Count of indices in an array that satisfy the given condition

Last Updated : 01 Mar, 2022

Given an array arr[] of N positive integers, the task is to find the count of indices i such that all the elements from arr[0] to arr[i - 1] are smaller than arr[i].

Examples:

Input: arr[] = {1, 2, 3, 4}
Output: 4
All indices satisfy the given condition.
Input: arr[] = {4, 3, 2, 1}
Output: 1
Only i = 0 is the valid index.

Approach: The idea is to traverse the array from left to right and keep track of the current maximum, whenever this maximum changes then the current index is a valid index so increment the resulting counter.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count
// of indices that satisfy
// the given condition
int countIndices(int arr[], int n)
{

    // To store the result
    int cnt = 0;

    // To store the current maximum
    // Initialized to 0 since there are only
    // positive elements in the array
    int max = 0;
    for (int i = 0; i < n; i++) {

        // i is a valid index
        if (max < arr[i]) {

            // Update the maximum so far
            max = arr[i];

            // Increment the counter
            cnt++;
        }
    }

    return cnt;
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(int);

    cout << countIndices(arr, n);

    return 0;
}

Java

// Java implementation of the approach
class GFG 
{

// Function to return the count
// of indices that satisfy
// the given condition
static int countIndices(int arr[], int n)
{

    // To store the result
    int cnt = 0;

    // To store the current maximum
    // Initialized to 0 since there are only
    // positive elements in the array
    int max = 0;
    for (int i = 0; i < n; i++) 
    {

        // i is a valid index
        if (max < arr[i]) 
        {

            // Update the maximum so far
            max = arr[i];

            // Increment the counter
            cnt++;
        }
    }
    return cnt;
}

// Driver code
public static void main(String[] args) 
{
    int arr[] = { 1, 2, 3, 4 };
    int n = arr.length;

    System.out.println(countIndices(arr, n));
}
}

// This code is contributed by Rajput-Ji

Python3

# Python implementation of the approach

# Function to return the count
# of indices that satisfy
# the given condition
def countIndices(arr, n):

    # To store the result
    cnt = 0;

    # To store the current maximum
    # Initialized to 0 since there are only
    # positive elements in the array
    max = 0;
    for i in range(n):
        # i is a valid index
        if (max < arr[i]): 

            # Update the maximum so far
            max = arr[i];

            # Increment the counter
            cnt += 1;

    return cnt;

# Driver code
if __name__ == '__main__':
    arr = [ 1, 2, 3, 4 ];
    n = len(arr);

    print(countIndices(arr, n));

# This code is contributed by 29AjayKumar

// C# implementation of the approach
using System;
    
class GFG 
{

// Function to return the count
// of indices that satisfy
// the given condition
static int countIndices(int []arr, int n)
{

    // To store the result
    int cnt = 0;

    // To store the current maximum
    // Initialized to 0 since there are only
    // positive elements in the array
    int max = 0;
    for (int i = 0; i < n; i++) 
    {

        // i is a valid index
        if (max < arr[i]) 
        {

            // Update the maximum so far
            max = arr[i];

            // Increment the counter
            cnt++;
        }
    }
    return cnt;
}

// Driver code
public static void Main(String[] args) 
{
    int []arr = { 1, 2, 3, 4 };
    int n = arr.Length;

    Console.WriteLine(countIndices(arr, n));
}
}

// This code is contributed by PrinciRaj1992

JavaScript

<script>
// javascript implementation of the approach    
// Function to return the count
    // of indices that satisfy
    // the given condition
    function countIndices(arr , n) {

        // To store the result
        var cnt = 0;

        // To store the current maximum
        // Initialized to 0 since there are only
        // positive elements in the array
        var max = 0;
        for (i = 0; i < n; i++) {

            // i is a valid index
            if (max < arr[i]) {

                // Update the maximum so far
                max = arr[i];

                // Increment the counter
                cnt++;
            }
        }
        return cnt;
    }

    // Driver code
    
        var arr = [ 1, 2, 3, 4 ];
        var n = arr.length;

        document.write(countIndices(arr, n));

// This code contributed by aashish1995 
</script>

Output:

Time Complexity: O(n)

Auxiliary Space: O(1)

Count of indices in an array that satisfy the given condition

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Count of indices in an array that satisfy the given condition

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