Find a triplet that sum to a given value

2.6

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.

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);
    }
}


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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