product = INT_MIN;
 
    // to store maximum element on left of array
    int max_sum = arr[0];
 
    // to store minimum element on left of array
    int min_sum = arr[0];
 
    // leftMax[i] will contain max element
    // on left of arr[i] excluding arr[i].
    // leftMin[i] will contain min element
    // on left of arr[i] excluding arr[i].
    for (int i = 1; i < n; i++)
    {
        leftMax[i] = max_sum;
        if (arr[i] > max_sum)
            max_sum = arr[i];
 
        leftMin[i] = min_sum;
        if (arr[i] < min_sum)
            min_sum = arr[i];
    }
 
    // reset max_sum to store maximum element on
    // right of array
    max_sum = arr[n - 1];
 
    // reset min_sum to store minimum element on
    // right of array
    min_sum = arr[n - 1];
 
    // rightMax[i] will contain max element
    // on right of arr[i] excluding arr[i].
    // rightMin[i] will contain min element
    // on right of arr[i] excluding arr[i].
    for (int j = n - 2; j >= 0; j--)
    {
        rightMax[j] = max_sum;
        if (arr[j] > max_sum)
            max_sum = arr[j];
 
        rightMin[j] = min_sum;
        if (arr[j] < min_sum)
            min_sum = arr[j];
    }
 
    // For all array indexes i except first and
    // last, compute maximum of arr[i]*x*y where
    // x can be leftMax[i] or leftMin[i] and
    // y can be rightMax[i] or rightMin[i].
    for (int i = 1; i < n - 1; i++)
    {
        int max1 = max(arr[i] * leftMax[i] * rightMax[i],
                    arr[i] * leftMin[i] * rightMin[i]);
 
        int max2 = max(arr[i] * leftMax[i] * rightMin[i],
                    arr[i] * leftMin[i] * rightMax[i]);
 
        max_product = max(max_product, max(max1, max2));
    }
 
    return max_product;
}

Approach 3: O(nlogn) Time, O(1) Space
  1. Sort the array using some efficient in-place sorting algorithm in ascending order.
  2. Return the maximum of product of last three elements of the array and product of first two elements and last element.
int maxProduct(int arr[], int n)
{
    // if size is less than 3, no triplet exists
    if (n < 3)
        return -1;
    // Sort the array in ascending order
    sort(arr, arr + n);
    // Return the maximum of product of last three
    // elements and product of first two elements
    // and last element
    return max(arr[0] * arr[1] * arr[n - 1],
               arr[n - 1] * arr[n - 2] * arr[n - 3]);
}

Approach 1 (Naive, O(n3) time, O(1) Space)
    1. Print the triplet that has maximum product.
    2. Find a minimum product of a triplet in array.
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