“piles geekforgeeks” Réponses codées

piles geekforgeeks

// 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;
}