Given an array of distinct integers and a sum value. Find count of triplets with sum smaller than 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)
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 the triplet sum is smaller than the given sum.
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
The time complexity of the 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 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.
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
Please refer complete article on Count triplets with sum smaller than a given value for more details!
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