Find all triplets in a list with given sum in Python

In a list of numbers we want to find out which three elements can join to give a certain sum. We call it a triplet. And in the list there can be many such triplets. For example, the sum 10 can be generated from numbers 1,6,3 as well as 1,5,4. In this article we will see how to find out all such triplets from a given list of numbers.

Using Nested Loops with Set

This approach uses nested loops with a set to efficiently find triplets. We iterate through the list, and for each element, we look for pairs in the remaining elements that sum to the target value.

def find_triplets(numbers, target_sum):
    result = []
    
    for i in range(len(numbers) - 2):
        seen = set()
        remaining_sum = target_sum - numbers[i]
        
        for j in range(i + 1, len(numbers)):
            complement = remaining_sum - numbers[j]
            
            if complement in seen:
                triplet = (numbers[i], complement, numbers[j])
                result.append(triplet)
            
            seen.add(numbers[j])
    
    return result

numbers = [11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
triplets = find_triplets(numbers, 40)

print("Required triplets:")
for triplet in triplets:
    print(triplet)

Required triplets:
(11, 15, 14)
(11, 16, 13)
(11, 17, 12)
(12, 15, 13)

Using itertools.combinations

The itertools.combinations function generates all possible combinations of three elements. We then filter these combinations to find those that sum to our target value.

from itertools import combinations

def find_triplets_combinations(numbers, target_sum):
    all_triplets = combinations(numbers, 3)
    return [triplet for triplet in all_triplets if sum(triplet) == target_sum]

numbers = [11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
triplets = find_triplets_combinations(numbers, 40)

print("Required triplets:")
for triplet in triplets:
    print(triplet)

Required triplets:
(11, 12, 17)
(11, 13, 16)
(11, 14, 15)
(12, 13, 15)

Comparison

Finding Triplets with Duplicates

If your list contains duplicate values and you want unique triplets, you can use a set to store results ?

def find_unique_triplets(numbers, target_sum):
    unique_triplets = set()
    
    for i in range(len(numbers) - 2):
        seen = set()
        remaining_sum = target_sum - numbers[i]
        
        for j in range(i + 1, len(numbers)):
            complement = remaining_sum - numbers[j]
            
            if complement in seen:
                triplet = tuple(sorted([numbers[i], complement, numbers[j]]))
                unique_triplets.add(triplet)
            
            seen.add(numbers[j])
    
    return list(unique_triplets)

numbers = [1, 2, 3, 4, 5, 6, 3, 2]
triplets = find_unique_triplets(numbers, 10)

print("Unique triplets that sum to 10:")
for triplet in sorted(triplets):
    print(triplet)

Unique triplets that sum to 10:
(1, 3, 6)
(1, 4, 5)
(2, 2, 6)
(2, 3, 5)

Conclusion

Use the nested loop approach for better performance with large datasets. Use itertools.combinations for simpler, more readable code with smaller datasets. For duplicate handling, sort triplets before adding to a set.