Given an array, count number of pairs that can be formed from all possible contiguous sub-arrays containing distinct numbers. The array contains positive numbers between 0 to n-1 where n is the size of the array.
Examples:
Input: [1, 4, 2, 4, 3, 2] Output: 8 The subarrays with distinct elements are [1, 4, 2], [2, 4, 3] and [4, 3, 2]. There are 8 pairs that can be formed from above array. (1, 4), (1, 2), (4, 2), (2, 4), (2, 3), (4, 3), (4, 2), (3, 2) Input: [1, 2, 2, 3] Output: 2 There are 2 pairs that can be formed from above array. (1, 2), (2, 3)
The idea is to use Sliding Window for the given array. Let us use a window covering from index left to index right and an Boolean array visited to mark elements in current window. The window invariant is that all elements inside the window are distinct. We keep on expanding the window to the right and if a duplicate is found, we shrink the window from left till all elements are distinct again. We update the count of pairs in current window along the way. An observation showed that in an expanding window, number of pairs can be incremented by value equal to window size – 1.
For example,
Expanding Window Count
[1] Count = 0
[1, 2] Count += 1 pair
i.e. (1, 2)
[1, 2, 3] Count += 2 pairs
i.e. (1, 3) and (2, 3)
[1, 2, 3, 4] Count += 3 pairs
i.e. (1, 4), (2, 4)
and (3, 4)
So, total Count for above window of size 4 containing distinct elements is 6. Nothing need to be done when the window is shrinking.
Below is the implementation of the
C++
// C++ program to count number of distinct pairs// that can be formed from all possible contiguous// sub-arrays containing distinct numbers#include <bits/stdc++.h>using namespace std;int countPairs(int arr[], int n){ // initialize number of pairs to zero int count = 0; //Left and right indexes of current window int right = 0, left = 0; // Boolean array visited to mark elements in // current window. Initialized as false vector<bool> visited(n, false); // While right boundary of current window // doesn't cross right end while (right < n) { // If current window contains all distinct // elements, widen the window toward right while (right < n && !visited[arr[right]]) { count += (right - left); visited[arr[right]] = true; right++; } // If duplicate is found in current window, // then reduce the window from left while (left < right && (right != n && visited[arr[right]])) { visited[arr[left]] = false; left++; } } return count;}// Driver codeint main(){ int arr[] = {1, 4, 2, 4, 3, 2}; int n = sizeof arr / sizeof arr[0]; cout << countPairs(arr, n); return 0;} |
Java
// Java program to count number of distinct pairs // that can be formed from all possible contiguous // sub-arrays containing distinct numbers class GFG{static int countPairs(int arr[], int n) { // initialize number of pairs to zero int count = 0; //Left and right indexes of current window int right = 0, left = 0; // Boolean array visited to mark elements in // current window. Initialized as false boolean visited[] = new boolean[n]; for(int i = 0; i < n; i++) visited[i] = false; // While right boundary of current window // doesn't cross right end while (right < n) { // If current window contains all distinct // elements, widen the window toward right while (right < n && !visited[arr[right]]) { count += (right - left); visited[arr[right]] = true; right++; } // If duplicate is found in current window, // then reduce the window from left while (left < right && (right != n && visited[arr[right]])) { visited[arr[left]] = false; left++; } } return count; } // Driver code public static void main(String args[]){ int arr[] = {1, 4, 2, 4, 3, 2}; int n = arr.length; System.out.println( countPairs(arr, n)); }} // This code is contributed by Arnab Kundu |
Python3
# Python 3 program to count number of distinct # pairs that can be formed from all possible # contiguous sub-arrays containing distinct numbersdef countPairs(arr, n): # initialize number of pairs to zero count = 0 # Left and right indexes of # current window right = 0 left = 0 # Boolean array visited to mark elements # in current window. Initialized as false visited = [False for i in range(n)] # While right boundary of current # window doesn't cross right end while (right < n): # If current window contains all distinct # elements, widen the window toward right while (right < n and visited[arr[right]] == False): count += (right - left) visited[arr[right]] = True right += 1 # If duplicate is found in current window, # then reduce the window from left while (left < right and (right != n and visited[arr[right]] == True)): visited[arr[left]] = False left += 1 return count# Driver codeif __name__ == '__main__': arr = [1, 4, 2, 4, 3, 2] n = len(arr) print(countPairs(arr, n))# This code is contributed by# Sanjit_Prasad |
C#
// C# program to count number of distinct pairs // that can be formed from all possible contiguous // sub-arrays containing distinct numbers using System;class GFG{static int countPairs(int []arr, int n) { // initialize number of pairs to zero int count = 0; //Left and right indexes of current window int right = 0, left = 0; // Boolean array visited to mark elements in // current window. Initialized as false bool [] visited = new bool[n]; for(int i = 0; i < n; i++) visited[i] = false; // While right boundary of current window // doesn't cross right end while (right < n) { // If current window contains all distinct // elements, widen the window toward right while (right < n && !visited[arr[right]]) { count += (right - left); visited[arr[right]] = true; right++; } // If duplicate is found in current window, // then reduce the window from left while (left < right && (right != n && visited[arr[right]])) { visited[arr[left]] = false; left++; } } return count; } // Driver code public static void Main(){ int [] arr = {1, 4, 2, 4, 3, 2}; int n = arr.Length; Console.Write( countPairs(arr, n)); }} // This code is contributed by mohit kumar 29 |
Javascript
<script>// Javascript program to count number of distinct pairs // that can be formed from all possible contiguous // sub-arrays containing distinct numbers function countPairs(arr,n) { // initialize number of pairs to zero let count = 0; //Left and right indexes of current window let right = 0, left = 0; // Boolean array visited to mark elements in // current window. Initialized as false let visited = new Array(n); for(let i = 0; i < n; i++) visited[i] = false; // While right boundary of current window // doesn't cross right end while (right < n) { // If current window contains all distinct // elements, widen the window toward right while (right < n && !visited[arr[right]]) { count += (right - left); visited[arr[right]] = true; right++; } // If duplicate is found in current window, // then reduce the window from left while (left < right && (right != n && visited[arr[right]])) { visited[arr[left]] = false; left++; } } return count; } // Driver code let arr=[1, 4, 2, 4, 3, 2]; let n = arr.length; document.write( countPairs(arr, n)); // This code is contributed by unknown2108</script> |
Output
8
Time Complexity: The complexity might look O(n^2) as 2 while loop are involved but note that left and right index are changing from 0 to N-1. So overall complexity is O(n + n) = O(n). Auxiliary space required in above solution is O(n) as we are using visited array to mark elements of the current window.
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