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Python Program to Find a triplet such that sum of two equals to third element

  • Last Updated : 11 Jan, 2022

Given an array of integers, you have to find three numbers such that the sum of two elements equals the third element.
Examples:

Input: {5, 32, 1, 7, 10, 50, 19, 21, 2}
Output: 21, 2, 19

Input: {5, 32, 1, 7, 10, 50, 19, 21, 0}
Output: no such triplet exist

Question source: Arcesium Interview Experience | Set 7 (On campus for Internship)

Simple approach: Run three loops and check if there exists a triplet such that sum of two elements equals the third element.
Time complexity: O(n^3)
Efficient approach: The idea is similar to Find a triplet that sum to a given value.

  • Sort the given array first.
  • Start fixing the greatest element of three from the back and traverse the array to find the other two numbers which sum up to the third element.
  • Take two pointers j(from front) and k(initially i-1) to find the smallest of the two number and from i-1 to find the largest of the two remaining numbers
  • If the addition of both the numbers is still less than A[i], then we need to increase the value of the summation of two numbers, thereby increasing the j pointer, so as to increase the value of A[j] + A[k].
  • If the addition of both the numbers is more than A[i], then we need to decrease the value of the summation of two numbers, thereby decrease the k pointer so as to decrease the overall value of A[j] + A[k].

Below image is a dry run of the above approach:

Below is the implementation of the above approach:

Python




# Python program to find three numbers
# such that sum of two makes the
# third element in array
  
# Utility function for finding
# triplet in array
def findTriplet(arr, n):
      
    # Sort the array
    arr.sort()
   
    # For every element in arr check 
    # if a pair exist(in array) whose
    # sum is equal to arr element
    i = n - 1
    while(i >= 0):
        j = 0
        k = i - 1
        while (j < k):
            if (arr[i] == arr[j] + arr[k]):
                 
                # Pair found
                print "numbers are ", arr[i], 
                       arr[j], arr[k]
                return
            elif (arr[i] > arr[j] + arr[k]):
                j += 1
            else:
                k -= 1
        i -= 1
          
    # No such triplet is found in array
    print "No such triplet exists"
   
# Driver code
arr = [5, 32, 1, 7, 10, 50, 19, 21, 2]
n = len(arr)
findTriplet(arr, n)
# This code is contributed by Sachin Bisht

Output:  

numbers are 21 2 19

Time complexity: O(N^2)
Please refer complete article on Find a triplet such that sum of two equals to third element for more details!


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