Next Greater Frequency Element

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++

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// 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 (freq[a[s.top()]] < freq[a[i]]
                   && !s.empty()) {
 
                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;
}

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Java

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// Java program of Next Greater Frequency Element
import java.util.*;
 
class GFG {
 
    /*NFG function to find the next greater frequency
    element for each element in the array*/
    static void NFG(int a[], int n, int freq[])
    {
 
        // stack data structure to store the position
        // of array element
        Stack<Integer> s = new Stack<Integer>();
        s.push(0);
 
        // res to store the value of next greater
        // frequency element for each element
        int res[] = new int[n];
        for (int i = 0; i < n; i++)
            res[i] = 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.peek()]] > 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 (freq[a[s.peek()]] < freq[a[i]]
                       && s.size() > 0) {
                    res[s.peek()] = a[i];
                    s.pop();
                }
 
                // now push the current element
                s.push(i);
            }
        }
 
        while (s.size() > 0) {
            res[s.peek()] = -1;
            s.pop();
        }
 
        for (int i = 0; i < n; i++) {
            // Print the res list containing next
            // greater frequency element
            System.out.print(res[i] + " ");
        }
    }
 
    // Driver code
    public static void main(String args[])
    {
 
        int a[] = { 1, 1, 2, 3, 4, 2, 1 };
        int len = 7;
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < len; i++) {
            // Getting the max element of the array
            if (a[i] > max) {
                max = a[i];
            }
        }
        int freq[] = new int[max + 1];
 
        for (int i = 0; i < max + 1; i++)
            freq[i] = 0;
 
        // Calculating frequency of each element
        for (int i = 0; i < len; i++) {
            freq[a[i]]++;
        }
        // Function call
        NFG(a, len, freq);
    }
}
 
// This code is contributed by Arnab Kundu

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Python3

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'''NFG function to find the next greater frequency
   element for each element in the array'''
 
 
def NFG(a, n):
 
    if (n <= 0):
        print("List empty")
        return []
 
    # stack data structure to store the position
    # of array element
    stack = [0]*n
 
    # freq is a dictionary which maintains the
    # frequency of each element
    freq = {}
    for i in a:
        freq[a[i]] = 0
    for i in a:
        freq[a[i]] += 1
 
    # res to store the value of next greater
    # frequency element for each element
    res = [0]*n
 
    # initialize top of stack to -1
    top = -1
 
    # push the first position of array in the stack
    top += 1
    stack[top] = 0
 
    # now iterate for the rest of elements
    for i in range(1, n):
 
        ''' 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[stack[top]]] > freq[a[i]]):
            top += 1
            stack[top] = 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 (top > -1 and freq[a[stack[top]]] < freq[a[i]]):
                res[stack[top]] = a[i]
                top -= 1
 
            # now push the current element
            top += 1
            stack[top] = i
 
    '''After iterating over the loop, the remaining
    position of elements in stack do not have the
    next greater element, so print -1 for them'''
    while (top > -1):
        res[stack[top]] = -1
        top -= 1
 
    # return the res list containing next
    # greater frequency element
    return res
 
 
# Driver Code
print(NFG([1, 1, 2, 3, 4, 2, 1], 7))

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C#

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// C# program of Next Greater Frequency Element
using System;
using System.Collections;
 
class GFG {
 
    /*NFG function to find the
    next greater frequency
    element for each element
    in the array*/
    static void NFG(int[] a, int n, int[] freq)
    {
 
        // stack data structure to store
        // the position of array element
        Stack s = new Stack();
        s.Push(0);
 
        // res to store the value of next greater
        // frequency element for each element
        int[] res = new int[n];
        for (int i = 0; i < n; i++)
            res[i] = 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[(int)s.Peek()]] > 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 (freq[a[(int)(int)s.Peek()]]
                           < freq[a[i]]
                       && s.Count > 0) {
                    res[(int)s.Peek()] = a[i];
                    s.Pop();
                }
 
                // now Push the current element
                s.Push(i);
            }
        }
 
        while (s.Count > 0) {
            res[(int)s.Peek()] = -1;
            s.Pop();
        }
 
        for (int i = 0; i < n; i++) {
 
            // Print the res list containing next
            // greater frequency element
            Console.Write(res[i] + " ");
        }
    }
 
    // Driver code
    public static void Main(String[] args)
    {
 
        int[] a = { 1, 1, 2, 3, 4, 2, 1 };
        int len = 7;
        int max = int.MinValue;
        for (int i = 0; i < len; i++) {
            // Getting the max element of the array
            if (a[i] > max) {
                max = a[i];
            }
        }
        int[] freq = new int[max + 1];
 
        for (int i = 0; i < max + 1; i++)
            freq[i] = 0;
 
        // Calculating frequency of each element
        for (int i = 0; i < len; i++) {
            freq[a[i]]++;
        }
        NFG(a, len, freq);
    }
}
 
// This code is contributed by Arnab Kundu

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Output:

[-1, -1, 1, 2, 2, 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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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