Next Greater Frequency Element - GeeksforGeeks

Sruti Rai

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Next Greater Frequency Element

Difficulty Level : Medium
Last Updated : 02 Sep, 2021

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, 3Lets calls Next Greater Frequency element as NGF1. 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 > 24. For element a[3] = 3 which has frequency = 1,NGF element is 2 at position = 5 with frequency   of 2 > 15. For element a[4] = 4 which has frequency = 1,NGF element is 2 at position = 5 with frequency   of 2 > 16. For element a[5] = 2 which has frequency = 2,NGF element is 1 at position = 6 with frequencyof 3 > 27. For element a[6] = 1 there is no element to its   right, hence -1Input : 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++
Java
Python3
C#
Javascript

// 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).

Space Efficient Approach: using a hash map instead of a list as mentioned in the above approach.

Steps:

Create a class pair to store pair<int, int> with pair<element, frequency>.
Create a hasp map with pair as generics to store keys as the element and values as the frequency of every element.
Iterate the array and save the element and its frequency in the hashmap.
Create a res array that stores the resultant array.
Initially make res[n-1] = -1 and push the element in the end along with its frequency into the stack.
Iterate through the array in reverse order.
If the frequency of the element which is pointed at the top of the stack is less than the frequency of the current element and the stack is not empty then pop.
Continue till the loop fails.
If the stack is empty, it means that there is no element with a higher frequency. So, place -1 as the next higher frequency element in the resultant array.
If the stack is not empty, it means that the top of the stack has a higher frequency element. Put it in the resultant array as the next higher frequency.
Push the current element along with its frequency.

C++
Java
Python3
C#
Javascript

// C++ program of Next Greater Frequency Element

#include <bits/stdc++.h>

using namespace std;

stack<pair<int,int>> mystack;

map<int, int> mymap;

/*NFG function to find the next greater frequency

element for each element and for placing it in the

resultant array */

void NGF(int arr[], int res[], int n) {

// Initially store the frequencies of all elements

// in a hashmap

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

mymap[arr[i]] += 1;

}

// Get the frequency of the last element

int curr_freq = mymap[arr[n-1]];

// push it to the stack

mystack.push({arr[n-1], curr_freq});

// place -1 as next greater freq for the last

// element as it does not have next greater.

res[n-1] = -1;

// iterate through array in reverse order

for(int i = n-2;i>=0;i--) {

curr_freq = mymap[arr[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*/

while(mystack.size() > 0 && curr_freq >= mystack.top().second)

mystack.pop();

// If the stack is empty, place -1. If it is not empty

// then we will have next higher freq element at the top of the stack.

res[i] = (mystack.size() == 0) ? -1 : mystack.top().first;

// push the element at current position

mystack.push({arr[i], mymap[arr[i]]});

}

int main()

{

int arr[] = {1, 1, 1, 2, 2, 2, 2, 11, 3, 3};

int n = sizeof(arr) / sizeof(arr[0]);

int res[n];

NGF(arr, res, n);

cout << "[";

for(int i = 0; i < n - 1; i++)

{

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

}

cout << res[n - 1] << "]";

return 0;

}

// This code is contributed by divyeshrabadiya07.

Output

[2, 2, 2, -1, -1, -1, -1, 3, -1, -1]

Time Complexity: O(n).

This article is contributed by Sruti Rai . Thank you Koustav for your valuable support. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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