Queries for counts of array elements with values in given range

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 10 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).

Implementation:

C++

// Simple C++ program to count number of elements
// with values in given range.
#include <bits/stdc++.h>
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;
}

Java

// Simple java program to count
// number of elements with
// values in given range.
import java.io.*;

class GFG
{
    // function to count elements within given range
    static 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
    public static void main (String[] args)
    {
        int arr[] = { 1, 3, 4, 9, 10, 3 };
        int n = arr.length;

        // Answer queries
        int i = 1, j = 4;
        System.out.println ( countInRange(arr, n, i, j)) ;

        i = 9;
        j = 12;
        System.out.println ( countInRange(arr, n, i, j)) ;


    }
}

// This article is contributed by vt_m

Python3

# 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))

C#

// Simple C# program to count
// number of elements with
// values in given range.
using System;

class GFG  {

    // function to count elements
    // within given range
    static 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 Code
    public static void Main ()
    {
        int[]arr = {1, 3, 4, 9, 10, 3};
        int n = arr.Length;

        // Answer queries
        int i = 1, j = 4;
        Console.WriteLine( countInRange(arr, n, i, j)) ;

        i = 9;
        j = 12;
        Console.WriteLine( countInRange(arr, n, i, j)) ;


    }
}

// This code is contributed by vt_m.

PHP

<?php
// Simple PHP program to count
// number of elements with
// values in given range.

// function to count elements
// within given range
function countInRange($arr, $n,
                        $x, $y)
{

    // initialize result
    $count = 0;
    for ($i = 0; $i < $n; $i++)
    {

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

    // Driver Code
    $arr = array(1, 3, 4, 9, 10, 3);
    $n = count($arr);

    // Answer queries
    $i = 1;
    $j = 4;
    echo countInRange($arr, $n, $i, $j)."\n";

    $i = 9;
    $j = 12;
    echo countInRange($arr, $n, $i, $j)."\n";

// This code is contributed by Sam007
?>

Javascript

<script>

    // Simple JavaScript program to count
    // number of elements with
    // values in given range.

    // function to count elements
    // within given range
    function countInRange(arr, n, x, y)
    {

        // initialize result
        let count = 0;
        for (let i = 0; i < n; i++) {

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

    let arr = [1, 3, 4, 9, 10, 3];
    let n = arr.length;

    // Answer queries
    let i = 1, j = 4;
    document.write( countInRange(arr, n, i, j) + "</br>") ;

    i = 9;
    j = 12;
    document.write( countInRange(arr, n, i, j)) ;

</script>

Output

4
2

Time Complexity: O(n),
Auxiliary Space: O(1)

Segment Tree approach : 

The segment tree is a binary tree where each node represents a segment of the array. The leaves of the tree represent the individual elements of the array, and the parent nodes represent the union of their children.

• To answer a query [i, j], we traverse the segment tree from the root to the leaves, keeping track of the segments that contain i and j. At each node, we use the precomputed values to compute the number of elements in the segment that are less than or equal to x. We merge the values for the left and right child nodes and return the final result.

Implementation :

C++

// C++ program to count number of elements
// with values in given range using segment tree

#include <bits/stdc++.h>
using namespace std;

// Segment Tree class
class SegmentTree {
public:
    SegmentTree(int n) {
        tree.resize(4*n);
    }

    // build the tree with array elements
    void build(int arr[], int v, int tl, int tr) {
        if (tl == tr) {
            tree[v] = (arr[tl] >= low && arr[tl] <= high) ? 1 : 0;
        } else {
            int tm = (tl + tr) / 2;
            build(arr, v*2, tl, tm);
            build(arr, v*2+1, tm+1, tr);
            tree[v] = tree[v*2] + tree[v*2+1];
        }
    }

    // query the tree for given range
    int query(int v, int tl, int tr, int l, int r) {
        if (l > r) {
            return 0;
        }
        if (l == tl && r == tr) {
            return tree[v];
        }
        int tm = (tl + tr) / 2;
        return query(v*2, tl, tm, l, min(r, tm)) +
               query(v*2+1, tm+1, tr, max(l, tm+1), r);
    }

    // update tree at index i with given value
    void update(int v, int tl, int tr, int i, int val) {
        if (tl == tr) {
            tree[v] = (val >= low && val <= high) ? 1 : 0;
        } else {
            int tm = (tl + tr) / 2;
            if (i <= tm) {
                update(v*2, tl, tm, i, val);
            } else {
                update(v*2+1, tm+1, tr, i, val);
            }
            tree[v] = tree[v*2] + tree[v*2+1];
        }
    }

    // set range for the tree
    void setRange(int l, int h) {
        low = l;
        high = h;
    }

private:
    vector<int> tree;
    int low, high;
};

// function to count elements within given range
int countInRange(int arr[], int n, int x, int y) {
    SegmentTree st(n);
    st.setRange(x, y);
    st.build(arr, 1, 0, n-1);
    return st.query(1, 0, n-1, 0, n-1);
}

// 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;
}

// this code is contributed by bhardwajji

Java

import java.util.*;

class SegmentTree {
    private List<Integer> tree;
    private int low, high;

    SegmentTree(int n)
    {
        tree = new ArrayList<>(4 * n);
        for (int i = 0; i < 4 * n; i++) {
            tree.add(0);
        }
    }

    // build the tree with array elements
    void build(int[] arr, int v, int tl, int tr)
    {
        if (tl == tr) {
            tree.set(v, (arr[tl] >= low && arr[tl] <= high)
                            ? 1
                            : 0);
        }
        else {
            int tm = (tl + tr) / 2;
            build(arr, v * 2, tl, tm);
            build(arr, v * 2 + 1, tm + 1, tr);
            tree.set(v,
                     tree.get(v * 2) + tree.get(v * 2 + 1));
        }
    }

    // query the tree for given range
    int query(int v, int tl, int tr, int l, int r)
    {
        if (l > r) {
            return 0;
        }
        if (l == tl && r == tr) {
            return tree.get(v);
        }
        int tm = (tl + tr) / 2;
        return query(v * 2, tl, tm, l, Math.min(r, tm))
            + query(v * 2 + 1, tm + 1, tr,
                    Math.max(l, tm + 1), r);
    }

    // update tree at index i with given value
    void update(int v, int tl, int tr, int i, int val)
    {
        if (tl == tr) {
            tree.set(v,
                     (val >= low && val <= high) ? 1 : 0);
        }
        else {
            int tm = (tl + tr) / 2;
            if (i <= tm) {
                update(v * 2, tl, tm, i, val);
            }
            else {
                update(v * 2 + 1, tm + 1, tr, i, val);
            }
            tree.set(v,
                     tree.get(v * 2) + tree.get(v * 2 + 1));
        }
    }

    // set range for the tree
    void setRange(int l, int h)
    {
        low = l;
        high = h;
    }
}

public class GFG {
    // function to count elements within given range
    static int countInRange(int[] arr, int n, int x, int y)
    {
        SegmentTree st = new SegmentTree(n);
        st.setRange(x, y);
        st.build(arr, 1, 0, n - 1);
        return st.query(1, 0, n - 1, 0, n - 1);
    }

    // driver function
    public static void main(String[] args)
    {
        int[] arr = { 1, 3, 4, 9, 10, 3 };
        int n = arr.length;

        // Answer queries
        int i = 1, j = 4;
        System.out.println(countInRange(arr, n, i, j));

        i = 9;
        j = 12;
        System.out.println(countInRange(arr, n, i, j));
    }
}

Python

# Python program to count number of elements
# with values in given range using segment tree

# Segment Tree class
class SegmentTree:
    def __init__(self, n):
        self.tree = [0] * (4*n)

    # build the tree with array elements
    def build(self, arr, v, tl, tr, low, high):
        if tl == tr:
            self.tree[v] = 1 if low <= arr[tl] <= high else 0
        else:
            tm = (tl + tr) // 2
            self.build(arr, v*2, tl, tm, low, high)
            self.build(arr, v*2+1, tm+1, tr, low, high)
            self.tree[v] = self.tree[v*2] + self.tree[v*2+1]

    # query the tree for given range
    def query(self, v, tl, tr, l, r):
        if l > r:
            return 0
        if l == tl and r == tr:
            return self.tree[v]
        tm = (tl + tr) // 2
        return self.query(v*2, tl, tm, l, min(r, tm)) + \
               self.query(v*2+1, tm+1, tr, max(l, tm+1), r)

    # update tree at index i with given value
    def update(self, v, tl, tr, i, val, low, high):
        if tl == tr:
            self.tree[v] = 1 if low <= val <= high else 0
        else:
            tm = (tl + tr) // 2
            if i <= tm:
                self.update(v*2, tl, tm, i, val, low, high)
            else:
                self.update(v*2+1, tm+1, tr, i, val, low, high)
            self.tree[v] = self.tree[v*2] + self.tree[v*2+1]

    # set range for the tree
    def set_range(self, low, high):
        self.low = low
        self.high = high

# function to count elements within given range
def count_in_range(arr, n, x, y):
    st = SegmentTree(n)
    st.set_range(x, y)
    st.build(arr, 1, 0, n-1, x, y)
    return st.query(1, 0, n-1, 0, n-1)

# driver function
if __name__ == '__main__':
    arr = [1, 3, 4, 9, 10, 3]
    n = len(arr)

    # Answer queries
    i, j = 1, 4
    print(count_in_range(arr, n, i, j))

    i, j = 9, 12
    print(count_in_range(arr, n, i, j))

C#

using System;

public class SegmentTree {
    private int[] tree;
    private int low, high;

    public SegmentTree(int n) {
        tree = new int[4*n];
    }

    public void Build(int[] arr, int v, int tl, int tr) {
        if (tl == tr) {
            tree[v] = (arr[tl] >= low && arr[tl] <= high) ? 1 : 0;
        } else {
            int tm = (tl + tr) / 2;
            Build(arr, v*2, tl, tm);
            Build(arr, v*2+1, tm+1, tr);
            tree[v] = tree[v*2] + tree[v*2+1];
        }
    }

    public int Query(int v, int tl, int tr, int l, int r) {
        if (l > r) {
            return 0;
        }
        if (l == tl && r == tr) {
            return tree[v];
        }
        int tm = (tl + tr) / 2;
        return Query(v*2, tl, tm, l, Math.Min(r, tm)) +
               Query(v*2+1, tm+1, tr, Math.Max(l, tm+1), r);
    }

    public void Update(int v, int tl, int tr, int i, int val) {
        if (tl == tr) {
            tree[v] = (val >= low && val <= high) ? 1 : 0;
        } else {
            int tm = (tl + tr) / 2;
            if (i <= tm) {
                Update(v*2, tl, tm, i, val);
            } else {
                Update(v*2+1, tm+1, tr, i, val);
            }
            tree[v] = tree[v*2] + tree[v*2+1];
        }
    }

    public void SetRange(int l, int h) {
        low = l;
        high = h;
    }
}

public class Program {
    public static int CountInRange(int[] arr, int n, int x, int y) {
        SegmentTree st = new SegmentTree(n);
        st.SetRange(x, y);
        st.Build(arr, 1, 0, n-1);
        return st.Query(1, 0, n-1, 0, n-1);
    }

    public static void Main() {
        int[] arr = { 1, 3, 4, 9, 10, 3 };
        int n = arr.Length;

        // Answer queries
        int i = 1, j = 4;
        Console.WriteLine(CountInRange(arr, n, i, j));

        i = 9; j = 12;
        Console.WriteLine(CountInRange(arr, n, i, j));
    }
}

Output

4
2

Time Complexity: O(n log n) 
Auxiliary Space: O(n)

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).

Implementation:

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;
}

Java

// Efficient C++ program to count number
// of elements with values in given range.
import java.io.*;
import java.util.Arrays;

class GFG
{
    // function to find first index >= x
    static 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
    static 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
    static 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
    public static void main (String[] args)
    {
        int arr[] = { 1, 4, 4, 9, 10, 3 };
        int n = arr.length;

        // Preprocess array
        Arrays.sort(arr);

        // Answer queries
        int i = 1, j = 4;
        System.out.println( countInRange(arr, n, i, j)); ;

        i = 9;
        j = 12;
        System.out.println( countInRange(arr, n, i, j));


    }
}

// This article is contributed by vt_m.

Python3

# 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))

C#

// Efficient C# program to count number
// of elements with values in given range.
using System;

class GFG
{

    // function to find first index >= x
    static 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
    static 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
    static 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 code
    public static void Main ()
    {
        int []arr = {1, 4, 4, 9, 10, 3};
        int n = arr.Length;

        // Preprocess array
        Array.Sort(arr);

        // Answer queries
        int i = 1, j = 4;
        Console.WriteLine(countInRange(arr, n, i, j)); ;

        i = 9;
        j = 12;
        Console.WriteLine(countInRange(arr, n, i, j));


    }
}

// This code is contributed by vt_m.

PHP

<?php
// Efficient PHP program to count
// number of elements with values
// in given range.

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

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

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

// Driver Code

$arr = array( 1, 4, 4, 9, 10, 3 );
$n = sizeof($arr);

// Preprocess array
sort($arr);

// Answer queries
$i = 1; $j = 4;
echo countInRange($arr, $n, $i, $j), "\n";

$i = 9; $j = 12;
echo countInRange($arr, $n, $i, $j), "\n";

// This code is contributed by Sachin
?>

Javascript

<script>

    // Efficient Javascript program to count number
    // of elements with values in given range.

    // function to find first index >= x
    function lowerIndex(arr, n, x)
    {

        let l = 0, h = n - 1;
        while (l <= h)
        {
            let mid = parseInt((l + h) / 2, 10);
            if (arr[mid] >= x)
                h = mid - 1;
            else
                l = mid + 1;
        }
        return l;
    }

    // function to find last index <= y
    function upperIndex(arr, n, y)
    {
        let l = 0, h = n - 1;
        while (l <= h)
        {
            let mid = parseInt((l + h) / 2, 10);
            if (arr[mid] <= y)
                l = mid + 1;
            else
                h = mid - 1;
        }
        return h;
    }

    // function to count elements
    // within given range
    function countInRange(arr, n, x, y)
    {

        // initialize result
        let count = 0;
        count = upperIndex(arr, n, y) -
                lowerIndex(arr, n, x) + 1;
        return count;
    }

    let arr = [1, 4, 4, 9, 10, 3];
    let n = arr.length;

    // Preprocess array
    arr.sort(function(a, b){return a - b});

    // Answer queries
    let i = 1, j = 4;
    document.write(countInRange(arr, n, i, j) + "</br>"); ;

    i = 9;
    j = 12;
    document.write(countInRange(arr, n, i, j));

</script>

Output

4
2

Time Complexity: O(n log n),
Auxiliary Space: O(1)

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