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| # Given an array of integers, find the pair of adjacent elements | |
| # that has the largest product and return that product. | |
| # Approach 1: (Brute Force) - Check all the pairs in the list and then return the maximum pair | |
| # Time Complexity: O(N^2) | |
| def adjacentElementProductBF(inputArray): | |
| largestProduct = -999999 | |
| # for sanity check, assert if array contains at least 2 elements | |
| if len(inputArray) < 2: | |
| print("No pairs exists") | |
| return -1 | |
| for i in range(0, len(inputArray)): | |
| for j in range(i+1, len(inputArray)): | |
| currentProduct = inputArray[i]*inputArray[j] | |
| if currentProduct > largestProduct: | |
| largestProduct = currentProduct | |
| return largestProduct | |
| # Approach 2: (Sort & Pick Last Pair) - Sort the list and then pick the last two numbers | |
| # Caveat: All elements must be positive | |
| # Time Complexity: O(Nlog(N)) | |
| def adjacentElementsProductSort(inputArray): | |
| size = len(inputArray) | |
| if size < 2: | |
| print("No Pairs exist") | |
| return -1 | |
| sortedArray = sorted(inputArray) | |
| return sortedArray[-1] * sortedArray[-2] | |
| def adjacentElementsProduct(inputArray): | |
| length = int(len(inputArray)) | |
| maxm = inputArray[0]*inputArray[1] | |
| product = 1 | |
| for i in range(1, length-1): | |
| product = inputArray[i]*inputArray[i+1] | |
| if product>maxm: | |
| maxm = product | |
| return maxm | |
| # print(adjacentElementsProduct([3,6,7,5])) | |
| print(adjacentElementsProduct([3, 6, -2, -5, 7, 3])) | |
| #Alternate solution | |
| #return max([inputArray[i]*inputArray[i+1] for i in range(0, int(len(inputArray)-1))]) |
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