Find three element from different three arrays such that that a + b + c = sum in Python

Finding three elements from different arrays that sum to a target value is a common programming problem. We need to check if there exist elements a, b, c from arrays A, B, C respectively such that a + b + c equals the given sum.

So, if the input is like A = [2,3,4,5,6], B = [3,4,7,2,3], C = [4,3,5,6,7], sum = 12, then the output will be True as 4+2+6 = 12, where 4, 2, 6 are taken from A, B, C respectively.

Method 1: Brute Force Approach

The straightforward approach uses three nested loops to check all possible combinations ?

Algorithm

• Iterate through each element in array A

• For each element in A, iterate through array B

• For each pair from A and B, iterate through array C

• Check if the sum of three elements equals the target

• Return True if found, False otherwise

Example

def is_sum_from_three_arr(A, B, C, target_sum):
    for i in range(len(A)):
        for j in range(len(B)):
            for k in range(len(C)):
                if (A[i] + B[j] + C[k] == target_sum):
                    return True
    return False

A = [2, 3, 4, 5, 6]
B = [3, 4, 7, 2, 3]
C = [4, 3, 5, 6, 7]
target_sum = 12

result = is_sum_from_three_arr(A, B, C, target_sum)
print(f"Can find three elements that sum to {target_sum}: {result}")

Can find three elements that sum to 12: True

Method 2: Optimized Approach Using Set

We can optimize by using a set to store elements from the third array for faster lookup ?

def is_sum_from_three_arr_optimized(A, B, C, target_sum):
    # Convert C to set for O(1) lookup
    set_C = set(C)
    
    for a in A:
        for b in B:
            required = target_sum - a - b
            if required in set_C:
                return True
    return False

A = [2, 3, 4, 5, 6]
B = [3, 4, 7, 2, 3]
C = [4, 3, 5, 6, 7]
target_sum = 12

result = is_sum_from_three_arr_optimized(A, B, C, target_sum)
print(f"Optimized approach result: {result}")

Optimized approach result: True

Method 3: Finding Actual Elements

To return the actual triplet instead of just True/False ?

def find_triplet_with_sum(A, B, C, target_sum):
    for a in A:
        for b in B:
            for c in C:
                if a + b + c == target_sum:
                    return (a, b, c)
    return None

A = [2, 3, 4, 5, 6]
B = [3, 4, 7, 2, 3]
C = [4, 3, 5, 6, 7]
target_sum = 12

triplet = find_triplet_with_sum(A, B, C, target_sum)
if triplet:
    print(f"Found triplet: {triplet[0]} + {triplet[1]} + {triplet[2]} = {target_sum}")
else:
    print("No triplet found")

Found triplet: 2 + 4 + 6 = 12

Comparison

Conclusion

The brute force approach is simple but has O(n³) complexity. The optimized set-based approach reduces time complexity to O(n²) by eliminating one loop. Choose the method based on your array sizes and whether you need the actual triplet values.