Print the elements of an array in the decreasing frequency if 2 numbers have same frequency then print the one which came first.

Examples:

Input: arr[] = {2, 5, 2, 8, 5, 6, 8, 8} Output: arr[] = {8, 8, 8, 2, 2, 5, 5, 6} Input: arr[] = {2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8} Output: arr[] = {8, 8, 8, 2, 2, 5, 5, 6, -1, 9999999}

**METHOD 1 (Use Sorting)**

1) Use a sorting algorithm to sort the elements O(nlogn) 2) Scan the sorted array and construct a 2D array of element and count O(n). 3) Sort the 2D array according to count O(nlogn).

**Example:**

Input 2 5 2 8 5 6 8 8 After sorting we get 2 2 5 5 6 8 8 8 Now construct the 2D array as 2, 2 5, 2 6, 1 8, 3 Sort by count 8, 3 2, 2 5, 2 6, 1

**How to maintain order of elements if frequency is same?**

The above approach doesn’t make sure order of elements if frequency is same. To handle this, we should use indexes in step 3, if two counts are same then we should first process(or print) the element with lower index. In step 1, we should store the indexes instead of elements.

Input 5 2 2 8 5 6 8 8 After sorting we get Element 2 2 5 5 6 8 8 8 Index 1 2 0 4 5 3 6 7 Now construct the 2D array as Index, Count 1, 2 0, 2 5, 1 3, 3 Sort by count (consider indexes in case of tie) 3, 3 0, 2 1, 2 5, 1 Print the elements using indexes in the above 2D array.

Below is C++ implementation of above approach.

`// Sort elements by frequency. If two elements have same ` `// count, then put the elements that appears first ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Used for sorting ` `struct` `ele ` `{ ` ` ` `int` `count, index, val; ` `}; ` ` ` `// Used for sorting by value ` `bool` `mycomp(` `struct` `ele a, ` `struct` `ele b) { ` ` ` `return` `(a.val < b.val); ` `} ` ` ` `// Used for sorting by frequency. And if frequency is same, ` `// then by appearance ` `bool` `mycomp2(` `struct` `ele a, ` `struct` `ele b) { ` ` ` `if` `(a.count != b.count) ` `return` `(a.count < b.count); ` ` ` `else` `return` `a.index > b.index; ` `} ` ` ` `void` `sortByFrequency(` `int` `arr[], ` `int` `n) ` `{ ` ` ` `struct` `ele element[n]; ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `{ ` ` ` `element[i].index = i; ` `/* Fill Indexes */` ` ` `element[i].count = 0; ` `/* Initialize counts as 0 */` ` ` `element[i].val = arr[i]; ` `/* Fill values in structure ` ` ` `elements */` ` ` `} ` ` ` ` ` `/* Sort the structure elements according to value, ` ` ` `we used stable sort so relative order is maintained. */` ` ` `stable_sort(element, element+n, mycomp); ` ` ` ` ` `/* initialize count of first element as 1 */` ` ` `element[0].count = 1; ` ` ` ` ` `/* Count occurrences of remaining elements */` ` ` `for` `(` `int` `i = 1; i < n; i++) ` ` ` `{ ` ` ` `if` `(element[i].val == element[i-1].val) ` ` ` `{ ` ` ` `element[i].count += element[i-1].count+1; ` ` ` ` ` `/* Set count of previous element as -1 , we are ` ` ` `doing this because we'll again sort on the ` ` ` `basis of counts (if counts are equal than on ` ` ` `the basis of index)*/` ` ` `element[i-1].count = -1; ` ` ` ` ` `/* Retain the first index (Remember first index ` ` ` `is always present in the first duplicate we ` ` ` `used stable sort. */` ` ` `element[i].index = element[i-1].index; ` ` ` `} ` ` ` ` ` `/* Else If previous element is not equal to current ` ` ` `so set the count to 1 */` ` ` `else` `element[i].count = 1; ` ` ` `} ` ` ` ` ` `/* Now we have counts and first index for each element so now ` ` ` `sort on the basis of count and in case of tie use index ` ` ` `to sort.*/` ` ` `stable_sort(element, element+n, mycomp2); ` ` ` `for` `(` `int` `i = n-1, index=0; i >= 0; i--) ` ` ` `if` `(element[i].count != -1) ` ` ` `for` `(` `int` `j=0; j<element[i].count; j++) ` ` ` `arr[index++] = element[i].val; ` `} ` ` ` `// Driver program ` `int` `main() ` `{ ` ` ` `int` `arr[] = {2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8}; ` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]); ` ` ` ` ` `sortByFrequency(arr, n); ` ` ` ` ` `for` `(` `int` `i=0; i<n; i++) ` ` ` `cout << arr[i] << ` `" "` `; ` ` ` `return` `0; ` `} ` |

Output:

8 8 8 2 2 5 5 6 -1 9999999

Thanks to Gaurav Ahirwar for providing above implementation.

**METHOD 2(Use BST and Sorting)**

1. Insert elements in BST one by one and if an element is already present then increment the count of the node. Node of the Binary Search Tree (used in this approach) will be as follows.

`struct` `tree ` `{ ` ` ` `int` `element; ` ` ` `int` `first_index ` `/*To handle ties in counts*/` ` ` `int` `count; ` `}BST; ` |

2.Store the first indexes and corresponding counts of BST in a 2D array.

3 Sort the 2D array according to counts (and use indexes in case of tie).

**Time Complexity:** O(nlogn) if a Self Balancing Binary Search Tree is used. This is implemented in Set 2.

**METHOD 3(Use Hashing and Sorting)**

Using a hashing mechanism, we can store the elements (also first index) and their counts in a hash. Finally, sort the hash elements according to their counts.

**Set 2:**

Sort elements by frequency | Set 2

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