Given an array arr, the task is to print the elements of the array in descending order along with their frequencies.
Examples:
Input: arr[] = {1, 3, 3, 3, 4, 4, 5}
Output: 5 occurs 1 times
4 occurs 2 times
3 occurs 3 times
1 occurs 1 timesInput: arr[] = {1, 1, 1, 2, 3, 4, 9, 9, 10}
Output: 10 occurs 1 times
9 occurs 2 times
4 occurs 1 times
3 occurs 1 times
2 occurs 1 times
1 occurs 3 times
Naive approach: Use some Data-Structure (e.g. multiset) which stores elements in decreasing order and then print the elements one by one with it’s count and then erase it from the Data-structure. The time complexity will be O(N log N) and the auxiliary space will be O(N) for the Data-structure used.
Below is the implementation of the above approach:
// C++ program to print the elements in // descending along with their frequencies #include <bits/stdc++.h> using namespace std; // Function to print the elements in descending // along with their frequencies void printElements(int a[], int n) { // A multiset to store elements in decreasing order multiset<int, greater<int> > ms; // Insert elements in the multiset for (int i = 0; i < n; i++) { ms.insert(a[i]); } // Print the elements along with their frequencies while (!ms.empty()) { // Find the maximum element int maxel = *ms.begin(); // Number of times it occurs int times = ms.count(maxel); cout << maxel << " occurs " << times << " times\n"; // Erase the maxel ms.erase(maxel); } } // Driver Code int main() { int a[] = { 1, 1, 1, 2, 3, 4, 9, 9, 10 }; int n = sizeof(a) / sizeof(a[0]); printElements(a, n); return 0; } |
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Output:
10 occurs 1 times 9 occurs 2 times 4 occurs 1 times 3 occurs 1 times 2 occurs 1 times 1 occurs 3 times
Efficient Approach: Sort the array in descending order and then start printing the elements from the beginning along with their frequencies.
Below is the implementation of the above approach:
C++
// C++ program to print the elements in // descending along with their frequencies #include <bits/stdc++.h> using namespace std; // Function to print the elements in descending // along with their frequencies void printElements(int a[], int n) { // Sorts the element in decreasing order sort(a, a + n, greater<int>()); int cnt = 1; // traverse the array elements for (int i = 0; i < n - 1; i++) { // Prints the number and count if (a[i] != a[i + 1]) { cout << a[i] << " occurs " << cnt << " times\n"; cnt = 1; } else cnt += 1; } // Prints the last step cout << a[n - 1] << " occurs " << cnt << " times\n"; } // Driver Code int main() { int a[] = { 1, 1, 1, 2, 3, 4, 9, 9, 10 }; int n = sizeof(a) / sizeof(a[0]); printElements(a, n); return 0; } |
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Java
// Java program to print the elements in // descending along with their frequencies import java.util.*; class GFG { // Function to print the elements in descending // along with their frequencies static void printElements(int a[], int n) { // Sorts the element in decreasing order Arrays.sort(a); a = reverse(a); int cnt = 1; // traverse the array elements for (int i = 0; i < n - 1; i++) { // Prints the number and count if (a[i] != a[i + 1]) { System.out.print(a[i]+ " occurs " + cnt + " times\n"); cnt = 1; } else cnt += 1; } // Prints the last step System.out.print(a[n - 1]+ " occurs " + cnt + " times\n"); } static int[] reverse(int a[]) { int i, n = a.length, t; for (i = 0; i < n / 2; i++) { t = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = t; } return a; } // Driver Code public static void main(String[] args) { int a[] = { 1, 1, 1, 2, 3, 4, 9, 9, 10 }; int n = a.length; printElements(a, n); } } // This code is contributed by PrinciRaj1992 |
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Python3
# Python3 program to print the elements in # descending along with their frequencies # Function to print the elements in # descending along with their frequencies def printElements(a, n) : # Sorts the element in decreasing order a.sort(reverse = True) cnt = 1 # traverse the array elements for i in range(n - 1) : # Prints the number and count if (a[i] != a[i + 1]) : print(a[i], " occurs ", cnt, "times") cnt = 1 else : cnt += 1 # Prints the last step print(a[n - 1], "occurs", cnt, "times") # Driver Code if __name__ == "__main__" : a = [ 1, 1, 1, 2, 3, 4, 9, 9, 10 ] n = len(a) printElements(a, n) # This code is contributed by Ryuga |
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C#
// C# program to print the elements in // descending along with their frequencies using System; class GFG { // Function to print the elements in descending // along with their frequencies static void printElements(int []a, int n) { // Sorts the element in decreasing order Array.Sort(a); a = reverse(a); int cnt = 1; // traverse the array elements for (int i = 0; i < n - 1; i++) { // Prints the number and count if (a[i] != a[i + 1]) { Console.Write(a[i]+ " occurs " + cnt + " times\n"); cnt = 1; } else cnt += 1; } // Prints the last step Console.Write(a[n - 1]+ " occurs " + cnt + " times\n"); } static int[] reverse(int []a) { int i, n = a.Length, t; for (i = 0; i < n / 2; i++) { t = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = t; } return a; } // Driver Code public static void Main(String[] args) { int []a = { 1, 1, 1, 2, 3, 4, 9, 9, 10 }; int n = a.Length; printElements(a, n); } } // This code is contributed by PrinciRaj1992 |
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PHP
<?php // PHP program to print the elements in // descending along with their frequencies // Function to print the elements in // descending along with their frequencies function printElements(&$a, $n) { // Sorts the element in // decreasing order rsort($a); $cnt = 1; // traverse the array elements for ($i = 0; $i < $n - 1; $i++) { // Prints the number and count if ($a[$i] != $a[$i + 1]) { echo ($a[$i]); echo (" occurs " ); echo $cnt ; echo (" times\n"); $cnt = 1; } else $cnt += 1; } // Prints the last step echo ($a[$n - 1]); echo (" occurs "); echo $cnt; echo (" times\n"); } // Driver Code $a = array(1, 1, 1, 2, 3, 4, 9, 9, 10 ); $n = sizeof($a); printElements($a, $n); // This code is contributed // by Shivi_Aggarwal ?> |
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Output:
10 occurs 1 times 9 occurs 2 times 4 occurs 1 times 3 occurs 1 times 2 occurs 1 times 1 occurs 3 times
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