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Count number of triplets in an array having sum in the range [a, b]
• Difficulty Level : Medium
• Last Updated : 22 Apr, 2021

Given an array of distinct integers and a range [a, b], the task is to count the number of triplets having a sum in the range [a, b].
Examples:

```Input : arr[] = {8, 3, 5, 2}
range = [7, 11]
Output : 1
There is only one triplet {2, 3, 5}
having sum 10 in range [7, 11].

Input : arr[] = {2, 7, 5, 3, 8, 4, 1, 9}
range = [8, 16]
Output : 36```

A naive approach is to run three loops to consider all the triplets one by one. Find the sum of each triplet and increment the count if the sum lies in a given range [a, b].
Below is the implementation of the above approach:

## C++

 `// C++ program to count triplets with``// sum that lies in given range [a, b].``#include ` `using` `namespace` `std;` `// Function to count triplets``int` `countTriplets(``int` `arr[], ``int` `n, ``int` `a, ``int` `b)``{``    ``// 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] >= a``                    ``&& arr[i] + arr[j] + arr[k] <= b)``                    ``ans++;``        ``}``    ``}` `    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 7, 5, 3, 8, 4, 1, 9 };``    ``int` `n = ``sizeof` `arr / ``sizeof` `arr;``    ``int` `a = 8, b = 16;``    ``cout << countTriplets(arr, n, a, b) << endl;``    ``return` `0;``}`

## Java

 `// Java program to count triplets``// with sum that lies in given``// range [a, b].``import` `java.util.*;` `class` `GFG``{``    ` `// Function to count triplets``public` `static` `int` `countTriplets(``int` `[]arr, ``int` `n,``                                ``int` `a, ``int` `b)``{``    ``// 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] >= a &&``                    ``arr[i] + arr[j] + arr[k] <= b)``                    ``{ans++;}``            ``}``        ``}``    ``}` `    ``return` `ans;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int``[] arr = { ``2``, ``7``, ``5``, ``3``, ``8``, ``4``, ``1``, ``9` `};``    ``int` `n = arr.length;``    ``int` `a = ``8``, b = ``16``;``    ``System.out.println(``""` `+ countTriplets(arr, n,``                                        ``a, b));``}``}` `// This code is contributed``// by Harshit Saini`

## Python3

 `# Python3 program to count``# triplets with sum that``# lies in given range [a, b].` `# Function to count triplets``def` `countTriplets(arr, n, a, b):``    ` `    ``# 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] >``=` `a)``                ``and` `(arr[i] ``+` `arr[j] ``+` `arr[k] <``=` `b)):``                        ``ans ``+``=` `1``                        ` `    ``return` `ans` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``arr ``=` `[ ``2``, ``7``, ``5``, ``3``, ``8``, ``4``, ``1``, ``9` `]``    ``n ``=` `len``(arr)``    ``a ``=` `8``; b ``=` `16``    ``print``(countTriplets(arr, n, a, b))` `# This code is contributed``# by Harshit Saini`

## C#

 `// C# program to count triplets``// with sum that lies in given``// range [a, b].``using` `System;` `class` `GFG``{``    ` `// Function to count triplets``public` `static` `int` `countTriplets(``int` `[]arr, ``int` `n,``                                ``int` `a, ``int` `b)``{``    ``// 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] >= a &&``                    ``arr[i] + arr[j] + arr[k] <= b)``                    ``{ans++;}``            ``}``        ``}``    ``}` `    ``return` `ans;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int``[] arr = {2, 7, 5, 3, 8, 4, 1, 9};``    ``int` `n = arr.Length;``    ``int` `a = 8, b = 16;``    ``Console.WriteLine(``""` `+ countTriplets(arr, n,``                                        ``a, b));``}``}` `// This code is contributed``// by Akanksha Rai(Abby_akku)`

## PHP

 `= ``\$a` `&&``                    ``\$arr``[``\$i``] + ``\$arr``[``\$j``] + ``\$arr``[``\$k``] <= ``\$b``)``                    ``\$ans``++;``        ``}``    ``}` `    ``return` `\$ans``;``}` `// Driver Code``\$arr` `= ``array``( 2, 7, 5, 3, 8, 4, 1, 9 );``\$n` `= sizeof(``\$arr``);``\$a` `= 8; ``\$b` `= 16;``echo` `countTriplets(``\$arr``, ``\$n``, ``\$a``, ``\$b``) . ``"\n"``;` `// This code is contributed``// by Akanksha Rai(Abby_akku)``?>`

## Javascript

 ``
Output:
`36`

Time complexity: O(n3)
An efficient solution is to first find the count of triplets having a sum less than or equal to upper limit b in the range [a, b]. This count of triplets will also include triplets having a sum less than the lower limit a. Subtract the count of triplets having a sum less than a. The final result is the count of triplets having a sum in the range [a, b].
The algorithm is as follows:

• Find count of triplets having a sum less than or equal to b. Let this count be x.
• Find count of triplets having a sum less than a. Let this count be y.
• Final result is x-y.

To find the count of triplets having a sum less than or equal to the given value, refer Count triplets with sum smaller than a given value
Below is the implementation of the above approach:

## C++

 `// C++ program to count triplets with``// sum that lies in given range [a, b].``#include ` `using` `namespace` `std;` `// Function to find count of triplets having``// sum less than or equal to val.``int` `countTripletsLessThan(``int` `arr[], ``int` `n, ``int` `val)``{``    ``// sort the input array.``    ``sort(arr, arr + n);` `    ``// Initialize result``    ``int` `ans = 0;` `    ``int` `j, k;` `    ``// to store sum``    ``int` `sum;` `    ``// Fix the first element``    ``for` `(``int` `i = 0; i < n - 2; i++) {` `        ``// 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) {``            ``sum = arr[i] + arr[j] + arr[k];` `            ``// If sum of current triplet``            ``// is greater, then to reduce it``            ``// decrease k.``            ``if` `(sum > val)``                ``k--;` `            ``// If sum is less than or equal``            ``// to given value, then add``            ``// possible triplets (k-j) to result.``            ``else` `{``                ``ans += (k - j);``                ``j++;``            ``}``        ``}``    ``}` `    ``return` `ans;``}` `// Function to return count of triplets having``// sum in range [a, b].``int` `countTriplets(``int` `arr[], ``int` `n, ``int` `a, ``int` `b)``{` `    ``// to store count of triplets.``    ``int` `res;` `    ``// Find count of triplets having sum less``    ``// than or equal to b and subtract count``    ``// of triplets having sum less than or``    ``// equal to a-1.``    ``res = countTripletsLessThan(arr, n, b) -``        ``countTripletsLessThan(arr, n, a - 1);` `    ``return` `res;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 7, 5, 3, 8, 4, 1, 9 };``    ``int` `n = ``sizeof` `arr / ``sizeof` `arr;``    ``int` `a = 8, b = 16;``    ``cout << countTriplets(arr, n, a, b) << endl;``    ``return` `0;``}`

## Java

 `// Java program to count triplets``// with sum that lies in given``// range [a, b].``import` `java.util.*;` `class` `GFG``{``// Function to find count of``// triplets having sum less``// than or equal to val.``public` `static` `int` `countTripletsLessThan(``int` `[]arr,``                                        ``int` `n, ``int` `val)``{``    ``// sort the input array.``    ``Arrays.sort(arr);` `    ``// Initialize result``    ``int` `ans = ``0``;` `    ``int` `j, k;` `    ``// to store sum``    ``int` `sum;` `    ``// Fix the first element``    ``for` `(``int` `i = ``0``; i < n - ``2``; i++)``    ``{` `        ``// 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)``        ``{``            ``sum = arr[i] + arr[j] + arr[k];` `            ``// If sum of current triplet``            ``// is greater, then to reduce it``            ``// decrease k.``            ``if` `(sum > val)``                ``k--;` `            ``// If sum is less than or``            ``// equal to given value,``            ``// then add possible``            ``// triplets (k-j) to result.``            ``else``            ``{``                ``ans += (k - j);``                ``j++;``            ``}``        ``}``    ``}` `    ``return` `ans;``}` `    ``// Function to return count``    ``// of triplets having sum``    ``// in range [a, b].``    ``public` `static` `int` `countTriplets(``int` `arr[], ``int` `n,``                                    ``int` `a, ``int` `b)``    ``{``    ` `        ``// to store count``        ``// of triplets.``        ``int` `res;``    ` `        ``// Find count of triplets``        ``// having sum less than or``        ``// equal to b and subtract``        ``// count of triplets having``        ``// sum less than or equal``        ``// to a-1.``        ``res = countTripletsLessThan(arr, n, b) -``            ``countTripletsLessThan(arr, n, a - ``1``);``    ` `        ``return` `res;``    ``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int``[] arr = {``2``, ``7``, ``5``, ``3``,``                ``8``, ``4``, ``1``, ``9``};``    ``int` `n = arr.length;``    ``int` `a = ``8``, b = ``16``;``    ``System.out.println(``""` `+ countTriplets(arr, n,``                                        ``a, b));``}``}` `// This code is contributed``// by Harshit Saini`

## Python3

 `# Python program to count``# triplets with sum that``# lies in given range [a, b].` `# Function to find count of``# triplets having sum less``# than or equal to val.``def` `countTripletsLessThan(arr, n, val):` `    ``# sort the input array.``    ``arr.sort()` `    ``# Initialize result``    ``ans ``=` `0` `    ``j ``=` `0``; k ``=` `0` `    ``# to store sum``    ``sum` `=` `0` `    ``# Fix the first element``    ``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 :``            ``sum` `=` `arr[i] ``+` `arr[j] ``+` `arr[k]``            ` `            ``# If sum of current triplet``            ``# is greater, then to reduce it``            ``# decrease k.``            ``if` `sum` `> val:``                ``k``-``=``1` `            ``# If sum is less than or``            ``# equal to given value,``            ``# then add possible``            ``# triplets (k-j) to result.``            ``else` `:``                ``ans ``+``=` `(k ``-` `j)``                ``j ``+``=` `1``    ``return` `ans` `# Function to return``# count of triplets having``# sum in range [a, b].``def` `countTriplets(arr, n, a, b):``    ` `    ``# to store count of triplets.``    ``res ``=` `0` `    ``# Find count of triplets``    ``# having sum less than or``    ``# equal to b and subtract``    ``# count of triplets having``    ``# sum less than or equal to a-1.``    ``res ``=` `(countTripletsLessThan(arr, n, b) ``-``        ``countTripletsLessThan(arr, n, a ``-` `1``))` `    ``return` `res` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``arr ``=` `[ ``2``, ``7``, ``5``, ``3``, ``8``, ``4``, ``1``, ``9` `]``    ``n ``=` `len``(arr)``    ``a ``=` `8``; b ``=` `16``    ``print``(countTriplets(arr, n, a, b))``    ` `# This code is contributed by``# Harshit Saini`

## C#

 `// C# program to count triplets``// with sum that lies in given``// range [a, b].``using` `System;` `class` `GFG``{``// Function to find count of``// triplets having sum less``// than or equal to val.``public` `static` `int` `countTripletsLessThan(``int``[] arr,``                                        ``int` `n, ``int` `val)``{``    ``// sort the input array.``    ``Array.Sort(arr);` `    ``// Initialize result``    ``int` `ans = 0;` `    ``int` `j, k;` `    ``// to store sum``    ``int` `sum;` `    ``// Fix the first element``    ``for` `(``int` `i = 0; i < n - 2; i++)``    ``{` `        ``// 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)``        ``{``            ``sum = arr[i] + arr[j] + arr[k];` `            ``// If sum of current triplet``            ``// is greater, then to reduce it``            ``// decrease k.``            ``if` `(sum > val)``                ``k--;` `            ``// If sum is less than or``            ``// equal to given value,``            ``// then add possible``            ``// triplets (k-j) to result.``            ``else``            ``{``                ``ans += (k - j);``                ``j++;``            ``}``        ``}``    ``}` `    ``return` `ans;``}` `    ``// Function to return count``    ``// of triplets having sum``    ``// in range [a, b].``    ``public` `static` `int` `countTriplets(``int``[] arr, ``int` `n,``                                    ``int` `a, ``int` `b)``    ``{``    ` `        ``// to store count``        ``// of triplets.``        ``int` `res;``    ` `        ``// Find count of triplets``        ``// having sum less than or``        ``// equal to b and subtract``        ``// count of triplets having``        ``// sum less than or equal``        ``// to a-1.``        ``res = countTripletsLessThan(arr, n, b) -``            ``countTripletsLessThan(arr, n, a - 1);``    ` `        ``return` `res;``    ``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int``[] arr = {2, 7, 5, 3,``                ``8, 4, 1, 9};``    ``int` `n = arr.Length;``    ``int` `a = 8, b = 16;``    ``Console.WriteLine(``""` `+ countTriplets(arr, n,``                                        ``a, b));``}``}` `// This code is contributed``// by Akanksha Rai(Abby_akku)`

## PHP

 ` ``\$val``)``                ``\$k``--;` `            ``// If sum is less than or equal``            ``// to given value, then add possible``            ``// triplets (k-j) to result.``            ``else``            ``{``                ``\$ans` `+= (``\$k` `- ``\$j``);``                ``\$j``++;``            ``}``        ``}``    ``}` `    ``return` `\$ans``;``}` `// Function to return count of triplets``// having sum in range [a, b].``function` `countTriplets(``\$arr``, ``\$n``, ``\$a``, ``\$b``)``{` `    ``// to store count of triplets.``    ``\$res``;` `    ``// Find count of triplets having sum less``    ``// than or equal to b and subtract count``    ``// of triplets having sum less than or``    ``// equal to a-1.``    ``\$res` `= countTripletsLessThan(``\$arr``, ``\$n``, ``\$b``) -``           ``countTripletsLessThan(``\$arr``, ``\$n``, ``\$a` `- 1);` `    ``return` `\$res``;``}` `// Driver Code``\$arr` `= ``array``( 2, 7, 5, 3, 8, 4, 1, 9 );``\$n` `= sizeof(``\$arr``);``\$a` `= 8;``\$b` `= 16;``echo` `countTriplets(``\$arr``, ``\$n``, ``\$a``, ``\$b``), ``"\n"``;``    ` `// This code is contributed by Sachin``?>`
Output:
`36`

Time complexity: O(n2
Auxiliary space: O(1)

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