# Next Greater Frequency Element

Hello Everyone,

Given an array, for each element find the value of the nearest element to the right which is having a frequency greater than as that of the current element. If there does not exist an answer for a position, then make the value ‘-1’.

Examples:

Input : a[] = [1, 1, 2, 3, 4, 2, 1] Output : [-1, -1, 1, 2, 2, 1, -1] Explanation: Given array a[] = [1, 1, 2, 3, 4, 2, 1] Frequency of each element is: 3, 3, 2, 1, 1, 2, 3 Lets calls Next Greater Frequency element as NGF 1. For element a[0] = 1 which has a frequency = 3, As it has frequency of 3 and no other next element has frequency more than 3 so ‘-1’ 2. For element a[1] = 1 it will be -1 same logic like a[0] 3. For element a[2] = 2 which has frequency = 2, NGF element is 1 at position = 6 with frequency of 3 > 2 4. For element a[3] = 3 which has frequency = 1, NGF element is 2 at position = 5 with frequency of 2 > 1 5. For element a[4] = 4 which has frequency = 1, NGF element is 2 at position = 5 with frequency of 2 > 1 6. For element a[5] = 2 which has frequency = 2, NGF element is 1 at position = 6 with frequency of 3 > 2 7. For element a[6] = 1 there is no element to its right, hence -1 Input : a[] = [1, 1, 1, 2, 2, 2, 2, 11, 3, 3] Output : [2, 2, 2, -1, -1, -1, -1, 3, -1, -1]

Naive approach:
A simple hashing technique is to use values as the index is being used to store the frequency of each element. Create a list suppose to store the frequency of each number in the array. (Single traversal is required). Now use two loops.
The outer loop picks all the elements one by one.
The inner loop looks for the first element whose frequency is greater than the frequency of the current element.
If a greater frequency element is found then that element is printed, otherwise -1 is printed.
Time complexity: O(n*n)

Efficient approach:
We can use hashing and stack data structure to efficiently solve for many cases. A simple hashing technique is to use values as index and frequency of each element as value. We use the stack data structure to store the position of elements in the array.

1. Create a list to use values as index to store frequency of each element.
2. Push the position of first element to stack.
3. Pick rest of the position of elements one by one and follow following steps in loop.
…….a) Mark the position of current element as ‘i’ .
……. b) If the frequency of the element which is pointed by the top of stack is greater than frequency of the current element, push the current position i to the stack
……. c) If the frequency of the element which is pointed by the top of stack is less than frequency of the current element and the stack is not empty then follow these steps:
…….i) continue popping the stack
…….ii) if the condition in step c fails then push the current position i to the stack
4. After the loop in step 3 is over, pop all the elements from stack and print -1 as next greater frequency element for them does not exist.

Below is the implementation of the above problem.

`// C++ program of Next Greater Frequency Element`

`#include <iostream>`

`#include <stack>`

`#include <stdio.h>`

`using` `namespace` `std;`

`/*NFG function to find the next greater frequency`

`element for each element in the array*/`

`void` `NFG(` `int` `a[], ` `int` `n, ` `int` `freq[])`

`{`

` ` `// stack data structure to store the position`

` ` `// of array element`

` ` `stack<` `int` `> s;`

` ` `s.push(0);`

` ` `// res to store the value of next greater`

` ` `// frequency element for each element`

` ` `int` `res[n] = { 0 };`

` ` `for` `(` `int` `i = 1; i < n; i++)`

` ` `{`

` ` `/* If the frequency of the element which is`

` ` `pointed by the top of stack is greater`

` ` `than frequency of the current element`

` ` `then push the current position i in stack*/`

` ` `if` `(freq[a[s.top()]] > freq[a[i]])`

` ` `s.push(i);`

` ` `else` `{`

` ` `/*If the frequency of the element which`

` ` `is pointed by the top of stack is less`

` ` `than frequency of the current element, then`

` ` `pop the stack and continuing popping until`

` ` `the above condition is true while the stack`

` ` `is not empty*/`

` ` `while` `( !s.empty()`

` ` `&& freq[a[s.top()]] < freq[a[i]])`

` ` `{`

` ` `res[s.top()] = a[i];`

` ` `s.pop();`

` ` `}`

` ` `// now push the current element`

` ` `s.push(i);`

` ` `}`

` ` `}`

` ` `while` `(!s.empty()) {`

` ` `res[s.top()] = -1;`

` ` `s.pop();`

` ` `}`

` ` `for` `(` `int` `i = 0; i < n; i++)`

` ` `{`

` ` `// Print the res list containing next`

` ` `// greater frequency element`

` ` `cout << res[i] << ` `" "` `;`

` ` `}`

`}`

`// Driver code`

`int` `main()`

`{`

` ` `int` `a[] = { 1, 1, 2, 3, 4, 2, 1 };`

` ` `int` `len = 7;`

` ` `int` `max = INT16_MIN;`

` ` `for` `(` `int` `i = 0; i < len; i++)`

` ` `{`

` ` `// Getting the max element of the array`

` ` `if` `(a[i] > max) {`

` ` `max = a[i];`

` ` `}`

` ` `}`

` ` `int` `freq[max + 1] = { 0 };`

` ` `// Calculating frequency of each element`

` ` `for` `(` `int` `i = 0; i < len; i++)`

` ` `{`

` ` `freq[a[i]]++;`

` ` `}`

` ` `// Function call`

` ` `NFG(a, len, freq);`

` ` `return` `0;`

`}`

Output:

[-1, -1, 1, 2, 2, 1, -1]

Time complexity: O(n).