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Count array elements exceeding all previous elements as well as the next array element
  • Last Updated : 22 Jan, 2021

Given an array arr[], the task is to find the count of array elements satisfying the following conditions:

  • The array elements should be strictly greater than all the previously occurring array elements.
  • Either it is the last array element or the integer should be strictly larger than the next array element.

Note: The first integer of the array can also be considered.

Examples:

Input: arr[] = {1, 2, 0, 7, 2, 0, 2, 0}
Output: 2
Explanation: arr[1] (= 2) and arr[3] ( = 7) are the array elements satisfying the given condition.

Input: arr[] = {4, 8, 15, 16, 23, 42}
Output: 1

Approach: The idea is to linearly traverse the array and check for each array element, if it satisfies the given condition or not. Follow the steps below to solve this problem:



  • Traverse the array.
  • Starting from the first element of the array, keep track of the maximum array element encountered so far.
  • Update the maximum array element if it’s greater than the previous maximum array element encountered.
  • After updating the current maximum, check if the next array element is greater than the current array element or not. If found to be true, increment the count.
  • Repeat this process until the last array element is traversed.
  • Finally, print the count obtained.

Below is the implementation for the above approach :

C++

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// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count array elements
// satisfying the given condition
int numberOfIntegers(int arr[], int N)
{
    int cur_max = 0, count = 0;
 
    // If there is only one
    // array element
    if (N == 1) {
        count = 1;
    }
    else {
 
        // Traverse the array
        for (int i = 0; i < N - 1; i++) {
 
            // Update the maximum element
            // encountered so far
            if (arr[i] > cur_max) {
                cur_max = arr[i];
 
                // Count the number of array elements
                // strictly greater than all previous
                // and immediately next elements
                if (arr[i] > arr[i + 1]) {
 
                    count++;
                }
            }
        }
        if (arr[N - 1] > cur_max)
            count++;
    }
 
    // Print the count
    cout << count;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 1, 2, 0, 7, 2, 0, 2, 0 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    numberOfIntegers(arr, N);
 
    return 0;
}

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Java

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// Java program for the above approach
import java.io.*;
class GFG {
 
    // Function to count array elements
    // satisfying the given condition
    static void numberOfIntegers(int[] arr, int N)
    {
        int cur_max = 0, count = 0;
 
        // If there is only one
        // array element
        if (N == 1) {
            count = 1;
        }
        else
        {
 
            // Traverse the array
            for (int i = 0; i < N - 1; i++)
            {
 
                // Update the maximum element
                // encountered so far
                if (arr[i] > cur_max)
                {
                    cur_max = arr[i];
 
                    // Count the number of array elements
                    // strictly greater than all previous
                    // and immediately next elements
                    if (arr[i] > arr[i + 1])
                    {
                        count++;
                    }
                }
            }
            if (arr[N - 1] > cur_max)
                count++;
        }
 
        // Print the count
        System.out.println(count);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        // Given array
        int[] arr = new int[] { 1, 2, 0, 7, 2, 0, 2, 0 };
 
        // Size of the array
        int N = arr.length;
        numberOfIntegers(arr, N);
    }
}
 
// This code is contributed by dharanendralv23

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Python3

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# Python program for the above approach
 
# Function to count array elements
# satisfying the given condition
def numberOfIntegers(arr, N) :  
    cur_max = 0
    count = 0
 
    # If there is only one
    # array element
    if (N == 1) :
        count = 1   
    else :
 
        # Traverse the array
        for i in range(N - 1):
 
            # Update the maximum element
            # encountered so far
            if (arr[i] > cur_max) :
                cur_max = arr[i]
 
                # Count the number of array elements
                # strictly greater than all previous
                # and immediately next elements
                if (arr[i] > arr[i + 1]) :
                    count += 1                  
        if (arr[N - 1] > cur_max) :
            count += 1
     
    # Prthe count
    print(count)
 
# Driver Code
 
# Given array
arr = [ 1, 2, 0, 7, 2, 0, 2, 0 ]
 
# Size of the array
N = len(arr)
numberOfIntegers(arr, N)
 
# This code is contributed by sanjoy_62.

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C#

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// C# program for the above approach
using System;
class GFG
{
 
    // Function to count array elements
    // satisfying the given condition
    static void numberOfIntegers(int[] arr, int N)
    {
        int cur_max = 0, count = 0;
 
        // If there is only one
        // array element
        if (N == 1) {
            count = 1;
        }
        else {
 
            // Traverse the array
            for (int i = 0; i < N - 1; i++) {
 
                // Update the maximum element
                // encountered so far
                if (arr[i] > cur_max) {
                    cur_max = arr[i];
 
                    // Count the number of array elements
                    // strictly greater than all previous
                    // and immediately next elements
                    if (arr[i] > arr[i + 1]) {
 
                        count++;
                    }
                }
            }
            if (arr[N - 1] > cur_max)
                count++;
        }
 
        // Print the count
        Console.WriteLine(count);
    }
 
    // Driver Code
    static public void Main()
    {
 
        // Given array
        int[] arr = new int[] { 1, 2, 0, 7, 2, 0, 2, 0 };
 
        // Size of the array
        int N = arr.Length;
 
        numberOfIntegers(arr, N);
    }
}
 
// This code is contributed by dharavendralv23

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Output: 

2

 

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

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