Count number of triplets in an array having sum in the range [a, b]

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
```

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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[0]; ` `    ``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) ` `?> `

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 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[0]; ` `    ``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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A Programmer and A Machine learning Enthusiast

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