Given an array, for each element find the value of nearest element to the right which is having frequency greater than as that of 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 index is be used to store frequency of each element. Create a list suppose to store 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 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 stack data structure to store position of elements in the array.
1) Create a list to 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.
Time complexity is O(n).
Below is the implementation of the above problem.
C++
// C++ program of Next Greater Frequency Element #include <stack> #include <iostream> #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_MAX; 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]]++; } NFG(a, len, freq); return 0; } |
Java
// 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]]++; } NFG(a, len, freq); } } // This code is contributed by Arnab Kundu |
Python3
'''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 program to test the function print(NFG([1,1,2,3,4,2,1],7)) |
C#
// 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 |
Output:
[-1, -1, 1, 2, 2, 1, -1]
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