Javascript Program to Count triplets with sum smaller than a given value

Last Updated : 16 Sep, 2024

Given an array of distinct integers and a sum value. Find the count of triplets with a sum smaller than the given sum value. The 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)

Native Approach:

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

Example:

JavaScript

// A Simple Javascript program to count triplets with sum smaller // than a given value let arr = [5, 1, 3, 4, 7];  function countTriplets(n, sum) {      // Initialize result     let ans = 0;      // Fix the first element as A[i]     for (let i = 0; i < n - 2; i++) {         // Fix the second element as A[j]         for (let j = i + 1; j < n - 1; j++) {             // Now look for the third number             for (let k = j + 1; k < n; k++)                 if (arr[i] + arr[j] + arr[k] < sum)                     ans++;         }     }     return ans; }  // Driver method to test the above function let sum = 12; console.log(countTriplets(arr.length, sum));

Output

Complexity Analysis:

Time Complexity: O(n³)
Auxiliary Space: O(1)

Efficient Solution:

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 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 the above idea.

JavaScript

// A Simple Javascript program to count triplets with sum smaller // than a given value let arr = [5, 1, 3, 4, 7];  function countTriplets(n, sum) {      // Sort input array     arr.sort(function (a, b) { return b - a });      // Initialize result     let ans = 0;      // Every iteration of loop counts triplet with     // first element as arr[i].     for (let i = 0; i < n - 2; i++) {          // Initialize other two elements as corner elements         // of subarray arr[j+1..k]         let 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 let sum = 12; console.log(countTriplets(arr.length, sum));

Output

Complexity Analysis:

Time Complexity: O(n²)
Auxiliary Space: O(1)

Please refer complete article on Count triplets with sum smaller than a given value for more details!

Javascript Program to Find a triplet that sum to a given value

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Javascript Program to Count triplets with sum smaller than a given value

Native Approach:

Complexity Analysis:

Efficient Solution:

Complexity Analysis:

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