Count of array elements which are greater than all elements on its left
Given an array arr[] of size N, the task is to count the number of array elements such that all the elements to its left are strictly smaller than it.
Note: The first element of the array will be considered to be always satisfying the condition.
Examples :
Input: arr[] = { 2, 4, 5, 6 }
Output: 4
Explanation:
Since the array is in increasing order, all the array elements satisfy the condition.
Hence, the count of such elements is equal to the size of the array, i.e. equal to 4.
Input: { 3, 3, 3, 3, 3, 3 }
Output: 1
Explanation: The first array element is the only element satisfying the condition.
Approach:
Follow the steps below to solve the problem:
- Set count = 1, as the first array element will be considered to be satisfying the condition.
- Set the first array element(i.e. arr[0]) as the maximum.
- Traverse the array starting from i =1, and compare every array element with the current maximum.
- If an array element is found to be greater than the current maximum, that element satisfies the condition as all the elements on its left are smaller than it. Hence, increase the count, and update the maximum.
- Finally, print the count.
Below is the implementation of the above approach:
C++
#include<bits/stdc++.h>
using namespace std;
int count_elements( int arr[], int n)
{
int count = 1;
int max = arr[0];
for ( int i = 1; i < n; i++)
{
if (arr[i] > max)
{
count += 1;
max = arr[i];
}
}
return count;
}
int main()
{
int arr[] = { 2, 1, 4, 6, 3 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << (count_elements(arr, n));
}
|
Java
import java.util.*;
class GFG{
static int count_elements( int arr[], int n)
{
int count = 1 ;
int max = arr[ 0 ];
for ( int i = 1 ; i < n; i++)
{
if (arr[i] > max)
{
count += 1 ;
max = arr[i];
}
}
return count;
}
public static void main(String s[])
{
int arr[] = { 2 , 1 , 4 , 6 , 3 };
int n = arr.length;
System.out.print(count_elements(arr, n));
}
}
|
Python3
def count_elements(arr):
count = 1
max = arr[ 0 ]
for i in range ( 1 , len (arr)):
if arr[i] > max :
count + = 1
max = arr[i]
return count
arr = [ 2 , 1 , 4 , 6 , 3 ]
print (count_elements(arr))
|
C#
using System;
class GFG{
static int count_elements( int [] arr, int n)
{
int count = 1;
int max = arr[0];
for ( int i = 1; i < n; i++)
{
if (arr[i] > max)
{
count += 1;
max = arr[i];
}
}
return count;
}
public static void Main()
{
int [] arr = { 2, 1, 4, 6, 3 };
int n = arr.Length;
Console.Write(count_elements(arr, n));
}
}
|
Javascript
<script>
function count_elements(arr, n)
{
let count = 1;
let max = arr[0];
for (let i = 1; i < n; i++)
{
if (arr[i] > max)
{
count += 1;
max = arr[i];
}
}
return count;
}
let arr = [ 2, 1, 4, 6, 3 ];
let n = arr.length;
document.write(count_elements(arr, n));
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Last Updated :
09 Jan, 2023
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