Queries for counts of array elements with values in given range

2.5

Given an unsorted array of size n, find no of elements between two elements i and j (both inclusive).

Examples:

Input :  arr = [1 3 3 9 10 4] 
         i1 = 1, j1 = 4
         i2 = 9, j2 = 12
Output : 4
         2
The numbers are: 1 3 3 4 for first query
The numbers are: 9 12 for second query

Source: Amazon Interview Experience

A simple approach will be to run a for loop to check if each element is in the given range and maintain their count. Time complexity for running each query will be O(n).

C++

// Simple C++ program to count number of elements
// with values in given range.
#include <iostream>
using namespace std;

// function to count elements within given range
int countInRange(int arr[], int n, int x, int y)
{
    // initialize result
    int count = 0;
    for (int i = 0; i < n; i++) {

        // check if element is in range
        if (arr[i] >= x && arr[i] <= y)
            count++;
    }
    return count;
}

// driver function
int main()
{
    int arr[] = { 1, 3, 4, 9, 10, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Answer queries
    int i = 1, j = 4;
    cout << countInRange(arr, n, i, j) << endl;

    i = 9, j = 12;
    cout << countInRange(arr, n, i, j) << endl;
    return 0;
}

Python

# function to count elements within given range
def countInRange(arr, n, x, y):

    # initialize result
    count = 0;

    for i in range(n):

        # check if element is in range
        if (arr[i] >= x and arr[i] <= y):
            count += 1
    return count

# driver function
arr = [1, 3, 4, 9, 10, 3]
n = len(arr)

# Answer queries
i = 1
j = 4
print(countInRange(arr, n, i, j))
i = 9
j = 12
print(countInRange(arr, n, i, j))

Output:

4
2

 

An Efficient Approach will be to first sort the array and then using a modified binary search function find two indices, one of first element greater than or equal to lower bound of range and the other of the last element less than or equal to upperbound. Time for running each query will be O(logn) and for sorting the array once will be O(nlogn).

C++

// Efficient C++ program to count number of elements
// with values in given range.
#include <bits/stdc++.h>
using namespace std;

// function to find first index >= x
int lowerIndex(int arr[], int n, int x)
{
    int l = 0, h = n - 1;
    while (l <= h) {
        int mid = (l + h) / 2;
        if (arr[mid] >= x)
            h = mid - 1;
        else
            l = mid + 1;
    }
    return l;
}

// function to find last index <= y
int upperIndex(int arr[], int n, int y)
{
    int l = 0, h = n - 1;
    while (l <= h) {
        int mid = (l + h) / 2;
        if (arr[mid] <= y)
            l = mid + 1;
        else
            h = mid - 1;
    }
    return h;
}

// function to count elements within given range
int countInRange(int arr[], int n, int x, int y)
{
    // initialize result
    int count = 0;
    count = upperIndex(arr, n, y) - lowerIndex(arr, n, x) + 1;
    return count;
}

// driver function
int main()
{
    int arr[] = { 1, 4, 4, 9, 10, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Preprocess array
    sort(arr, arr + n);

    // Answer queries
    int i = 1, j = 4;
    cout << countInRange(arr, n, i, j) << endl;

    i = 9, j = 12;
    cout << countInRange(arr, n, i, j) << endl;
    return 0;
}

Python

# function to find first index >= x
def lowerIndex(arr, n, x):
  l = 0
  h = n-1
  while (l <= h):
    mid = int((l + h)/2)
    if (arr[mid] >= x):
      h = mid - 1
    else:
      l = mid + 1
  return l


# function to find last index <= x
def upperIndex(arr, n, x):
  l = 0
  h = n-1
  while (l <= h):
    mid = int((l + h)/2)
    if (arr[mid] <= x):
      l = mid + 1
    else:
      h = mid - 1
  return h


# function to count elements within given range
def countInRange(arr, n, x, y):
  # initialize result
  count = 0;
  count = upperIndex(arr, n, y) - lowerIndex(arr, n, x) + 1;
  return count

# driver function
arr = [1, 3, 4, 9, 10, 3]

# Preprocess array
arr.sort()
n = len(arr)

# Answer queries
i = 1
j = 4
print(countInRange(arr, n, i, j))
i = 9
j = 12
print(countInRange(arr, n, i, j))


Output:

4
2

This article is contributed by Aditi Sharma. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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