Count triplets with sum smaller than a given value

2.7

Given an array of distinct integers and a sum value. Find count of triplets with sum smaller than given sum value. Expected Time Complexity is O(n2).

Examples:

Input : arr[] = {-2, 0, 1, 3}
        sum = 2.
Output : 2
Explanation :  Below are triplets with sum less than 2
               (-2, 0, 1) and (-2, 0, 3) 

Input : arr[] = {5, 1, 3, 4, 7}
        sum = 12.
Output : 4
Explanation :  Below are triplets with sum less than 4
               (1, 3, 4), (1, 3, 5), (1, 3, 7) and 
               (1, 4, 5)

A Simple Solution is to run three loops to consider all triplets one by one. For every triplet, compare the sums and increment count if triplet sum is smaller than given sum.

C++

// A Simple C++ program to count triplets with sum smaller
// than a given value
#include<bits/stdc++.h>
using namespace std;

int countTriplets(int arr[], int n, int sum)
{
    // Initialize result
    int ans = 0;

    // Fix the first element as A[i]
    for (int i = 0; i < n-2; i++)
    {
       // Fix the second element as A[j]
       for (int j = i+1; j < n-1; j++)
       {
           // Now look for the third number
           for (int k = j+1; k < n; k++)
               if (arr[i] + arr[j] + arr[k] < sum)
                   ans++;
       }
    }

    return ans;
}

// Driver program
int main()
{
    int arr[] = {5, 1, 3, 4, 7};
    int n = sizeof arr / sizeof arr[0];
    int sum = 12;
    cout << countTriplets(arr, n, sum) << endl;
    return 0;
}

Java

// A Simple Java program to count triplets with sum smaller
// than a given value

class Test
{
	static int arr[] = new int[]{5, 1, 3, 4, 7};
	
	static int countTriplets(int n, int sum)
	{
	    // Initialize result
	    int ans = 0;
	 
	    // Fix the first element as A[i]
	    for (int i = 0; i < n-2; i++)
	    {
	       // Fix the second element as A[j]
	       for (int j = i+1; j < n-1; j++)
	       {
	           // Now look for the third number
	           for (int k = j+1; k < n; k++)
	               if (arr[i] + arr[j] + arr[k] < sum)
	                   ans++;
	       }
	    }
	 
	    return ans;
	}
	
	// Driver method to test the above function
	public static void main(String[] args) 
	{
		int sum = 12; 
		System.out.println(countTriplets(arr.length, sum));
    }
}


Output:
4

Time complexity of above solution is O(n3). An Efficient Solution can count triplets in O(n2) by sorting the array first, and then using method 1 of this post in a loop.

1) Sort the input array in increasing order.
2) Initialize result as 0.
3) Run a loop from i = 0 to n-2.  An iteration of this loop finds all
   triplets with arr[i] as first element.
     a) Initialize other two elements as corner elements of subarray
        arr[i+1..n-1], i.e., j = i+1 and k = n-1
     b) Move j and k toward each other until they meet, i.e., while (j < k)
            (i) if (arr[i] + arr[j] + arr[k] >= sum), then do k--

            // Else for current i and j, there can (k-j) possible third elements
            // that satisfy the constraint.
            (ii) Else Do ans += (k - j) followed by j++ 

Below is C++ and Java implementation of above idea.

C++

// C++ program to count triplets with sum smaller than a given value
#include<bits/stdc++.h>
using namespace std;

int countTriplets(int arr[], int n, int sum)
{
    // Sort input array
    sort(arr, arr+n);

    // Initialize result
    int ans = 0;

    // Every iteration of loop counts triplet with
    // first element as arr[i].
    for (int i = 0; i < n - 2; i++)
    {
        // Initialize other two elements as corner elements
        // of subarray arr[j+1..k]
        int j = i + 1, k = n - 1;

        // Use Meet in the Middle concept
        while (j < k)
        {
            // If sum of current triplet is more or equal,
            // move right corner to look for smaller values
            if (arr[i] + arr[j] + arr[k] >= sum)
                k--;

            // Else move left corner
            else
            {
                // This is important. For current i and j, there
                // can be total k-j third elements.
                ans += (k - j);
                j++;
            }
        }
    }
    return ans;
}

// Driver program
int main()
{
    int arr[] = {5, 1, 3, 4, 7};
    int n = sizeof arr / sizeof arr[0];
    int sum = 12;
    cout << countTriplets(arr, n, sum) << endl;
    return 0;
}

Java

// A Simple Java program to count triplets with sum smaller
// than a given value

import java.util.Arrays;

class Test
{
	static int arr[] = new int[]{5, 1, 3, 4, 7};
	
	static int countTriplets(int n, int sum)
	{
	    // Sort input array
	    Arrays.sort(arr);
	 
	    // Initialize result
	    int ans = 0;
	 
	    // Every iteration of loop counts triplet with
	    // first element as arr[i].
	    for (int i = 0; i < n - 2; i++)
	    {
	        // Initialize other two elements as corner elements
	        // of subarray arr[j+1..k]
	        int j = i + 1, k = n - 1;
	 
	        // Use Meet in the Middle concept
	        while (j < k)
	        {
	            // If sum of current triplet is more or equal,
	            // move right corner to look for smaller values
	            if (arr[i] + arr[j] + arr[k] >= sum)
	                k--;
	 
	            // Else move left corner
	            else
	            {
	                // This is important. For current i and j, there
	                // can be total k-j third elements.
	                ans += (k - j);
	                j++;
	            }
	        }
	    }
	    return ans;
	}
	
	// Driver method to test the above function
	public static void main(String[] args) 
	{
		int sum = 12; 
		System.out.println(countTriplets(arr.length, sum));
    }
}


Output:
4

Thanks to Gaurav Ahirwar for suggesting this solution.

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