Given a sorted array of distinct positive integers, print all triplets that form AP (or Arithmetic Progression)
Examples :
Input : arr[] = { 2, 6, 9, 12, 17, 22, 31, 32, 35, 42 };
Output :
6 9 12
2 12 22
12 17 22
2 17 32
12 22 32
9 22 35
2 22 42
22 32 42
Input : arr[] = { 3, 5, 6, 7, 8, 10, 12};
Output :
3 5 7
5 6 7
6 7 8
6 8 10
8 10 12
A simple solution is to run three nested loops to generate all triplets and for every triplet, check if it forms AP or not. Time complexity of this solution is O(n3)
A better solution is to use hashing. We traverse array from left to right. We consider every element as middle and all elements after it as next element. To search the previous element, we use a hash table.
Python3
# Python program to print all # triplets in given array # that form Arithmetic # Progression # Function to print # all triplets in # given sorted array # that forms AP def printAllAPTriplets(arr, n) : s = []; for i in range(0, n - 1) : for j in range(i + 1, n) : # Use hash to find if # there is a previous # element with difference # equal to arr[j] - arr[i] diff = arr[j] - arr[i]; if ((arr[i] - diff) in arr) : print ("{} {} {}" . format((arr[i] - diff), arr[i], arr[j]), end = " "); s.append(arr[i]); # Driver code arr = [2, 6, 9, 12, 17, 22, 31, 32, 35, 42]; n = len(arr); printAllAPTriplets(arr, n); # This code is contributed by # Manish Shaw(manishshaw1) |
Output :
6 9 12 2 12 22 12 17 22 2 17 32 12 22 32 9 22 35 2 22 42 22 32 42
Time Complexity : O(n2)
Auxiliary Space : O(n)
An efficient solution is based on the fact that the array is sorted. We use the same concept as discussed in GP triplet question. The idea is to start from the second element and fix every element as a middle element and search for the other two elements in a triplet (one smaller and one greater).
Below is the implementation of the above idea.
Python 3
# python 3 program to print all triplets in given # array that form Arithmetic Progression # Function to print all triplets in # given sorted array that forms AP def printAllAPTriplets(arr, n): for i in range(1, n - 1): # Search other two elements of # AP with arr[i] as middle. j = i - 1 k = i + 1 while(j >= 0 and k < n ): # if a triplet is found if (arr[j] + arr[k] == 2 * arr[i]): print(arr[j], "", arr[i], "", arr[k]) # Since elements are distinct, # arr[k] and arr[j] cannot form # any more triplets with arr[i] k += 1 j -= 1 # If middle element is more move to # higher side, else move lower side. elif (arr[j] + arr[k] < 2 * arr[i]): k += 1 else: j -= 1 # Driver code arr = [ 2, 6, 9, 12, 17, 22, 31, 32, 35, 42 ] n = len(arr) printAllAPTriplets(arr, n) # This article is contributed # by Smitha Dinesh Semwal |
Output :
6 9 12 2 12 22 12 17 22 2 17 32 12 22 32 9 22 35 2 22 42 22 32 42
Time Complexity : O(n2)
Auxiliary Space : O(1)
Please refer complete article on Print all triplets in sorted array that form AP for more details!
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