# Find a triplet that sum to a given value

Given an array and a value, find if there is a triplet in array whose sum is equal to the given value. If there is such a triplet present in array, then print the triplet and return true. Else return false. For example, if the given array is {12, 3, 4, 1, 6, 9} and given sum is 24, then there is a triplet (12, 3 and 9) present in array whose sum is 24.

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Method 1 (Naive)
A simple method is to generate all possible triplets and compare the sum of every triplet with the given value. The following code implements this simple method using three nested loops.

## C

```# include <stdio.h>

// returns true if there is triplet with sum equal
// to 'sum' present in A[]. Also, prints the triplet
bool find3Numbers(int A[], int arr_size, int sum)
{
int l, r;

// Fix the first element as A[i]
for (int i = 0; i < arr_size-2; i++)
{
// Fix the second element as A[j]
for (int j = i+1; j < arr_size-1; j++)
{
// Now look for the third number
for (int k = j+1; k < arr_size; k++)
{
if (A[i] + A[j] + A[k] == sum)
{
printf("Triplet is %d, %d, %d", A[i], A[j], A[k]);
return true;
}
}
}
}

// If we reach here, then no triplet was found
return false;
}

/* Driver program to test above function */
int main()
{
int A[] = {1, 4, 45, 6, 10, 8};
int sum = 22;
int arr_size = sizeof(A)/sizeof(A[0]);

find3Numbers(A, arr_size, sum);

return 0;
}
```

## Java

```class FindTriplet
{
// returns true if there is triplet with sum equal
// to 'sum' present in A[]. Also, prints the triplet
boolean find3Numbers(int A[], int arr_size, int sum)
{
int l, r;

// Fix the first element as A[i]
for (int i = 0; i < arr_size - 2; i++)
{
// Fix the second element as A[j]
for (int j = i + 1; j < arr_size - 1; j++)
{
// Now look for the third number
for (int k = j + 1; k < arr_size; k++)
{
if (A[i] + A[j] + A[k] == sum)
{
System.out.print("Triplet is " + A[i] + " ," + A[j]
+ " ," + A[k]);
return true;
}
}
}
}

// If we reach here, then no triplet was found
return false;
}

// Driver program to test above functions
public static void main(String[] args)
{
FindTriplet triplet = new FindTriplet();
int A[] = {1, 4, 45, 6, 10, 8};
int sum = 22;
int arr_size = A.length;

triplet.find3Numbers(A, arr_size, sum);
}
}

```

## Python3

```# Python3 program to find a triplet
# that sum to a given value

# returns true if there is triplet with
# sum equal to 'sum' present in A[].
# Also, prints the triplet
def find3Numbers(A, arr_size,sum):

# Fix the first element as A[i]
for i in range( 0,arr_size-2):

# Fix the second element as A[j]
for j in range(i+1, arr_size-1):

# Now look for the third number
for k in range(j + 1, arr_size):
if A[i] + A[j] + A[k] == sum:
print("Triplet is",A[i],
",",A[j],",",A[k])
return True

# If we reach here, then no
# triplet was found
return False

# Driver program to test above function
A = [1, 4, 45, 6, 10, 8]
sum = 22
arr_size = len(A)
find3Numbers(A, arr_size, sum)

# This code is contributed by Smitha Dinesh Semwal
```

Output:
```Triplet is 4, 10, 8
```

Time Complexity: O(n^3)

Method 2 (Use Sorting)
Time complexity of the method 1 is O(n^3). The complexity can be reduced to O(n^2) by sorting the array first, and then using method 1 of this post in a loop.
1) Sort the input array.
2) Fix the first element as A[i] where i is from 0 to array size – 2. After fixing the first element of triplet, find the other two elements using method 1 of this post.

## C++

```// C++ program to find a triplet
# include <bits/stdc++.h>
using namespace std;

// returns true if there is triplet with sum equal
// to 'sum' present in A[]. Also, prints the triplet
bool find3Numbers(int A[], int arr_size, int sum)
{
int l, r;

/* Sort the elements */
sort(A, A+arr_size);

/* Now fix the first element one by one and find the
other two elements */
for (int i=0; i<arr_size-2; i++)
{

// To find the other two elements, start two index
// variables from two corners of the array and move
// them toward each other
l = i + 1; // index of the first element in the
// remaining elements
r = arr_size-1; // index of the last element
while (l < r)
{
if( A[i] + A[l] + A[r] == sum)
{
printf("Triplet is %d, %d, %d", A[i],
A[l], A[r]);
return true;
}
else if (A[i] + A[l] + A[r] < sum)
l++;
else // A[i] + A[l] + A[r] > sum
r--;
}
}

// If we reach here, then no triplet was found
return false;
}

/* Driver program to test above function */
int main()
{
int A[] = {1, 4, 45, 6, 10, 8};
int sum = 22;
int arr_size = sizeof(A)/sizeof(A[0]);

find3Numbers(A, arr_size, sum);

return 0;
}
```

## Java

```class FindTriplet
{

// returns true if there is triplet with sum equal
// to 'sum' present in A[]. Also, prints the triplet
boolean find3Numbers(int A[], int arr_size, int sum)
{
int l, r;

/* Sort the elements */
quickSort(A, 0, arr_size - 1);

/* Now fix the first element one by one and find the
other two elements */
for (int i = 0; i < arr_size - 2; i++)
{
// To find the other two elements, start two index variables
// from two corners of the array and move them toward each
// other
l = i + 1; // index of the first element in the remaining elements
r = arr_size - 1; // index of the last element
while (l < r)
{
if (A[i] + A[l] + A[r] == sum)
{
System.out.print("Triplet is " + A[i] + " ," + A[l]
+ " ," + A[r]);
return true;
}
else if (A[i] + A[l] + A[r] < sum)
l++;

else // A[i] + A[l] + A[r] > sum
r--;
}
}

// If we reach here, then no triplet was found
return false;
}

int partition(int A[], int si, int ei)
{
int x = A[ei];
int i = (si - 1);
int j;

for (j = si; j <= ei - 1; j++)
{
if (A[j] <= x)
{
i++;
int temp = A[i];
A[i] = A[j];
A[j] = temp;
}
}
int temp = A[i + 1];
A[i + 1] = A[ei];
A[ei] = temp;
return (i + 1);
}

/* Implementation of Quick Sort
A[] --> Array to be sorted
si  --> Starting index
ei  --> Ending index
*/
void quickSort(int A[], int si, int ei)
{
int pi;

/* Partitioning index */
if (si < ei)
{
pi = partition(A, si, ei);
quickSort(A, si, pi - 1);
quickSort(A, pi + 1, ei);
}
}

// Driver program to test above functions
public static void main(String[] args)
{
FindTriplet triplet = new FindTriplet();
int A[] = {1, 4, 45, 6, 10, 8};
int sum = 22;
int arr_size = A.length;

triplet.find3Numbers(A, arr_size, sum);
}
}

```

Output:
```Triplet is 4, 8, 10
```

Time Complexity: O(n^2)

How to print all triplets with given sum?
Please refer Find all triplets with zero sum

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