Count of numbers in given Array greater than next K elements - GeeksforGeeks

21 Dec, 2021
Given an array arr[] of integers of size N, the task is to count the number of elements whose value is greater than all the K elements to its immediate right. If there are less than K numbers to the right of the ith element, then the value of all of them must be lesser than that of the ith person.
Examples:
Input: arr[] = {4, 3, 1, 2, 5}, N = 5, K = 1Output: 3Explanation: The 1st ,2nd and 5th are the elements whose values are greater than the element to its right(As k = 1 consider only the next). While the 3rd element can not be considered because of the 4th element whose value is greater than the 3rd element’s value. 
Input: arr[] = {9, 7, 7, 7, 4}, N = 5, K = 3Output: 3Explanation: The 1st ,4th and the 5th element will be the elements whose values are greater than all 3 elements elements after it.
Naive Approach: For each element check K elements to their immediate right and see if it is greater than all of those or not.
Time Complexity: O(N2)Auxiliary Space: O(1)
Efficient Approach: This problem can be solved using the following steps:
Consider a queue for storing K elements.
Add K elements from the last to the queue and keep the max of the last K values in a variable (say, max_value).
Iterate over the remaining array in the backward direction from the position (N – K).While iterating pop an element from the queue and push the current element in the queue.If the current element has a value greater than max_value, then increment the count and update max_value to the current element.Else If the popped value is the same as the max_value then find the new max from the queue.
While iterating pop an element from the queue and push the current element in the queue.
If the current element has a value greater than max_value, then increment the count and update max_value to the current element.
Else If the popped value is the same as the max_value then find the new max from the queue.
Return the count value as the count of the element whose value is greater than all the k elements to its immediate right.
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// C++ code to implement the above approach#include <bits/stdc++.h>using namespace std; // Function to find// maximum element in queueint find_max(queue<int> q){    int ans = INT_MIN;     // Loop to find maximum from queue    while (!q.empty()) {        ans = max(ans, q.front());        q.pop();    }    return ans;} // Function to count the elements// whose values are greater than// all the k elements to its immediate rightint solve(int n, int k, vector<int> arr){    int max_value = INT_MIN;    queue<int> q;    int count = 0;    int p = n - k;     // Checking base cases    if (n == 0)        return 0;    else if (k == 0)        return n;     // Traversing last k elements    for (int i = n - 1; i >= p; i--) {        q.push(arr[i]);        if (arr[i] > max_value) {            max_value = arr[i];            count++;        }    }     // Traversing rest of the elements    for (int i = p - 1; i >= 0; i--) {        int x = q.front();        q.pop();        q.push(arr[i]);        if (arr[i] > max_value) {            count++;            max_value = arr[i];        }        else {            if (x == max_value) {                // If popped value                // is same as max value                // then update max value                max_value = find_max(q);            }        }    }    return count;} // Driver Code Starts.int main(){    int N, K;    N = 5;    K = 1;    vector<int> arr = { 4, 3, 1, 2, 5};     cout << solve(N, K, arr) << "\n";    return 0;}
// Java code to implement the above approachimport java.util.ArrayList;import java.util.LinkedList;import java.util.Queue; class GFG {     // Function to find    // maximum element in queue    public static int find_max(Queue<Integer> q) {        int ans = Integer.MIN_VALUE;         // Loop to find maximum from queue        while (!q.isEmpty()) {            ans = Math.max(ans, q.peek());            q.remove();        }        return ans;    }     // Function to count the elements    // whose values are greater than    // all the k elements to its immediate right    public static int solve(int n, int k, ArrayList<Integer> arr) {        int max_value = Integer.MIN_VALUE;        Queue<Integer> q = new LinkedList<Integer>();        int count = 0;        int p = n - k;         // Checking base cases        if (n == 0)            return 0;        else if (k == 0)            return n;         // Traversing last k elements        for (int i = n - 1; i >= p; i--) {            q.add(arr.get(i));            if (arr.get(i) > max_value) {                max_value = arr.get(i);                count++;            }        }         // Traversing rest of the elements        for (int i = p - 1; i >= 0; i--) {            int x = 0;            if (q.size() > 0) {                x = q.peek();                q.remove();            }            q.add(arr.get(i));            if (arr.get(i) > max_value) {                count++;                max_value = arr.get(i);            } else {                if (x == max_value) {                    // If popped value                    // is same as max value                    // then update max value                    max_value = find_max(q);                }            }        }        return count;    }     // Driver Code Starts.    public static void main(String args[]) {        int N, K;        N = 5;        K = 1;        ArrayList<Integer> arr = new ArrayList<>();        arr.add(4);        arr.add(3);        arr.add(1);        arr.add(2);        arr.add(5);         System.out.println(solve(N, K, arr));    }} // This code is contributed by saurabh_jaiswal.
# Python3 code to implement the above approachfrom queue import Queueimport copyINT_MIN = -2147483648 # Function to find# maximum element in queuedef find_max(q):         ans = INT_MIN         # Loop to find maximum from queue    while (not q.empty()):        ans = max(ans, q.get())     return ans     # Function to count the elements# whose values are greater than# all the k elements to its immediate rightdef solve(n, k, arr):     max_value = INT_MIN    q = Queue()    count = 0    p = n - k     # Checking base cases    if (n == 0):        return 0    elif (k == 0):        return n     # Traversing last k elements    for i in range(n - 1, p - 1, -1):        q.put(arr[i])        if (arr[i] > max_value):            max_value = arr[i]            count += 1     # Traversing rest of the elements    for i in range(p - 1, -1, -1):        x = q.get()        q.put(arr[i])                 if (arr[i] > max_value):            count += 1            max_value = arr[i]         else:            if (x == max_value):                                 # If popped value is same                # as max value then update                # max value                temp = Queue()                for i in q.queue:                    temp.put(i)                                     max_value = find_max(temp)     return count # Driver codeif __name__ == "__main__":     N = 5    K = 1    arr = [ 4, 3, 1, 2, 5 ]     print(solve(N, K, arr)) # This code is contributed by rakeshsahni
// C# code to implement the above approachusing System;using System.Collections.Generic; public class GFG {     // Function to find    // maximum element in queue    public static int find_max(Queue<int> q) {        int ans = int.MinValue;         // Loop to find maximum from queue        while (q.Count != 0) {            ans = Math.Max(ans, q.Peek());            q.Dequeue();        }        return ans;    }     // Function to count the elements    // whose values are greater than    // all the k elements to its immediate right    public static int solve(int n, int k, List<int> arr) {        int max_value = int.MinValue;        Queue<int> q = new Queue<int>();        int count = 0;        int p = n - k;         // Checking base cases        if (n == 0)            return 0;        else if (k == 0)            return n;         // Traversing last k elements        for (int i = n - 1; i >= p; i--) {            q.Enqueue(arr[i]);            if (arr[i] > max_value) {                max_value = arr[i];                count++;            }        }         // Traversing rest of the elements        for (int i = p - 1; i >= 0; i--) {            int x = 0;            if (q.Count > 0) {                x = q.Peek();                q.Dequeue();            }            q.Enqueue(arr[i]);            if (arr[i] > max_value) {                count++;                max_value = arr[i];            } else {                if (x == max_value) {                    // If popped value                    // is same as max value                    // then update max value                    max_value = find_max(q);                }            }        }        return count;    }     // Driver Code Starts.    public static void Main(String []args) {        int N, K;        N = 5;        K = 1;        List<int> arr = new List<int>();        arr.Add(4);        arr.Add(3);        arr.Add(1);        arr.Add(2);        arr.Add(5);         Console.WriteLine(solve(N, K, arr));    }} // This code is contributed by shikhasingrajput
<script>// Javascript code to implement the above approach  // Function to find// maximum element in queuefunction find_max(q) {    let ans = Number.MIN_SAFE_INTEGER;     // Loop to find maximum from queue    while (q.length) {        ans = Math.max(ans, q[0]);        q.pop();    }    return ans;} // Function to count the elements// whose values are greater than// all the k elements to its immediate rightfunction solve(n, k, arr) {    let max_value = Number.MIN_SAFE_INTEGER;    let q = [];    let count = 0;    let p = n - k;     // Checking base cases    if (n == 0)        return 0;    else if (k == 0)        return n;     // Traversing last k elements    for (let i = n - 1; i >= p; i--) {        q.push(arr[i]);        if (arr[i] > max_value) {            max_value = arr[i];            count++;        }    }     // Traversing rest of the elements    for (let i = p - 1; i >= 0; i--) {        let x = q[0];        q.pop();        q.push(arr[i]);        if (arr[i] > max_value) {            count++;            max_value = arr[i];        }        else {            if (x == max_value) {                // If popped value                // is same as max value                // then update max value                max_value = find_max(q);            }        }    }    return count;} // Driver Code Starts. let N, K;N = 5;K = 1;let arr = [4, 3, 1, 2, 5]; document.write(solve(N, K, arr)) // This code is contributed by gfgking.</script>
3
Time Complexity: O(N)Auxiliary Space: O(K)
Space Optimized Approach: The problem can be solved with lesser space using two pointer approach. Follow the steps mentioned below. 
Initialize two-pointer (say “left” and “right”) pointing to the end of the array.
The left pointer points to the starting index of the K-sized window and the right pointer points to the maximum value in that window.
If the value at the left pointer is greater than the value at the right pointer, increase answer count by 1. (Because it means it is greater than all the K elements to its immediate right)
If at any point the window size exceeds K, decrement the right pointer by one and adjust it to point to the maximum value of the current window.
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// C++ code to implement the above approach#include <bits/stdc++.h>using namespace std; // Function to count the elements// whose values are greater than// all k elements to its immediate rightint solve(int n, int k, vector<int> arr){    int count = 1;    int left = n - 2, right = n - 1;     // Checking base cases    if (n == 0)        return 0;    else if (k == 0)        return n;         // Loop to implement two-pointer approach    for (; left >= 0; left--) {        if (right - left > k) {            right--;            while (arr[left] > arr[right])                right--;            if (right == left)                count++;        }        else if (arr[left] > arr[right]) {            count++;            right = left;        }    }     return count;} // Driver Code Starts.int main(){    int N, K;    N = 5;    K = 1;    vector<int> arr = { 4, 3, 1, 2, 5};    cout << solve(N, K, arr);    return 0;}
// Java code to implement the above approachimport java.io.*; class GFG{ // Function to count the elements// whose values are greater than// all k elements to its immediate rightstatic int solve(int n, int k, int[] arr){    int count = 1;    int left = n - 2, right = n - 1;     // Checking base cases    if (n == 0)        return 0;    else if (k == 0)        return n;     // Loop to implement two-pointer approach    for(; left >= 0; left--)    {        if (right - left > k)        {            right--;                         while (arr[left] > arr[right])                right--;            if (right == left)                count++;        }        else if (arr[left] > arr[right])        {            count++;            right = left;        }    }    return count;} // Driver Codepublic static void main(String[] args){    int N, K;    N = 5;    K = 1;    int[] arr = { 4, 3, 1, 2, 5 };     System.out.println(solve(N, K, arr));}} // This code is contributed by Potta Lokesh
# Python 3 code to implement the above approach # Function to count the elements# whose values are greater than# all k elements to its immediate rightdef solve(n,  k, arr):     count = 1    left = n - 2    right = n - 1     # Checking base cases    if (n == 0):        return 0    elif (k == 0):        return n     # Loop to implement two-pointer approach    while left >= 0:        if (right - left > k):            right -= 1            while (arr[left] > arr[right]):                right -= 1            if (right == left):                count += 1         elif (arr[left] > arr[right]):            count += 1            right = left         left -= 1     return count # Driver Codeif __name__ == "__main__":     N = 5    K = 1    arr = [4, 3, 1, 2, 5]    print(solve(N, K, arr))     # This code is contributed by ukasp.
// C# code to implement the above approachusing System; public class GFG{ // Function to count the elements// whose values are greater than// all k elements to its immediate rightstatic int solve(int n, int k, int[] arr){    int count = 1;    int left = n - 2, right = n - 1;     // Checking base cases    if (n == 0)        return 0;    else if (k == 0)        return n;     // Loop to implement two-pointer approach    for(; left >= 0; left--)    {        if (right - left > k)        {            right--;                         while (arr[left] > arr[right])                right--;            if (right == left)                count++;        }        else if (arr[left] > arr[right])        {            count++;            right = left;        }    }    return count;} // Driver Codepublic static void Main(String[] args){    int N, K;    N = 5;    K = 1;    int[] arr = { 4, 3, 1, 2, 5 };     Console.WriteLine(solve(N, K, arr));}} // This code is contributed by 29AjayKumar
<script>// javascript code to implement the above approach // Function to count the elements// whose values are greater than// all k elements to its immediate rightfunction solve(n , k, arr){    var count = 1;    var left = n - 2, right = n - 1;     // Checking base cases    if (n == 0)        return 0;    else if (k == 0)        return n;     // Loop to implement two-pointer approach    for(; left >= 0; left--)    {        if (right - left > k)        {            right--;                         while (arr[left] > arr[right])                right--;            if (right == left)                count++;        }        else if (arr[left] > arr[right])        {            count++;            right = left;        }    }    return count;} // Driver Codevar N, K;N = 5;K = 1;var arr = [ 4, 3, 1, 2, 5 ]; document.write(solve(N, K, arr)); // This code is contributed by 29AjayKumar</script>
3
Time Complexity: O(N)Auxiliary Space: O(1)
 
lokeshpotta20
rakeshsahni
29AjayKumar
gfgking
_saurabh_jaiswal
shikhasingrajput
ukasp
two-pointer-algorithm
Arrays
Queue
two-pointer-algorithm
Arrays
Queue
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