Next Greater Frequency Element

Software Development
#1

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).