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)
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; } |
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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)); } } |
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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 = 12print(countTriplets(arr, n, sum)) #Contributed by Smitha |
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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)); } } |
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Output:
4
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; } |
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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)); } } |
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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 |
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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 |
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Output:
4
Thanks to Gaurav Ahirwar for suggesting this solution.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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