# Wipro NTH Coding Sample Question- 4

## Wipro NTH Coding Question – 4

Ques: A Pythagorean triplet is a set of three integers a, b and c such that a2 + b2 = c2. Given a limit, generate all Pythagorean Triples with values smaller than given limit.

```Input : limit = 20
Output : 3 4 5
8 6 10
5 12 13
15 8 17
12 16 20```

Simple Solution is to generate these triplets smaller than given limit using three nested loop. For every triplet, check if Pythagorean condition is true, if true, then print the triplet. Time complexity of this solution is O(limit3) where ‘limit’ is given limit.

An Efficient Solution can print all triplets in O(k) time where k is number of triplets printed. The idea is to use square sum relation of Pythagorean triplet, i.e., addition of squares of a and b is equal to square of c, we can write these number in terms of m and n such that,

```       a = m2 - n2
b = 2 * m * n
c  = m2 + n2
because,
a2 = m4 + n4 – 2 * m2 * n2
b2 = 4 * m2 * n2
c2 = m4 + n4 + 2* m2 * n2
```

We can see that a2 + b2 = c2, so instead of iterating for a, b and c we can iterate for m and n and can generate these triplets

Run
```#include<stdio.h>
#include<math.h>
void pythagoreanTriplets (int limit)
{
// triplet:  a^2 + b^2 = c^2
int a, b, c = 0;
int m = 2;
// Limiting c would limit all a, b and c
while (c < limit)
{
// now loop on j from 1 to i-1
for (int n = 1; n < m; ++n)
{
// Evaluate and print triplets using
// the relation between a, b and c
a = m * m - n * n;
b = 2 * m * n;
c = m * m + n * n;
if (c > limit)
break;
printf ("%d %d %d \n", a, b, c);
}
m++;
}
}
int main ()
{
int limit = 20;
pythagoreanTriplets (limit);
return 0;
}
```
Run
```#include <bits/stdc++.h>
using namespace std;
void pythagoreanTriplets(int limit)
{
int a, b, c = 0;
int m = 2;

while (c < limit)
{
for (int n = 1; n < m; ++n)
{
a = m * m - n * n;
b = 2 * m * n;
c = m * m + n * n;
if (c > limit)
break;
cout << a << " " << b << " " << c << "\n";
}
m++;
}
}
int main()
{
int limit = 20;
pythagoreanTriplets(limit);
return 0;
}
```
Run
```import java.util.*;
public class Main {
public static void pythagoreanTriplets(int limit) {
int a, b, c = 0;
int m = 2;

while (c < limit) {
for (int n = 1; n < m; ++n) {
a = m * m - n * n;
b = 2 * m * n;
c = m * m + n * n;
if (c > limit)
break;
System.out.println(a + " " + b + " " + c);
}
m++;
}
}

public static void main(String[] args) {
int limit = 20;
pythagoreanTriplets(limit);
}
}
```
Run
```def pythagoreanTriplets(limit):
a, b, c = 0, 0, 0
m = 2

while c < limit:
for n in range(1, m):
a = m * m - n * n
b = 2 * m * n
c = m * m + n * n
if c > limit:
break
print(a, b, c)

m += 1

limit = 20
pythagoreanTriplets(limit)
```

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