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Count of indices in an array that satisfy the given condition

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  • Difficulty Level : Easy
  • 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:
All indices satisfy the given condition.
Input: arr[] = {4, 3, 2, 1} 
Output:
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#




// 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: 

4

 

Time Complexity: O(n)

Auxiliary Space: O(1)


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