**Problem**

A Pythagorean triplet is a set of three natural numbers, a < b < c , for which,a^{2} + b^{2} = c^{2}

For example, 3^{2} + 4^{2} = 9 + 16 = 25 = 5^{2}.

There exists exactly one Pythagorean triplet for which a+b+c = 1000.

Find the product abc.

**Solution Approach**

In this problem of Special Pythagorean triplet, we have to find three number which forms Pythagorean triplet. The numbers should be such that they add up to 1000. We have to find the product of the three numbers as the answer.

We can use a brute force algorithm to solve the problem. The maximum number that we have to search for is bounded by **2*(1000 ^{2})** which fits the integer data type. We can start with giving one value to

**a**, then giving subsequent values to

**b**, and finding

**c**by subtracting values of

**a**and

**b**from 1000. Then we place these in Pythagoras equation to find if they are Pythagorean triplet or not.

The time complexity of this approach is **O(n ^{2})** as we have to iterate through 1 to 1000 twice for a and b. The space complexity of this approach is constant i.e

**O(1)**as we are just using a variable to store the value of the product.

Coming to the code, we run two loops. The outer loop will initialize values of a from 1 to 1000. The inner loop will be used for values of b. C’s value can be calculated by the equation **100-a-b**. Secondly, we test if a,b,c are Pythagorean triplets. Lastly, if they are then we calculate the product and return that value as the answer.

**Solution Code**

**Java Solution**

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | class Solution { public static void main (String[] args) throws java.lang.Exception { int LIMIT = 1000; for(int a=1; a<=LIMIT; a++){ for(int b=a+1; b<=LIMIT; b++){ int c = LIMIT-a-b; if(a*a + b*b == c*c){ System.out.println(a*b*c); return; } } } } } |

**Python Solution**

1 2 3 4 5 6 7 8 | LIMIT = 1000 for a in range(1, LIMIT): for b in range(a+1, LIMIT): c = LIMIT - a - b if (a**2 + b**2 == c**2): print(a*b*c) break |

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