Given a sorted array arr[] consisting of N positive integers and an integer D, the task is to find the number of triplets (i, j, k) such that arr[j] – arr[i] = D and arr[k] – arr[j] = D and 0 ? i < j < k < N.
Examples:
Input: arr[] = {1, 2, 4, 5, 7, 8, 10}, D = 3
Output: 3
Explanation:
Following are the triplets having the difference between the adjacent elements is D(= 3) are:
- {1, 4, 7}
- {4, 7, 10}
- {2, 5, 8}
Therefore, the total count of triplets is 3.
Input: arr[] = {1, 2, 4, 5, 7, 8, 10}, D = 1
Output: 0
Naive Approach: The simplest approach to solve this problem is to generate all the triplets of the given array and count those triplets having the difference between the adjacent elements is D(= 3).
Time Complexity: O(N3)
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by considering every element of the array as the last element of the triplet and check for the previous two elements i.e., (arr[i] – D) and (arr[i] – 2 * D) exists in the array or not. Follow the steps below to solve the problem.
- Initialize a HashMap, say M that stores the frequency of the array elements.
- Initialize a variable, say ans as 0.
- Traverse the given array arr[] and perform the following steps:
- Increment the frequency of arr[i] by 1 in the HashMap M.
- Now, check if the element (arr[i] – D) and (arr[i] – 2 * D) are present in the HashMap or not. If found to be true, then increment the value of ans by freq[arr[i] – D] * freq[arr[i] – 2 * D].
- After completing the above steps, print the value of ans as the resultant count of triplets in the array.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count the number of// triplets having difference// between adjacent elements equal to Dint countTriplets(int D, vector<int>& arr){ // Stores the frequency // of array elements unordered_map<int, int> freq; // Stores the count of // resultant triplets int ans = 0; // Traverse the array for (int i = 0; i < arr.size(); i++) { // Check if arr[i] - D and // arr[i] - 2 * D exists // in the Hashmap or not if (freq.find(arr[i] - D) != freq.end() && freq.find(arr[i] - 2 * D) != freq.end()) { // Update the value of ans ans += freq[arr[i] - D] * freq[arr[i] - 2 * D]; } // Increase the frequency // of the current element freq[arr[i]]++; } // Return the resultant count return ans;}// Driver Codeint main(){ vector<int> arr{ 1, 2, 4, 5, 7, 8, 10 }; int D = 1; cout << countTriplets(D, arr); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to count the number of// triplets having difference// between adjacent elements equal to Dstatic int countTriplets(int D, int []arr){ // Stores the frequency // of array elements HashMap<Integer, Integer> freq = new HashMap<Integer, Integer>(); // Stores the count of // resultant triplets int ans = 0; // Traverse the array for(int i = 0; i < arr.length; i++) { // Check if arr[i] - D and // arr[i] - 2 * D exists // in the Hashmap or not if (freq.containsKey(arr[i] - D) && freq.containsKey(arr[i] - 2 * D)) { // Update the value of ans ans += freq.get(arr[i] - D) * freq.get(arr[i] - 2 * D); } // Increase the frequency // of the current element if (freq.containsKey(arr[i])) { freq.put(arr[i], freq.get(arr[i]) + 1); } else { freq.put(arr[i], 1); } } // Return the resultant count return ans;}// Driver Codepublic static void main(String[] args){ int []arr = { 1, 2, 4, 5, 7, 8, 10 }; int D = 1; System.out.print(countTriplets(D, arr));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approach# Function to count the number of# triplets having difference# between adjacent elements equal to Ddef countTriplets(D, arr): # Stores the frequency # of array elements freq = {} # Stores the count of # resultant triplets ans = 0 # Traverse the array for i in range(len(arr)): # Check if arr[i] - D and # arr[i] - 2 * D exists # in the Hashmap or not if (((arr[i] - D) in freq) and (arr[i] - 2 * D) in freq): # Update the value of ans ans += (freq[arr[i] - D] * freq[arr[i] - 2 * D]) # Increase the frequency # of the current element freq[arr[i]] = freq.get(arr[i], 0) + 1 # Return the resultant count return ans# Driver Codeif __name__ == '__main__': arr = [ 1, 2, 4, 5, 7, 8, 10 ] D = 1 print (countTriplets(D, arr))# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to count the number of// triplets having difference// between adjacent elements equal to Dstatic int countTriplets(int D, int[] arr){ // Stores the frequency // of array elements Dictionary<int, int> freq = new Dictionary<int, int>(); // Stores the count of // resultant triplets int ans = 0; // Traverse the array for(int i = 0; i < arr.Length; i++) { // Check if arr[i] - D and // arr[i] - 2 * D exists // in the Hashmap or not if (freq.ContainsKey(arr[i] - D) && freq.ContainsKey(arr[i] - 2 * D)) { // Update the value of ans ans += freq[arr[i] - D] * freq[arr[i] - 2 * D]; } // Increase the frequency // of the current element if (!freq.ContainsKey(arr[i])) freq[arr[i]] = 0; freq[arr[i]]++; } // Return the resultant count return ans;}// Driver Codepublic static void Main(){ int[] arr = { 1, 2, 4, 5, 7, 8, 10 }; int D = 1; Console.WriteLine(countTriplets(D, arr));}}// This code is contributed by ukasp |
Javascript
<script>// javascript program for the above approach// Function to count the number of// triplets having difference// between adjacent elements equal to Dfunction countTriplets(D, arr){ // Stores the frequency // of array elements var freq = new Map(); // Stores the count of // resultant triplets var ans = 0; var i; // Traverse the array for (i = 0; i < arr.length; i++) { // Check if arr[i] - D and // arr[i] - 2 * D exists // in the Hashmap or not if (freq.has(arr[i] - D) && freq.has(arr[i] - 2 * D)) { // Update the value of ans ans += freq.get(arr[i] - D) * freq.get(arr[i] - 2 * D); } // Increase the frequency // of the current element freq.set(arr[i],freq.get(arr[i])+1); } // Return the resultant count return ans;}// Driver Code var arr = [1, 2, 4, 5, 7, 8, 10]; var D = 1; document.write(countTriplets(D, arr));</script> |
Output:
0
Time Complexity: O(N)
Auxiliary Space: O(N)
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