Given an array of **n** positive integers with many repeating elements. The task is to find maximum difference between the frequency of any two different elements, such that the element with greater frequency is also greater in value than the second integer.

Examples:

Input :arr[] = { 3, 1, 3, 2, 3, 2 }.Output :2 Frequency of 3 = 3. Frequency of 2 = 2. Frequency of 1 = 1. Here difference of frequency of element 3 and 1 is = 3 - 1 = 2. Also 3 > 1.

**Method 1 (Use Hashing):**

The naive approach can be, find the frequency of each element and for each element find the element having lesser value and lesser frequency than the current element.

Below is C++ implementation of this approach:

// C++ program to find maximum difference // between frequency of any two element // such that element with greater frequency // is also greater in value. #include<bits/stdc++.h> using namespace std; // Return the maximum difference between // frequencies of any two elements such that // element with greater frequency is also // greater in value. int maxdiff(int arr[], int n) { unordered_map<int, int> freq; // Finding the frequency of each element. for (int i = 0; i < n; i++) freq[arr[i]]++; int ans = 0; for (int i=0; i<n; i++) { for (int j=0; j<n; j++) { // finding difference such that element // having greater frequency is also // greater in value. if (freq[arr[i]] > freq[arr[j]] && arr[i] > arr[j] ) ans = max(ans, freq[arr[i]]-freq[arr[j]]); else if (freq[arr[i]] < freq[arr[j]] && arr[i] < arr[j] ) ans = max(ans, freq[arr[j]]-freq[arr[i]]); } } return ans; } // Driven Program int main() { int arr[] = { 3, 1, 3, 2, 3, 2 }; int n = sizeof(arr)/sizeof(arr[0]); cout << maxdiff(arr, n) << endl; return 0; }

Output:

2

Time Complexity: O(n^{2}).

**Method 2 (Use Hashing and Sorting):**

The idea is to find all the distinct elements and store in an array, say dist[ ]. Sort the distinct element array dist[] in increasing order. Now for any distinct element at index i, for all index j such that i > j > 0, find the element between index 0 to i-1 having minimum frequency. We can find frequency of an element in same way as method 1, i.e., storing frequencies in a hash table.

So do this for all i and find the maximum difference. To find the minimum frequency for all i maintain a prefix minimum.

Below is C++ representation of this approach:

// Efficient C++ program to find maximum // difference between frequency of any two // elements such that element with greater // frequency is also greater in value. #include<bits/stdc++.h> using namespace std; // Return the maximum difference between // frequencies of any two elements such that // element with greater frequency is also // greater in value. int maxdiff(int arr[], int n) { unordered_map<int, int> freq; int dist[n]; // Finding the frequency of each element. int j = 0; for (int i = 0; i < n; i++) { if (freq.find(arr[i]) == freq.end()) dist[j++] = arr[i]; freq[arr[i]]++; } // Sorting the distinct element sort(dist, dist + j); int min_freq = n+1; // Iterate through all sorted distinct elements. // For each distinct element, maintaining the // element with minimum frequency than that // element and also finding the maximum // frequency difference int ans = 0; for (int i=0; i<j; i++) { int cur_freq = freq[dist[i]]; ans = max(ans, cur_freq - min_freq); min_freq = min(min_freq, cur_freq); } return ans; } // Driven Program int main() { int arr[] = { 3, 1, 3, 2, 3, 2 }; int n = sizeof(arr)/sizeof(arr[0]); cout << maxdiff(arr, n) << endl; return 0; }

Output:

2

**Time Complexity : **O(n log n).

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