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Count of Triplets With Sum Less than Given Value


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We have given an array containing N number of elements. In the given array, Count the number of triplets with a sum less than the given value.

Example

Input

a[] = {1, 2, 3, 4, 5, 6, 7, 8}
Sum = 10

Output

7
Possible triplets are : (1,2,3), (1,2,4), (1,2,5), (1,2,6), (1,3,4), (1,3,5), (2,3,4)

Input

a[]  = {3, 7, 9, 1, 2, 5, 11, 4}
Sum = 10

Output

6
Possible triplets are : (3,1,2), (3,1,5), (3,1,4), (3,2,4), (1,2,5), (1,2,4)

Approach 1

Algorithm

1. Run 3 nested loops, selecting all triplets possible.

2. Initialize result = 0.

3. If the Sum of elements is less than Sum, increment count.

4. Print results as a count of triplets satisfying the condition.

C++ Program for Count of Triplets With Sum Less than Given Value

#include <bits/stdc++.h>
using namespace std;
 
//Main function
int main()
{
    int array[] = {3, 7, 9, 1, 2, 5, 11, 4};
    int N = sizeof(array)/sizeof(array[0]);
    int sum = 10;
    int result = 0;
    //First for loop for selecting first element
    //Second loop for second element
    //Third loop for third element.
    //check for triplets satisfing the condition, and increment result
    for (int i = 0; i < N-2; i++)
    {
       for (int j = i+1; j < N-1; j++)
       {
           for (int k = j+1; k < N; k++)
           {
               if(array[i]+array[j]+array[k] < sum)
                   {
                      result++;
                   }
           }
       }
    }
    cout<<"Number of Triplets found = ";
    cout<<result;
    return 0;
}
Number of Triplets found = 6

Approach 2

Algorithm

1. Sort the given array, initialize result = 0

2. Iterate from i to N -2 (N is the size of the array), and take array[i] and the first element of the triplet.

3. Initialize the other two elements as corner elements. array[i+1] and array[N-1]

4. move j and k until they meet to do,

  1. If the sum is greater than the given sum, then move the pointer of the last element. (array[N-1]).
  2. Else if the sum is less than the given sum, then there can k -j possible tird elements satisfying, so add (k-j) to result. And move array[i+1] pointer.

Step 5: Print the result after ending the loop.

C++ Program for Count of Triplets With Sum Less than Given Value

#include <bits/stdc++.h>
using namespace std;
 
// Main function
int main()
{
    int array[] ={1, 2,3,5,6, 3, 2, 5, 7};//input array
    int N = sizeof(array)/sizeof(array[0]);//size of the input array(N)
    int Sum = 9;//input Sum
    int result = 0;//initilizing count of triplets
    sort(array,array+N);//sorting the input array
    //selecting first element
    for(int i = 0; i < N - 2; i++)
    {
        int j=i+1,k=N-1;//second and third elements as last two elements
        while(j<k)
        {
            // Sum of triplet is greater than or equalto given sum, move to find smaller values
            if(array[i] + array[j] + array[k] >= Sum)
            {
                k--;
            }
            // Sum of triplet is less than given sum
            else
            {
                //for current i and j there can be k-j elements, move j pointer
                result += (k - j);
                j++;
            }
        }
    }
    cout<<"Number of Triplets found = ";
    cout<<result;
}
Number of Triplets found = 14

Complexity Analysis for Count of Triplets With Sum Less than Given Value

Time Complexity

O(n*n) where n is the number of elements present in the array. Here we fix one element of the triplet and then using two pointer methods that run O(N) in the worst case for one element.

READ  Construction of Longest Increasing Subsequence (N log N)

Space Complexity

O(1) because we don’t use any auxiliary space here.

References

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