Given an array arr of n elements, the task is to find the number of the sub-arrays of the given array that contain at least one duplicate element.
Examples:
Input: arr[] = {1, 2, 3}
Output: 0
There is no sub-array with duplicate elements.Input: arr[] = {4, 3, 4, 3}
Output: 3
Possible sub-arrays are {4, 3, 4}, {4, 3, 4, 3} and {3, 4, 3}
Approach:
- First, find the total number of sub-arrays that can be formed from the array and denote this by total then total = (n*(n+1))/2.
- Now find the sub-arrays that have all the elements distinct (can be found out using window sliding technique) and denote this by unique.
- Finally, the number of sub-arrays that have at least one element duplicate are (total – unique)
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>#define ll long long intusing namespace std;// Function to return the count of the// sub-arrays that have at least one duplicatell count(ll arr[], ll n){ ll unique = 0; // two pointers ll i = -1, j = 0; // to store frequencies of the numbers unordered_map<ll, ll> freq; for (j = 0; j < n; j++) { freq[arr[j]]++; // number is not distinct if (freq[arr[j]] >= 2) { i++; while (arr[i] != arr[j]) { freq[arr[i]]--; i++; } freq[arr[i]]--; unique = unique + (j - i); } else unique = unique + (j - i); } ll total = n * (n + 1) / 2; return total - unique;}// Driver codeint main(){ ll arr[] = { 4, 3, 4, 3 }; ll n = sizeof(arr) / sizeof(arr[0]); cout << count(arr, n) << endl; return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG {// Function to return the count of the// sub-arrays that have at least one duplicatestatic Integer count(Integer arr[], Integer n){ Integer unique = 0; // two pointers Integer i = -1, j = 0; // to store frequencies of the numbers Map<Integer, Integer> freq = new HashMap<>(); for (j = 0; j < n; j++) { if(freq.containsKey(arr[j])) { freq.put(arr[j], freq.get(arr[j]) + 1); } else { freq.put(arr[j], 1); } // number is not distinct if (freq.get(arr[j]) >= 2) { i++; while (arr[i] != arr[j]) { freq.put(arr[i], freq.get(arr[i]) - 1); i++; } freq.put(arr[i], freq.get(arr[i]) - 1); unique = unique + (j - i); } else unique = unique + (j - i); } Integer total = n * (n + 1) / 2; return total - unique;}// Driver codepublic static void main(String[] args){ Integer arr[] = { 4, 3, 4, 3 }; Integer n = arr.length; System.out.println(count(arr, n));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach from collections import defaultdict# Function to return the count of the # sub-arrays that have at least one duplicate def count(arr, n): unique = 0 # two pointers i, j = -1, 0 # to store frequencies of the numbers freq = defaultdict(lambda:0) for j in range(0, n): freq[arr[j]] += 1 # number is not distinct if freq[arr[j]] >= 2: i += 1 while arr[i] != arr[j]: freq[arr[i]] -= 1 i += 1 freq[arr[i]] -= 1 unique = unique + (j - i) else: unique = unique + (j - i) total = (n * (n + 1)) // 2 return total - unique # Driver Codeif __name__ == "__main__": arr = [4, 3, 4, 3] n = len(arr) print(count(arr, n))# This code is contributed # by Rituraj Jain |
C#
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG {// Function to return the count of the// sub-arrays that have at least one duplicatestatic int count(int []arr, int n){ int unique = 0; // two pointers int i = -1, j = 0; // to store frequencies of the numbers Dictionary<int, int> freq = new Dictionary<int, int>(); for (j = 0; j < n; j++) { if(freq.ContainsKey(arr[j])) { freq[arr[j]] = freq[arr[j]] + 1; } else { freq.Add(arr[j], 1); } // number is not distinct if (freq[arr[j]] >= 2) { i++; while (arr[i] != arr[j]) { freq[arr[i]] = freq[arr[i]] - 1; i++; } freq[arr[i]] = freq[arr[i]] - 1; unique = unique + (j - i); } else unique = unique + (j - i); } int total = n * (n + 1) / 2; return total - unique;}// Driver codepublic static void Main(String[] args){ int []arr = { 4, 3, 4, 3 }; int n = arr.Length; Console.WriteLine(count(arr, n));}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript implementation of the approach// Function to return the count of the// sub-arrays that have at least one duplicatefunction count(arr, n){ let unique = 0; // two pointers let i = -1, j = 0; // to store frequencies of the numbers let freq = new Map(); for (j = 0; j < n; j++) { if(freq.has(arr[j])){ freq.set(arr[j], freq.get(arr[j]) + 1) }else{ freq.set(arr[j], 1) } // number is not distinct if (freq.get(arr[j]) >= 2) { i++; while (arr[i] != arr[j]) { freq.set(arr[i], freq.get(arr[i]) - 1) i++; } freq.set(arr[i], freq.get(arr[i]) - 1) unique = unique + (j - i); } else unique = unique + (j - i); } let total = n *(n + 1) / 2; return total - unique;}// Driver code let arr = [ 4, 3, 4, 3 ]; let n = arr.length; document.write(count(arr, n) + "<br>");// This code is contributed by _saurabh_jaiswal</script> |
Output
3
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(N)
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