##### Short Problem Definition:

Maximize A[P] * A[Q] * A[R] for any triplet (P, Q, R).

##### Link

##### Complexity:

expected worst-case time complexity is O(N*log(N));

expected worst-case space complexity is O(1)

##### Execution:

After sorting the largest product can be found as a combination of the last three elements. Additionally, two negative numbers add to a positive, so by multiplying the two largest negatives with the largest positive, we get another candidate. If all numbers are negative, the three largest (closest to 0) still get the largest element!

##### Solution:

def solution(A): if len(A) < 3: raise Exception("Invalid input") A.sort() return max(A[0] * A[1] * A[-1], A[-1] * A[-2] * A[-3])

A better solution has been suggested in the commends by

def betterSolution(A): if len(A) < 3: raise Exception("Invalid input") minH = [] maxH = [] for val in A: if len(minH) < 2: heapq.heappush(minH, -val) else: heapq.heappushpop(minH, -val) if len(maxH) < 3: heapq.heappush(maxH, val) else: heapq.heappushpop(maxH, val) max_val = heapq.heappop(maxH) * heapq.heappop(maxH) top_ele = heapq.heappop(maxH) max_val *= top_ele min_val = -heapq.heappop(minH) * -heapq.heappop(minH) * top_ele return max(max_val, min_val)

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