Given an array arr[] of N integers, the task is to choose the two subsequences of equal lengths such that the first subsequence must have all the unique elements and the second subsequence must have all the same elements. Print the maximum length of the subsequence pair.
Examples:
Input: arr[] = {1, 2, 3, 1, 2, 3, 3, 3}
Output: 3
Explanation:
The first subsequence consists of elements {1, 2, 3}.
The second subsequence consists of elements {3, 3, 3}.Input: arr[] = {2, 2, 2, 3}
Output: 2
Explanation:
The first subsequence consists of elements {2, 3}.
The second subsequence consists of elements {2, 2}.
Approach:
- Count the maximum frequency(say f) of the element in the given array.
- Count the distinct elements(say d) present in the array.
- To make both the subsequence of equal length with the given property, then the size of the first subsequence cannot exceed d, and the size of the second subsequence cannot exceed f.
- The maximum value of the subsequence is given on the basis of two cases below:
- The subsequence with distinct elements must have an element with maximum frequency. Therefore, the minimum of (d, f – 1) is the possible length of a subsequence with the given property.
- If the length of a subsequence with a distinct element is greater than maximum frequency, then, the minimum of (d – 1, f) is the possible length of a subsequence with the given property.
- The maximum value in the above two cases is the maximum length of a subsequence with the given property.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the maximum length// of subsequences with given propertyint maximumSubsequence(int arr[], int N){ // To store the frequency unordered_map<int, int> M; // Traverse the array to store the // frequency for (int i = 0; i < N; i++) { M[arr[i]]++; } // M.size() given count of distinct // element in arr[] int distinct_size = M.size(); int maxFreq = 1; // Traverse map to find max frequency for (auto& it : M) { maxFreq = max(maxFreq, it.second); } // Find the maximum length on the basis // of two cases in the approach cout << max(min(distinct_size, maxFreq - 1), min(distinct_size - 1, maxFreq));}// Driver Codeint main(){ int arr[] = { 1, 2, 3, 4, 4, 4, 4, 4, 5 }; int N = sizeof(arr) / sizeof(arr[0]); // Function call maximumSubsequence(arr, N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to find the maximum length// of subsequences with given propertystatic void maximumSubsequence(int arr[], int N){ // To store the frequency HashMap<Integer, Integer> M = new HashMap<Integer, Integer>(); // Traverse the array to store the // frequency for(int i = 0; i < N; i++) { if(M.containsKey(arr[i])) { M.put(arr[i], M.get(arr[i]) + 1); } else { M.put(arr[i], 1); } } // M.size() given count of distinct // element in arr[] int distinct_size = M.size(); int maxFreq = 1; // Traverse map to find max frequency for(Map.Entry<Integer, Integer> it : M.entrySet()) { maxFreq = Math.max(maxFreq, it.getValue()); } // Find the maximum length on the basis // of two cases in the approach System.out.print(Math.max( Math.min(distinct_size, maxFreq - 1), Math.min(distinct_size - 1, maxFreq)));}// Driver Codepublic static void main(String[] args){ int arr[] = { 1, 2, 3, 4, 4, 4, 4, 4, 5 }; int N = arr.length; // Function call maximumSubsequence(arr, N);}}// This code is contributed by amal kumar choubey |
Python3
# Python 3 program for the above approach# Function to find the maximum length# of subsequences with given propertydef maximumSubsequence(arr, N): # To store the frequency M = {i : 0 for i in range(100)} # Traverse the array to store the # frequency for i in range(N): M[arr[i]] += 1 # M.size() given count of distinct # element in arr[] distinct_size = len(M) maxFreq = 1 # Traverse map to find max frequency for value in M.values(): maxFreq = max(maxFreq, value) # Find the maximum length on the basis # of two cases in the approach print(max(min(distinct_size, maxFreq - 1), min(distinct_size - 1, maxFreq)))# Driver Codeif __name__ == '__main__': arr = [ 1, 2, 3, 4, 4, 4, 4, 5 ] N = len(arr) # Function call maximumSubsequence(arr, N)# This code is contributed by Samarth |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to find the maximum length// of subsequences with given propertystatic void maximumSubsequence(int []arr, int N){ // To store the frequency Dictionary<int, int> M = new Dictionary<int, int>(); // Traverse the array to store the // frequency for(int i = 0; i < N; i++) { if(M.ContainsKey(arr[i])) { M[arr[i]] = M[arr[i]] + 1; } else { M.Add(arr[i], 1); } } // M.Count given count of distinct // element in []arr int distinct_size = M.Count; int maxFreq = 1; // Traverse map to find max frequency foreach(KeyValuePair<int, int> m in M) { maxFreq = Math.Max(maxFreq, m.Value); } // Find the maximum length on the basis // of two cases in the approach Console.Write(Math.Max( Math.Min(distinct_size, maxFreq - 1), Math.Min(distinct_size - 1, maxFreq)));}// Driver Codepublic static void Main(String[] args){ int []arr = { 1, 2, 3, 4, 4, 4, 4, 4, 5 }; int N = arr.Length; // Function call maximumSubsequence(arr, N);}}// This code is contributed by Rohit_ranjan |
Javascript
<script>// Javascript program for the above approach// Function to find the maximum length// of subsequences with given propertyfunction maximumSubsequence(arr, N) { // To store the frequency let M = new Map(); // Traverse the array to store the // frequency for (let i = 0; i < N; i++) { if (M.has(arr[i])) { M.set(arr[i], M.get(arr[i]) + 1); } else { M.set(arr[i], 1); } } // M.size() given count of distinct // element in arr[] let distinct_size = M.size; let maxFreq = 1; // Traverse map to find max frequency for (let it of M) { maxFreq = Math.max(maxFreq, it[1]); } // Find the maximum length on the basis // of two cases in the approach document.write(Math.max(Math.min(distinct_size, maxFreq - 1), Math.min(distinct_size - 1, maxFreq)));}// Driver Codelet arr = [1, 2, 3, 4, 4, 4, 4, 4, 5];let N = arr.length;// Function callmaximumSubsequence(arr, N);// This code is contributed by _saurabh_jaiswal</script> |
Output:
4
Time Complexity: O(N), where N is the number of elements in the array.
Auxiliary Space Complexity: O(N), where N is the number of elements in the array.
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