Count elements less than or equal to a given value in a sorted rotated array

Last Updated : 14 Feb, 2023

Given a sorted array of n distinct integers rotated at some point. Given a value x. The problem is to count all the elements in the array which are less than or equal to x.

Examples:

Input : arr[] = {4, 5, 8, 1, 3}, 
            x = 6
Output : 4

Input : arr[] = {6, 10, 12, 15, 2, 4, 5}, 
            x = 14
Output : 6

Naive Approach: One by one traverse all the elements of the array and count the one's which are less than or equal to x. Time Complexity O(n).

Efficient Approach: Prerequisite of a modified binary search which can return the index of the largest element smaller than or equal to a given value in a sorted range of arr[l...h].

Refer this post for the required modified binary search:

Steps:

Find the index of the smallest element in rotated sorted array. Refer this post. Let it be min_index.
If x <= arr[n-1], find the index of the largest element smaller than or equal to x in the sorted range arr[min_index...n-1] with the help of modified binary search.
Let it be index1. Now, count = index1 + 1 - min_index.
If 0 <= (min_index -1) && x <= arr[min_index - 1], find the index of the largest element smaller than or equal to x in the sorted range arr[0...min_index-1] with the help of modified binary search. Let it be index2. Now, count = n - min_index + index2 + 1.
Else count = n.

C++

// C++ implementation to count elements less than or 
// equal to a given value in a sorted rotated array
#include <bits/stdc++.h>

using namespace std;

// function to find the minimum element's index
int findMinIndex(int arr[], int low, int high)
{
    // This condition is needed to handle the case when 
    // array is not rotated at all
    if (high < low)  return 0;
 
    // If there is only one element left
    if (high == low) return low;
 
    // Find mid
    int mid = (low + high) / 2;
 
    // Check if element (mid+1) is minimum element. Consider
    // the cases like {3, 4, 5, 1, 2}
    if (mid < high && arr[mid+1] < arr[mid])
       return (mid + 1);
 
    // Check if mid itself is minimum element
    if (mid > low && arr[mid] < arr[mid - 1])
       return mid;
 
    // Decide whether we need to go to left half or right half
    if (arr[high] > arr[mid])
       return findMinIndex(arr, low, mid-1);
    return findMinIndex(arr, mid+1, high);
}

// function returns the index of largest element 
// smaller than equal to 'x' in 'arr[l...h]'. 
// If no such element exits in the given range, 
// then it returns l-1. 
int binary_search(int arr[], int l, int h, int x)
{
    while (l <= h)
    {
        int mid = (l+h) / 2;

        // if 'x' is less than or equal to arr[mid], 
        // then search in arr[mid+1...h]
        if (arr[mid] <= x)
            l = mid + 1;

        // else search in arr[l...mid-1]    
        else
            h = mid - 1;    
    }
    
    // required index
    return h;
}

// function to count elements less than 
// or equal to a given value
int countEleLessThanOrEqual(int arr[], int n, int x)
{
    // index of the smallest element in the array
    int min_index = findMinIndex(arr, 0, n-1);
    
    // if largest element smaller than or equal to 'x' lies
    // in the sorted range arr[min_index...n-1]
    if (x <= arr[n-1])
        return (binary_search(arr, min_index, n-1, x) + 1 - min_index);
    
    // if largest element smaller than or equal to 'x' lies
    // in the sorted range arr[0...min_index-1]
    if ((min_index - 1) >= 0 && x <= arr[min_index - 1])
        return (n - min_index + binary_search(arr, 0, min_index-1, x) + 1);
    
    // else all the elements of the array
    // are less than 'x'    
    return n;                
}

// Driver program to test above
int main()
{
    int arr[] = {6, 10, 12, 15, 2, 4, 5};
    int n = sizeof(arr) / sizeof(arr[0]);
    int x = 14;
    cout << "Count = "
         << countEleLessThanOrEqual(arr, n, x);
    return 0;
}

Java

// Java implementation to count elements
// less than or equal to a given
// value in a sorted rotated array

class GFG {
    
// function to find the minimum 
// element's index
static int findMinIndex(int arr[], int low, int high) 
{
    // This condition is needed to handle 
    // the case when array is not rotated at all
    if (high < low)
    return 0;

    // If there is only one element left
    if (high == low)
    return low;

    // Find mid
    int mid = (low + high) / 2;

    // Check if element (mid+1) is 
    // minimum element. Consider
    // the cases like {3, 4, 5, 1, 2}
    if (mid < high && arr[mid + 1] < arr[mid])
    return (mid + 1);

    // Check if mid itself is minimum element
    if (mid > low && arr[mid] < arr[mid - 1])
    return mid;

    // Decide whether we need to go to 
    // left half or right half
    if (arr[high] > arr[mid])
    return findMinIndex(arr, low, mid - 1);
    return findMinIndex(arr, mid + 1, high);
}

// function returns the index of largest element
// smaller than equal to 'x' in 'arr[l...h]'.
// If no such element exits in the given range,
// then it returns l-1.
static int binary_search(int arr[], int l, int h, int x) 
{
    while (l <= h) {
    int mid = (l + h) / 2;

    // if 'x' is less than or equal to arr[mid],
    // then search in arr[mid+1...h]
    if (arr[mid] <= x)
        l = mid + 1;

    // else search in arr[l...mid-1]
    else
        h = mid - 1;
    }

    // required index
    return h;
}

// function to count elements less than
// or equal to a given value
static int countEleLessThanOrEqual(int arr[], int n, int x) 
{
    // index of the smallest element in the array
    int min_index = findMinIndex(arr, 0, n - 1);

    // if largest element smaller than or 
    // equal to 'x' lies in the sorted
    // range arr[min_index...n-1]
    if (x <= arr[n - 1])
    return (binary_search(arr, min_index, n - 1, x) + 1 - min_index);

    // if largest element smaller than or 
    // equal to 'x' lies in the sorted
    // range arr[0...min_index-1]
    if ((min_index - 1) >= 0 && x <= arr[min_index - 1])
    return (n - min_index + binary_search(arr, 0, min_index - 1, x) + 1);

    // else all the elements of the array
    // are less than 'x'
    return n;
}

// Driver code
public static void main(String[] args) 
{
    int arr[] = {6, 10, 12, 15, 2, 4, 5};
    int n = arr.length;
    int x = 14;
    System.out.print("Count = " + 
                      countEleLessThanOrEqual(arr, n, x));
}
}

// This code is contributed by Anant Agarwal.

Python3

# Python implementation to
# count elements less than or 
# equal to a given value
# in a sorted rotated array

# function to find the
# minimum element's index
def findMinIndex(arr,low,high):

    # This condition is needed
    # to handle the case when 
    # array is not rotated at all
    if (high < low):
        return 0
  
    # If there is only one element left
    if (high == low):
        return low
  
    # Find mid
    mid = (low + high) // 2
  
    # Check if element (mid+1) is
    # minimum element. Consider
    # the cases like {3, 4, 5, 1, 2}
    if (mid < high and arr[mid+1] < arr[mid]):
       return (mid + 1)
  
    # Check if mid itself
    # is minimum element
    if (mid > low and arr[mid] < arr[mid - 1]):
       return mid
  
    # Decide whether we need to
    # go to left half or right half
    if (arr[high] > arr[mid]):
       return findMinIndex(arr, low, mid-1)
    return findMinIndex(arr, mid+1, high)

# function returns the 
# index of largest element 
# smaller than equal to 
# 'x' in 'arr[l...h]'. 
# If no such element exits
# in the given range, 
# then it returns l-1. 
def binary_search(arr,l,h,x):

    while (l <= h):
    
        mid = (l+h) // 2
 
        # if 'x' is less than
        # or equal to arr[mid], 
        # then search in arr[mid+1...h]
        if (arr[mid] <= x):
            l = mid + 1
 
        # else search in arr[l...mid-1]    
        else:
            h = mid - 1    
    
    # required index
    return h
 
# function to count
# elements less than 
# or equal to a given value
def countEleLessThanOrEqual(arr,n,x):

    # index of the smallest
    # element in the array
    min_index = findMinIndex(arr, 0, n-1)
     
    # if largest element smaller
    # than or equal to 'x' lies
    # in the sorted range arr[min_index...n-1]
    if (x <= arr[n-1]):
        return (binary_search(arr, min_index, n-1, x) + 1 - min_index)
     
    # if largest element smaller
    # than or equal to 'x' lies
    # in the sorted range arr[0...min_index-1]
    if ((min_index - 1) >= 0 and x <= arr[min_index - 1]):
        return (n - min_index + binary_search(arr, 0, min_index-1, x) + 1)
     
    # else all the elements of the array
    # are less than 'x'    
    return n                

# driver code
arr = [6, 10, 12, 15, 2, 4, 5]
n = len(arr)
x = 14

print("Count = ",end="")
print(countEleLessThanOrEqual(arr, n, x))

# This code is contributed
# by Anant Agarwal.

// C# implementation to count elements
// less than or equal to a given
// value in a sorted rotated array
using System;
        
public class GFG {
    
    // function to find the minimum 
    // element's index
    static int findMinIndex(int []arr, int low,
                                      int high) 
    {
        
        // This condition is needed to handle 
        // the case when array is not rotated
        // at all
        if (high < low)
            return 0;
    
        // If there is only one element left
        if (high == low)
            return low;
    
        // Find mid
        int mid = (low + high) / 2;
    
        // Check if element (mid+1) is 
        // minimum element. Consider
        // the cases like {3, 4, 5, 1, 2}
        if (mid < high && arr[mid + 1] < arr[mid])
            return (mid + 1);
    
        // Check if mid itself is minimum element
        if (mid > low && arr[mid] < arr[mid - 1])
            return mid;
    
        // Decide whether we need to go to 
        // left half or right half
        if (arr[high] > arr[mid])
            return findMinIndex(arr, low, mid - 1);
            
        return findMinIndex(arr, mid + 1, high);
    }
    
    // function returns the index of largest element
    // smaller than equal to 'x' in 'arr[l...h]'.
    // If no such element exits in the given range,
    // then it returns l-1.
    static int binary_search(int []arr, int l, 
                                       int h, int x) 
    {
        while (l <= h)
        {
            int mid = (l + h) / 2;
        
            // if 'x' is less than or equal to
            // arr[mid], then search in 
            // arr[mid+1...h]
            if (arr[mid] <= x)
                l = mid + 1;
        
            // else search in arr[l...mid-1]
            else
                h = mid - 1;
        }
    
        // required index
        return h;
    }
    
    // function to count elements less than
    // or equal to a given value
    static int countEleLessThanOrEqual(int []arr,
                                      int n, int x) 
    {
        
        // index of the smallest element in
        // the array
        int min_index = findMinIndex(arr, 0, n - 1);
    
        // if largest element smaller than or 
        // equal to 'x' lies in the sorted
        // range arr[min_index...n-1]
        if (x <= arr[n - 1])
            return (binary_search(arr, min_index,
                          n - 1, x) + 1 - min_index);
    
        // if largest element smaller than or 
        // equal to 'x' lies in the sorted
        // range arr[0...min_index-1]
        if ((min_index - 1) >= 0 && 
                             x <= arr[min_index - 1])
            return (n - min_index + binary_search(arr,
                             0, min_index - 1, x) + 1);
    
        // else all the elements of the array
        // are less than 'x'
        return n;
    }
    
    // Driver code
    public static void Main() 
    {
        int []arr = {6, 10, 12, 15, 2, 4, 5};
        int n = arr.Length;
        int x = 14;
        
        Console.Write("Count = " + 
                countEleLessThanOrEqual(arr, n, x));
    }
}

// This code is contributed by Sam007.

PHP

<?php 
// PHP implementation to count elements 
// less than or equal to a given value 
// in a sorted rotated array

// function to find the minimum 
// element's index
function findMinIndex(&$arr, $low, $high)
{
    // This condition is needed to
    // handle the case when array
    // is not rotated at all
    if ($high < $low) return 0;

    // If there is only one element left
    if ($high == $low) return $low;

    // Find mid
    $mid = intval(($low + $high) / 2);

    // Check if element (mid+1) is
    // minimum element. Consider
    // the cases like {3, 4, 5, 1, 2}
    if ($mid < $high && 
        $arr[$mid + 1] < $arr[$mid])
    return ($mid + 1);

    // Check if mid itself is minimum element
    if ($mid > $low &&
        $arr[$mid] < $arr[$mid - 1])
    return $mid;

    // Decide whether we need to go
    // to left half or right half
    if ($arr[$high] > $arr[$mid])
    return findMinIndex($arr, $low, $mid - 1);
    return findMinIndex($arr, $mid + 1, $high);
}

// function returns the index of largest
// element smaller than equal to 'x' in  
// 'arr[l...h]'. If no such element exits 
// in the given range, then it returns l-1. 
function binary_search(&$arr, $l, $h, $x)
{
    while ($l <= $h)
    {
        $mid = intval(($l + $h) / 2);

        // if 'x' is less than or equal 
        // to arr[mid], then search in
        // arr[mid+1...h]
        if ($arr[$mid] <= $x)
            $l = $mid + 1;

        // else search in arr[l...mid-1] 
        else
            $h = $mid - 1; 
    }
    
    // required index
    return $h;
}

// function to count elements less 
// than or equal to a given value
function countEleLessThanOrEqual(&$arr, $n, $x)
{
    // index of the smallest element 
    // in the array
    $min_index = findMinIndex($arr, 0, $n - 1);
    
    // if largest element smaller than 
    // or equal to 'x' lies in the sorted 
    // range arr[min_index...n-1]
    if ($x <= $arr[$n - 1])
        return (binary_search($arr, $min_index, 
                              $n - 1, $x) + 1 - $min_index);
    
    // if largest element smaller than or 
    // equal to 'x' lies in the sorted
    // range arr[0...min_index-1]
    if (($min_index - 1) >= 0 && 
         $x <= $arr[$min_index - 1])
        return ($n - $min_index + 
                binary_search($arr, 0, 
                              $min_index - 1, $x) + 1);
    
    // else all the elements of 
    // the array are less than 'x' 
    return $n;             
}

// Driver Code
$arr = array(6, 10, 12, 15, 2, 4, 5);
$n = sizeof($arr);
$x = 14;
echo "Count = " . countEleLessThanOrEqual($arr, $n, $x);

// This code is contributed 
// by ChitraNayal
?>

JavaScript

<script>
// Javascript implementation to count elements
// less than or equal to a given
// value in a sorted rotated array
    
    // function to find the minimum 
    // element's index
    function findMinIndex(arr,low,high)
    {
        // This condition is needed to handle 
    // the case when array is not rotated at all
    if (high < low)
        return 0;
  
    // If there is only one element left
    if (high == low)
        return low;
  
    // Find mid
    let mid = Math.floor((low + high) / 2);
  
    // Check if element (mid+1) is 
    // minimum element. Consider
    // the cases like {3, 4, 5, 1, 2}
    if (mid < high && arr[mid + 1] < arr[mid])
        return (mid + 1);
  
    // Check if mid itself is minimum element
    if (mid > low && arr[mid] < arr[mid - 1])
        return mid;
  
    // Decide whether we need to go to 
    // left half or right half
    if (arr[high] > arr[mid])
        return findMinIndex(arr, low, mid - 1);
    return findMinIndex(arr, mid + 1, high);
    }
    
    // function returns the index of largest element
// smaller than equal to 'x' in 'arr[l...h]'.
// If no such element exits in the given range,
// then it returns l-1.
    function binary_search(arr,l,h,x)
    {
        while (l <= h) {
    let mid = Math.floor((l + h) / 2);
  
    // if 'x' is less than or equal to arr[mid],
    // then search in arr[mid+1...h]
    if (arr[mid] <= x)
        l = mid + 1;
  
    // else search in arr[l...mid-1]
    else
        h = mid - 1;
    }
  
    // required index
    return h;
    }
    
    // function to count elements less than
// or equal to a given value
    function countEleLessThanOrEqual(arr,n,x)
    {
        // index of the smallest element in the array
    let min_index = findMinIndex(arr, 0, n - 1);
  
    // if largest element smaller than or 
    // equal to 'x' lies in the sorted
    // range arr[min_index...n-1]
    if (x <= arr[n - 1])
    return (binary_search(arr, min_index, n - 1, x) + 1 - min_index);
  
    // if largest element smaller than or 
    // equal to 'x' lies in the sorted
    // range arr[0...min_index-1]
    if ((min_index - 1) >= 0 && x <= arr[min_index - 1])
        return (n - min_index + binary_search(arr, 0, min_index - 1, x) + 1);
  
    // else all the elements of the array
    // are less than 'x'
    return n;
    }
    
    
    // Driver code
    let arr=[6, 10, 12, 15, 2, 4, 5];
    let n = arr.length;
    let x = 14;
    document.write("Count = " + 
                      countEleLessThanOrEqual(arr, n, x));
      
    
    // This code is contributed by rag2127
</script>

Output

Count = 6

Time Complexity: O(logn) as the binary search is used which takes logarithmic time.

The space complexity is O(1).

Count of smaller or equal elements in sorted array

Ayush Jauhari

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Count elements less than or equal to a given value in a sorted rotated array

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