Count triplets with sum smaller than a given value

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 12
               (1, 3, 4), (1, 3, 5), (1, 3, 7) and 
               (1, 4, 5)

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

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)); 
    } 
} 

Python 3

# A Simple Python 3 program to count triplets with sum smaller 
# than a given value 
#include<bits/stdc++.h> 
def countTriplets(arr, n, sum): 
  
    # Initialize result 
    ans = 0
  
    # Fix the first element as A[i] 
    for i in range( 0 ,n-2): 
      
        # Fix the second element as A[j] 
        for j in range( i+1 ,n-1): 
      
            # Now look for the third number 
            for k in range( j+1, n): 
                if (arr[i] + arr[j] + arr[k] < sum): 
                    ans+=1
      
    return ans 
  
# Driver program 
arr = [5, 1, 3, 4, 7] 
n = len(arr) 
sum = 12
print(countTriplets(arr, n, sum)) 
  
#Contributed by Smitha 

C#

// A Simple C# program to count triplets with sum smaller 
// than a given value 
   
using System; 
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()  
    { 
        int sum = 12;  
        Console.Write(countTriplets(arr.Length, sum)); 
    } 
} 

Output:

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 = 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 the 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)); 
    } 
} 

Python3

# Python3 program to count triplets with  
# sum smaller than a given value  
  
  
# Function to count triplets with sum smaller 
# than a given value          
def countTriplets(arr,n,sum): 
      
    # Sort input array 
    arr.sort() 
      
    # Initialize result  
    ans = 0
      
    # Every iteration of loop counts triplet with  
    # first element as arr[i]. 
    for i in range(0,n-2): 
          
        # Initialize other two elements as corner elements  
        # of subarray arr[j+1..k]  
        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 = k-1
              
            # 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 = j+1
      
    return ans 
  
# Driver program  
if __name__=='__main__': 
    arr = [5, 1, 3, 4, 7]  
    n = len(arr)  
    sum = 12
    print(countTriplets(arr, n, sum))  
      
# This code is contributed by  
# Yatin Gupta  

C#

// A Simple C# program to count  
// triplets with sum smaller 
// than a given value 
using System; 
  
class GFG 
{ 
    static int []arr = new int[]{5, 1, 3, 4, 7}; 
      
    static int countTriplets(int n, int sum) 
    { 
        // Sort input array 
        Array.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 Code 
    public static void Main()  
    { 
        int sum = 12;  
        Console.Write(countTriplets(arr.Length, sum)); 
    } 
} 
  
// This code is contributed by Smitha 

Output:

Thanks to Gaurav Ahirwar for suggesting this solution.

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My Personal Notes

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

C++

Java

Python 3

C#

C++

Java

Python3

C#

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