Find highest frequency of non-negative powers that are same as indices of elements in given Array

• Last Updated : 02 Dec, 2021

Given an array arr[] with N non-negative integers, find the maximum number of elements that are the same non-negative powers of their indices.

arr[i] = iX, where X is a non negative number.

The task is to return the maximum frequency of X.

Example:

Input: arr = [1, 1, 4, 17]
Output: 2
Explanation:
The element 1 at index 0, is a power of 0
The element 1 at index 1, is a power of 0
The element 4 at index 2, is a power of 2
The element 17 at index 3, is not a power of its index, so it is not considered
Therefore the maximum frequency is of power 0, which is 2

Input: arr = [0, 1, 1, 9, 1, 25]
Output: 4
Explanation:
The element 0 at index 0, is a power of 2
The element 1 at index 1, is a power of 2
The element 1 at index 2, is a power of 0
The element 9 at index 3, is a power of 2
The element 1 at index 4, is a power of 0
The element 25 at index 5, is a power of 2
Therefore the maximum frequency is of power 2, which is 4

Approach: Given problem can be solved by finding the powers of every index and checking if they are equal to the element present at that index.

Follow the steps below to solve the problem:

• Iterate the array arr from index 2 till the end and at every index:
• Use a loop to multiply the index by itself until the value is less than the maximum value of integer and less than or equal to the element present at that index
• If the power becomes equal to the element then check if its present in the hashmap:
• If the power is not present then add it in the hash map with value 1
• Else if the power is already present then increment its frequency by 1
• If arr = 1, then increment the frequency of 0 in the hashmap by 1
• Iterate the HashMap and find the value with the maximum frequency:
• If arr = 1, then the frequency of all values except 0 is increment by 1
• If arr = 1, then return, maxFreq +1, else return maxFreq

Below is the implementation of the above approach:

C++

 // C++ code for the above approach#include using namespace std; // Function to find count of elements which// are a non-negative power of their indicesint indPowEqualsEle(vector arr){     // Length of the array    int len = arr.size();     // Initialize the hashmap to store    // the frequency of elements    unordered_map map;     // Initialize maximum value    // of integer into long    long limit = INT_MAX;     // Iterate the array arr from index 2    for (int i = 2; i < len; i++)    {         // If current element is equal to 1        // then its equal to index power 0        if (arr[i] == 1)        {             // Increment the frequency of 0 by 1            map++;            continue;        }         // Initialize a variable to index        // which is to be multiplied        // by the index        long indPow = i;         // Initialize a variable to        // store the power of the index        int p = 1;         while (indPow <= limit && indPow <= arr[i])        {             // Element is equal to            // a power of its index            if (arr[i] == indPow)            {                 // Increment the frequency                // of p by 1                map[p]++;                 break;            }             // Increment power            p++;             // Multiply current value with            // index to get the next power            indPow *= i;        }    }     // If arr == 1, then increment    // the frequency of 0 in the hashmap    map++;     // Initialize maxFreq to 0 to calculate    // maximum frequency of powers    int maxFreq = 0;     // Iterate the hashmap    for (auto it = map.begin(); it != map.end(); it++)    {        int power = it->second;        // If arr == 0, then increment the        // frequency of all powers except 0        if (arr == 0 && power != 0)        {             maxFreq = max(maxFreq,                          map[power] + 1);        }        else        {             maxFreq = max(maxFreq,                          map[power]);        }    }     // Increment the maximum frequency by 1    // If arr is equal to 1    return arr == 1               ? maxFreq + 1               : maxFreq;} // Driver functionint main(){     // Initialize an array    vector arr = {0, 1, 1, 9, 1, 25};     // Call the function    // and print the answer    cout << (indPowEqualsEle(arr));} // This code is contributed by Potta Lokesh

Java

 // Java implementation for the above approach import java.io.*;import java.util.*; class GFG {     // Function to find count of elements which    // are a non-negative power of their indices    public static int indPowEqualsEle(int[] arr)    {         // Length of the array        int len = arr.length;         // Initialize the hashmap to store        // the frequency of elements        Map map            = new HashMap<>();         // Initialize maximum value        // of integer into long        long limit = (long)Integer.MAX_VALUE;         // Iterate the array arr from index 2        for (int i = 2; i < len; i++) {             // If current element is equal to 1            // then its equal to index power 0            if (arr[i] == 1) {                 // Increment the frequency of 0 by 1                map.put(0,                        map.getOrDefault(0, 0) + 1);                continue;            }             // Initialize a variable to index            // which is to be multiplied            // by the index            long indPow = i;             // Initialize a variable to            // store the power of the index            int p = 1;             while (indPow <= limit                   && indPow <= arr[i]) {                 // Element is equal to                // a power of its index                if (arr[i] == indPow) {                     // Increment the frequency                    // of p by 1                    map                        .put(p,                             map.getOrDefault(p, 0) + 1);                     break;                }                 // Increment power                p++;                 // Multiply current value with                // index to get the next power                indPow *= i;            }        }         // If arr == 1, then increment        // the frequency of 0 in the hashmap        map.put(0, map.getOrDefault(0, 0) + 1);         // Initialize maxFreq to 0 to calculate        // maximum frequency of powers        int maxFreq = 0;         // Iterate the hashmap        for (int power : map.keySet()) {             // If arr == 0, then increment the            // frequency of all powers except 0            if (arr == 0 && power != 0) {                 maxFreq                    = Math.max(maxFreq,                               map.get(power) + 1);            }            else {                 maxFreq = Math.max(maxFreq,                                   map.get(power));            }        }         // Increment the maximum frequency by 1        // If arr is equal to 1        return arr == 1            ? maxFreq + 1            : maxFreq;    }     // Driver function    public static void main(String[] args)    {         // Initialize an array        int[] arr = { 0, 1, 1, 9, 1, 25 };         // Call the function        // and print the answer        System.out.println(indPowEqualsEle(arr));    }}

Python3

 # Python 3 code for the above approachfrom collections import defaultdictimport sys # Function to find count of elements which# are a non-negative power of their indicesdef indPowEqualsEle(arr):     # Length of the array    length = len(arr)     # Initialize the hashmap to store    # the frequency of elements    map = defaultdict(int)     # Initialize maximum value    # of integer into long    limit = sys.maxsize     # Iterate the array arr from index 2    for i in range(2, length):         # If current element is equal to 1        # then its equal to index power 0        if (arr[i] == 1):             # Increment the frequency of 0 by 1            map += 1            continue         # Initialize a variable to index        # which is to be multiplied        # by the index        indPow = i         # Initialize a variable to        # store the power of the index        p = 1         while (indPow <= limit and indPow <= arr[i]):             # Element is equal to            # a power of its index            if (arr[i] == indPow):                 # Increment the frequency                # of p by 1                map[p] += 1                 break             # Increment power            p += 1             # Multiply current value with            # index to get the next power            indPow *= i     # If arr == 1, then increment    # the frequency of 0 in the hashmap    map += 1     # Initialize maxFreq to 0 to calculate    # maximum frequency of powers    maxFreq = 0     # Iterate the hashmap    for it in range(len(map)):         power = map[it]        # If arr == 0, then increment the        # frequency of all powers except 0        if (arr == 0 and power != 0):             maxFreq = max(maxFreq,                          map[power] + 1)        else:             maxFreq = max(maxFreq,                          map[power])     # Increment the maximum frequency by 1    # If arr is equal to 1    if(arr == 1):        return maxFreq + 1    return maxFreq # Driver functionif __name__ == "__main__":     # Initialize an array    arr = [0, 1, 1, 9, 1, 25]     # Call the function    # and print the answer    print(indPowEqualsEle(arr))     # This code is contributed by ukasp.

C#

 // C# implementation for the above approachusing System;using System.Collections.Generic; public class GFG {     // Function to find count of elements which    // are a non-negative power of their indices    public static int indPowEqualsEle(int[] arr)    {         // Length of the array        int len = arr.Length;         // Initialize the hashmap to store        // the frequency of elements        Dictionary map            = new Dictionary();         // Initialize maximum value        // of integer into long        long limit = (long)int.MaxValue;         // Iterate the array arr from index 2        for (int i = 2; i < len; i++) {             // If current element is equal to 1            // then its equal to index power 0            if (arr[i] == 1) {                 // Increment the frequency of 0 by 1                if(map.ContainsKey(0))                    map = map+1;                else                    map.Add(0, 1);                continue;            }             // Initialize a variable to index            // which is to be multiplied            // by the index            long indPow = i;             // Initialize a variable to            // store the power of the index            int p = 1;             while (indPow <= limit                   && indPow <= arr[i]) {                 // Element is equal to                // a power of its index                if (arr[i] == indPow) {                     // Increment the frequency                    // of p by 1                    if(map.ContainsKey(p))                    map[p] = map[p]+1;                else                    map.Add(p, 1);                     break;                }                 // Increment power                p++;                 // Multiply current value with                // index to get the next power                indPow *= i;            }        }         // If arr == 1, then increment        // the frequency of 0 in the hashmap        if(map.ContainsKey(0))            map = map+1;        else            map.Add(0, 1);         // Initialize maxFreq to 0 to calculate        // maximum frequency of powers        int maxFreq = 0;         // Iterate the hashmap        foreach (int power in map.Keys) {             // If arr == 0, then increment the            // frequency of all powers except 0            if (arr == 0 && power != 0) {                 maxFreq                    = Math.Max(maxFreq,                               map[power] + 1);            }            else {                 maxFreq = Math.Max(maxFreq,                                   map[power]);            }        }         // Increment the maximum frequency by 1        // If arr is equal to 1        return arr == 1            ? maxFreq + 1            : maxFreq;    }     // Driver function    public static void Main(String[] args)    {         // Initialize an array        int[] arr = { 0, 1, 1, 9, 1, 25 };         // Call the function        // and print the answer        Console.WriteLine(indPowEqualsEle(arr));    }} // This code is contributed by 29AjayKumar

Javascript



Output
4

Time Complexity: O(N * log N)
Auxiliary Space: O(1)

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