Given two arrays arr[] and q[] consisting of N integers, the task is for each query q[i] is to determine the number of times the maximum array element occurs in the subarray [q[i], arr[N – 1]].
Examples:
Input: arr[] = {5, 4, 5, 3, 2}, q[] = {1, 2, 3, 4, 5}
Output: 2 1 1 1 1
Explanation:
For the first query index start from 1. The subarray is [5, 4, 5, 3, 2], the maximum value is 5 and it’s occurrence in subarray is 2.
For the second query index start from 2. The subarray is [4, 5, 3, 2], the maximum value is 5 and it’s occurrence in subarray is 1.
For the third query index start from 3. The subarray is [5, 3, 2], the maximum value is 5 and it’s occurrence in subarray is 1.
For the forth query index start from 4. The subarray is [3, 2], the maximum value is 3 and it’s occurrence in subarray is 1.
For the fifth query index start from 5. The subarray is [2], the maximum value is 2 and it’s occurrence in subarray is 1.Input: arr[] = {2, 1, 2}, q[] = {1, 2, 3}
Output: 2 1 1
Naive Approach: The simplest approach is to traverse the array and find the maximum array element. Now, for each query q[i], traverse the subarray [q[i], arr[N – 1]] and print the count of occurrences of the maximum element in the subarray.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use Hashing. Below are the steps:
- Create an array maxFreqVec[] to store the occurrence of the max element from given index q[i] to N.
- Traverse the array arr[] from the right and keep track of the maximum element in the array and update the array maxFreqVec[] at that index with the occurrence of that maximum element.
- After the above steps, traverse over the array q[] and print the value maxFreqVec[q[i] – 1] as the maximum element in each subarray for each query.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include<bits/stdc++.h>using namespace std;// Function to find occurrence of// max element in given subarrayvoid FreqOfMaxElement(vector<int> arr, vector<int> q){ // Store the frequency of maximum // element vector<int> maxFreqVec(arr.size()); int maxSoFar = INT_MIN; int maxFreq = 0; // Traverse over the array arr[] // from right to left for(int i = arr.size() - 1; i >= 0; i--) { // If curr element is greater // than maxSofar if (arr[i] > maxSoFar) { // Reset maxSofar and maxFreq maxSoFar = arr[i]; maxFreq = 1; } // If curr is equal to maxSofar else if (arr[i] == maxSoFar) { // Increment the maxFreq maxFreq++; } // Update maxFreqVec[i] maxFreqVec[i] = maxFreq; } // Print occurrence of maximum // element for each query for(int k : q) { cout << maxFreqVec[k - 1] << " "; }}// Driver Codeint main(){ vector<int> arr = { 5, 4, 5, 3, 2 }; vector<int> q = { 1, 2, 3, 4, 5 }; // Function Call FreqOfMaxElement(arr, q);}// This code is contributed by mohit kumar 29 |
Java
// Java program for the above approachimport java.util.*;import java.lang.*;class GFG { // Function to find occurrence of // max element in given subarray static void FreqOfMaxElement( int[] arr, int[] q) { // Store the frequency of maximum // element int[] maxFreqVec = new int[arr.length]; int maxSoFar = Integer.MIN_VALUE; int maxFreq = 0; // Traverse over the array arr[] // from right to left for (int i = arr.length - 1; i >= 0; i--) { // If curr element is greater // than maxSofar if (arr[i] > maxSoFar) { // Reset maxSofar and maxFreq maxSoFar = arr[i]; maxFreq = 1; } // If curr is equal to maxSofar else if (arr[i] == maxSoFar) { // Increment the maxFreq maxFreq++; } // Update maxFreqVec[i] maxFreqVec[i] = maxFreq; } // Print occurrence of maximum // element for each query for (int k : q) { System.out.print( maxFreqVec[k - 1] + " "); } } // Driver Code public static void main(String[] args) { int[] arr = { 5, 4, 5, 3, 2 }; int[] q = { 1, 2, 3, 4, 5 }; // Function Call FreqOfMaxElement(arr, q); }} |
Python3
# Python3 program for # the above approachimport sys;# Function to find occurrence# of max element in given # subarraydef FreqOfMaxElement(arr, q): # Store the frequency of # maximum element maxFreqVec = [0] * (len(arr)); maxSoFar = -sys.maxsize; maxFreq = 0; # Traverse over the array # arr from right to left for i in range(len(arr)-1, -1, -1): # If curr element is # greater than maxSofar if (arr[i] > maxSoFar): # Reset maxSofar # and maxFreq maxSoFar = arr[i]; maxFreq = 1; # If curr is equal to # maxSofar elif (arr[i] == maxSoFar): # Increment the # maxFreq maxFreq += 1; # Update maxFreqVec[i] maxFreqVec[i] = maxFreq; # Proccurrence of maximum # element for each query for i in range(0, len(q)): print(maxFreqVec[q[i] - 1], end = " ");# Driver Codeif __name__ == '__main__': arr = [5, 4, 5, 3, 2]; q = [1, 2, 3, 4, 5]; # Function Call FreqOfMaxElement(arr, q);# This code is contributed by shikhasingrajput |
C#
// C# program for the // above approachusing System;class GFG{ // Function to find occurrence of// max element in given subarraystatic void FreqOfMaxElement(int[] arr, int[] q){ // Store the frequency of // maximum element int[] maxFreqVec = new int[arr.Length]; int maxSoFar = Int32.MinValue; int maxFreq = 0; // Traverse over the array arr[] // from right to left for (int i = arr.Length - 1; i >= 0; i--) { // If curr element is greater // than maxSofar if (arr[i] > maxSoFar) { // Reset maxSofar and maxFreq maxSoFar = arr[i]; maxFreq = 1; } // If curr is equal to maxSofar else if (arr[i] == maxSoFar) { // Increment the maxFreq maxFreq++; } // Update maxFreqVec[i] maxFreqVec[i] = maxFreq; } // Print occurrence of maximum // element for each query foreach (int k in q) { Console.Write(maxFreqVec[k - 1] + " "); }} // Driver codestatic void Main() { int[] arr = {5, 4, 5, 3, 2}; int[] q = {1, 2, 3, 4, 5}; // Function Call FreqOfMaxElement(arr, q); }}// This code is contributed by divyeshrabadiya07 |
2 1 1 1 1
Time Complexity: O(N)
Auxiliary Space: O(N)
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