Given an integer . The task is to find all factors of N and print the product of four factors of N such that:
- Sum of the four factors is equal to N.
- Product of the four factors is maximum.
If it is not possible to find 4 such factors then print “Not possible”.
Note: All the four factors can be equal to each other to maximize the product.
Input : N = 24 Output : Product -> 1296 All factors are -> 1 2 3 4 6 8 12 24 Choose the factor 6 four times, Therefore, 6+6+6+6 = 24 and product is maximum. Input : N = 100 Output : Product -> 390625 All the factors are -> 1 2 4 5 10 10 20 25 50 100 Choose the factor 25 four times.
An approach which takes a complexity of O(M^3), where M is the number of factors of N has been discussed in the previous post.
An efficient approach of time complexity O(N^2) can be obtained by following the below steps.
- Store all the factors of given number in a container.
- Iterate for all pairs and store their sum in a different container.
- Mark the index (element1 + element2) with pair(element1, element2 to get the elements by which the sum was obtained.
- Iterate for all the pair_sums, and check if n-pair_sum exists in the same container, then both the pairs form the quadruple.
- Use the pair hash array to get the elements by which the pair was formed.
- Store the maximum of all such quadruples, and print it at the end.
Below is the implementation of above approach:
The maximum product is 12500
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