Given an array arr[] of size N consisting of only 0s, 1s and 2s, the task is to find the count of triplets of indices (i, j, k) containing distinct array elements such that i < j < k and the array elements are not equidistant, i.e, (j – i )!= (k – j).
Examples:
Input: arr[] = { 0, 1, 2, 1 }
Output:1
Explanation:
Only triplet (0, 2, 3) contains distinct array elements and (2 – 0) != (3 – 2).
Therefore, the required output is 1.Input: arr[] = { 0, 1, 2 }
Output: 0
Explanation:
No triplet exists that satisfy the condition.
Therefore, the required output is 0.
Approach: The idea is to store the indices of array elements 0s, 1s and 2s in three separate arrays, then find the count triplets that satisfy the given conditions. Follow the steps below to solve the problem:
- Initialize two arrays, say zero_i[] and one_i[], to store the indices of 0s and 1s from the given array respectively.
- Initialize a map, say mp, to store the indices of 2s from the given array.
- Find the total count of all possible triplets by multiplying the size of zero_i[], one_i[] and mp.
- Now, subtract all those triplets which violate the given conditions.
- For finding such triplets, traverse both the arrays zero_i[] and one_i[] and try to find a third index in the Map that violates the condition.
- To find the third index which violates the conditions, following three cases arise:
- The third index is equidistant from both the indices and is present in between them.
- The third index is equidistant from both the indices and is present on the left side of the first index.
- The third index is equidistant from both the indices and is present on the right side of the second index.
- Remove all such triplets from the count of the total number of triplets.
- Finally, print the total count of triplets obtained.
Below is the implementation of the above approach:
C++14
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find the total count of// triplets (i, j, k) such that i < j < k// and (j - i) != (k - j)int countTriplets(int* arr, int N){ // Stores indices of 0s vector<int> zero_i; // Stores indices of 1s vector<int> one_i; // Stores indices of 2s unordered_map<int, int> mp; // Traverse the array for (int i = 0; i < N; i++) { // If current array element // is 0 if (arr[i] == 0) zero_i.push_back(i + 1); // If current array element is 1 else if (arr[i] == 1) one_i.push_back(i + 1); // If current array element // is 2 else mp[i + 1] = 1; } // Total count of triplets int total = zero_i.size() * one_i.size() * mp.size(); // Traverse the array zero_i[] for (int i = 0; i < zero_i.size(); i++) { // Traverse the array one_i[] for (int j = 0; j < one_i.size(); j++) { // Stores index of 0s int p = zero_i[i]; // Stores index of 1s int q = one_i[j]; // Stores third element of // triplets that does not // satisfy the condition int r = 2 * p - q; // If r present // in the map if (mp[r] > 0) total--; // Update r r = 2 * q - p; // If r present // in the map if (mp[r] > 0) total--; // Update r r = (p + q) / 2; // If r present in the map // and equidistant if (mp[r] > 0 && abs(r - p) == abs(r - q)) total--; } } // Print the obtained count cout << total;}// Driver Codeint main(){ int arr[] = { 0, 1, 2, 1 }; int N = sizeof(arr) / sizeof(arr[0]); countTriplets(arr, N); return 0;} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{// Function to find the total count of// triplets (i, j, k) such that i < j < k// and (j - i) != (k - j)static void countTriplets(int []arr, int N){ // Stores indices of 0s Vector<Integer> zero_i = new Vector<Integer>(); // Stores indices of 1s Vector<Integer> one_i = new Vector<Integer>(); // Stores indices of 2s HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>(); // Traverse the array for(int i = 0; i < N; i++) { // If current array element // is 0 if (arr[i] == 0) zero_i.add(i + 1); // If current array element is 1 else if (arr[i] == 1) one_i.add(i + 1); // If current array element // is 2 else mp.put(i + 1, 1); } // Total count of triplets int total = zero_i.size() * one_i.size() * mp.size(); // Traverse the array zero_i[] for(int i = 0; i < zero_i.size(); i++) { // Traverse the array one_i[] for(int j = 0; j < one_i.size(); j++) { // Stores index of 0s int p = zero_i.get(i); // Stores index of 1s int q = one_i.get(j); // Stores third element of // triplets that does not // satisfy the condition int r = 2 * p - q; // If r present // in the map if (mp.containsKey(r) && mp.get(r) > 0) total--; // Update r r = 2 * q - p; // If r present // in the map if (mp.containsKey(r) && mp.get(r) > 0) total--; // Update r r = (p + q) / 2; // If r present in the map // and equidistant if (mp.containsKey(r) && mp.get(r) > 0 && Math.abs(r - p) == Math.abs(r - q)) total--; } } // Print the obtained count System.out.print(total);}// Driver Codepublic static void main(String[] args){ int arr[] = { 0, 1, 2, 1 }; int N = arr.length; countTriplets(arr, N);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to implement# the above approach# Function to find the total count of# triplets (i, j, k) such that i < j < k# and (j - i) != (k - j)def countTriplets(arr, N): # Stores indices of 0s zero_i = [] # Stores indices of 1s one_i = [] # Stores indices of 2s mp = {} # Traverse the array for i in range(N): # If current array element # is 0 if (arr[i] == 0): zero_i.append(i + 1) # If current array element is 1 elif (arr[i] == 1): one_i.append(i + 1) # If current array element # is 2 else: mp[i + 1] = 1 # Total count of triplets total = len(zero_i) * len(one_i) * len(mp) # Traverse the array zero_i[] for i in range(len(zero_i)): # Traverse the array one_i[] for j in range(len(one_i)): # Stores index of 0s p = zero_i[i] # Stores index of 1s q = one_i[j] # Stores third element of # triplets that does not # satisfy the condition r = 2 * p - q # If r present # in the map if (r in mp): total -= 1 # Update r r = 2 * q - p # If r present # in the map if (r in mp): total -= 1 # Update r r = (p + q) // 2 # If r present in the map # and equidistant if ((r in mp) and abs(r - p) == abs(r - q)): total -= 1 # Print the obtained count print (total)# Driver Codeif __name__ == '__main__': arr = [0, 1, 2, 1] N = len(arr) countTriplets(arr, N) # This code is contributed by mohit kumar 29 |
C#
// C# program to implement// the above approachusing System;using System.Collections.Generic;class GFG{// Function to find the total count of// triplets (i, j, k) such that i < j < k// and (j - i) != (k - j)static void countTriplets(int []arr, int N){ // Stores indices of 0s List<int> zero_i = new List<int>(); // Stores indices of 1s List<int> one_i = new List<int>(); // Stores indices of 2s Dictionary<int, int> mp = new Dictionary<int, int>(); // Traverse the array for(int i = 0; i < N; i++) { // If current array element // is 0 if (arr[i] == 0) zero_i.Add(i + 1); // If current array element is 1 else if (arr[i] == 1) one_i.Add(i + 1); // If current array element // is 2 else mp.Add(i + 1, 1); } // Total count of triplets int total = zero_i.Count * one_i.Count * mp.Count; // Traverse the array zero_i[] for(int i = 0; i < zero_i.Count; i++) { // Traverse the array one_i[] for(int j = 0; j < one_i.Count; j++) { // Stores index of 0s int p = zero_i[i]; // Stores index of 1s int q = one_i[j]; // Stores third element of // triplets that does not // satisfy the condition int r = 2 * p - q; // If r present // in the map if (mp.ContainsKey(r) && mp[r] > 0) total--; // Update r r = 2 * q - p; // If r present // in the map if (mp.ContainsKey(r) && mp[r] > 0) total--; // Update r r = (p + q) / 2; // If r present in the map // and equidistant if (mp.ContainsKey(r) && mp[r] > 0 && Math.Abs(r - p) == Math.Abs(r - q)) total--; } } // Print the obtained count Console.Write(total);}// Driver Codepublic static void Main(String[] args){ int []arr = { 0, 1, 2, 1 }; int N = arr.Length; countTriplets(arr, N);}} // This code contributed by shikhasingrajput |
Javascript
<script>// JavaScript program to implement// the above approach// Function to find the total count of// triplets (i, j, k) such that i < j < k// and (j - i) != (k - j)function countTriplets(arr, N){ // Stores indices of 0s var zero_i = []; // Stores indices of 1s var one_i = []; // Stores indices of 2s var mp = new Map(); // Traverse the array for (var i = 0; i < N; i++) { // If current array element // is 0 if (arr[i] == 0) zero_i.push(i + 1); // If current array element is 1 else if (arr[i] == 1) one_i.push(i + 1); // If current array element // is 2 else mp.set(i + 1, 1); } // Total count of triplets var total = zero_i.length * one_i.length * mp.size; // Traverse the array zero_i[] for (var i = 0; i < zero_i.length; i++) { // Traverse the array one_i[] for (var j = 0; j < one_i.length; j++) { // Stores index of 0s var p = zero_i[i]; // Stores index of 1s var q = one_i[j]; // Stores third element of // triplets that does not // satisfy the condition var r = 2 * p - q; // If r present // in the map if (mp.has(r)) total--; // Update r r = 2 * q - p; // If r present // in the map if (mp.has(r)) total--; // Update r r = (p + q) / 2; // If r present in the map // and equidistant if (mp.has(r) && Math.abs(r - p) == Math.abs(r - q)) total--; } } // Print the obtained count document.write( total);}// Driver Codevar arr = [0, 1, 2, 1];var N = arr.length;countTriplets(arr, N);</script> |
Output:
1
Time Complexity: O(N2)
Auxiliary Space: O(N)
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