Find the frequency of a number in an array using divide and conquer
The problem at hand revolves around finding the frequency of a given number in an array using a divide and conquer approach. This technique breaks down complex problems into simpler sub-problems, solves them, and then combines the solutions to solve the original problem. In this case, the objective is to count how many times a specific number appears in a given array.
Problem Statement
Given an array arr
of integers and a target number x
, we need to determine how many times
x
occurs within the array. This count will be our desired frequency.
Example
Let's take the array arr1 = [4, 4, 10, 3, 12, 4, 6, 3]
. If we want to find the frequency of
x = 4
, the expected output should be 3
, as 4
appears three times in the
array.
Idea to Solve
The divide and conquer approach involves dividing the array into smaller parts, solving each part recursively, and then combining the results. To implement this, we can follow these steps:
- Divide the array into two halves, left and right.
- Recursively count the occurrences of
x
in the left and right halves. - Combine the counts obtained from the left and right halves to get the total frequency of
x
in the entire array.
Pseudocode
function countFrequency(arr, low, high, x):
if low > high:
return 0
if low == high:
if arr[low] == x:
return 1
else:
return 0
mid = low + (high - low) / 2
leftCount = countFrequency(arr, low, mid, x)
rightCount = countFrequency(arr, mid + 1, high, x)
return leftCount + rightCount
Algorithm Explanation
- The base cases are when the range
low
tohigh
is invalid (i.e.,low > high
) or when we are left with a single element in the range. - If there's only one element, we check whether it's equal to
x
or not. If it is, we return1
(indicating a match), otherwise0
. - If there are multiple elements in the range, we calculate the middle index
mid
to divide the range into two halves. - We then recursively count the occurrences of
x
in the left and right halves. - Finally, we sum up the counts from both halves and return the total count.
Code Solution
// C Program
// Find the frequency of a number in an array using divide and conquer
#include <stdio.h>
int countFrequency(int arr[], int low, int high, int x)
{
if (low > high)
{
return 0;
}
if (low == high)
{
if (x == arr[low])
{
// We get element
return 1;
}
return 0;
}
// Find middle element
int mid = low + ((high - low) / 2);
// Count element in left and right subtree using recursion
return countFrequency(arr, low, mid, x) +
countFrequency(arr, mid + 1, high, x);
}
int main()
{
int arr1[] = {
4 , 4 , 10 , 3 , 12 , 4 , 6 , 3
};
int arr2[] = {
8 , 20 , 16 , 10 , 6
};
/*
arr1 = [ 4, 4, 10 , 3 , 12 ,4 , 6 , 3]
*/
int n = sizeof(arr1) / sizeof(arr1[0]);
int x = 3;
printf("\n %d Occurring %d times", x,
countFrequency(arr1, 0, n - 1, x));
x = 4;
printf("\n %d Occurring %d times", x,
countFrequency(arr1, 0, n - 1, x));
n = sizeof(arr2) / sizeof(arr2[0]);
/*
arr2 = [8 , 20 , 16 , 10 , 6]
*/
x = 16;
printf("\n %d Occurring %d times", x,
countFrequency(arr2, 0, n - 1, x));
x = 9;
printf("\n %d Occurring %d times", x,
countFrequency(arr2, 0, n - 1, x));
return 0;
}
Output
3 Occurring 2 times
4 Occurring 3 times
16 Occurring 1 times
9 Occurring 0 times
// Java Program
// Find the frequency of a number in an array using divide and conquer
public class Counting
{
public int countFrequency(int[] arr, int low, int high, int x)
{
if (low > high)
{
return 0;
}
if (low == high)
{
if (x == arr[low])
{
// We get element
return 1;
}
return 0;
}
// Find middle element
int mid = low + ((high - low) / 2);
// Count element in left and right subtree using recursion
return countFrequency(arr, low, mid, x) +
countFrequency(arr, mid + 1, high, x);
}
public static void main(String args[])
{
Counting task = new Counting();
int[] arr1 = {
4 , 4 , 10 , 3 , 12 , 4 , 6 , 3
};
int[] arr2 = {
8 , 20 , 16 , 10 , 6
};
/*
arr1 = [ 4, 4, 10 , 3 , 12 ,4 , 6 , 3]
*/
int n = arr1.length;
int x = 3;
System.out.print("\n " + x + " Occurring " +
task.countFrequency(arr1, 0, n - 1, x) + " times");
x = 4;
System.out.print("\n " + x + " Occurring " +
task.countFrequency(arr1, 0, n - 1, x) + " times");
n = arr2.length;
/*
arr2 = [8 , 20 , 16 , 10 , 6]
*/
x = 16;
System.out.print("\n " + x + " Occurring " +
task.countFrequency(arr2, 0, n - 1, x) + " times");
x = 9;
System.out.print("\n " + x + " Occurring " +
task.countFrequency(arr2, 0, n - 1, x) + " times");
}
}
Output
3 Occurring 2 times
4 Occurring 3 times
16 Occurring 1 times
9 Occurring 0 times
// Include header file
#include <iostream>
using namespace std;
// C++ Program
// Find the frequency of a number in an array using divide and conquer
class Counting
{
public: int countFrequency(int arr[], int low, int high, int x)
{
if (low > high)
{
return 0;
}
if (low == high)
{
if (x == arr[low])
{
// We get element
return 1;
}
return 0;
}
// Find middle element
int mid = low + ((high - low) / 2);
// Count element in left and right subtree using recursion
return this->countFrequency(arr, low, mid, x) +
this->countFrequency(arr, mid + 1, high, x);
}
};
int main()
{
Counting *task = new Counting();
int arr1[] = {
4 , 4 , 10 , 3 , 12 , 4 , 6 , 3
};
int arr2[] = {
8 , 20 , 16 , 10 , 6
};
/*
arr1 = [ 4, 4, 10 , 3 , 12 ,4 , 6 , 3]
*/
int n = sizeof(arr1) / sizeof(arr1[0]);
int x = 3;
cout << "\n " << x << " Occurring "
<< task->countFrequency(arr1, 0, n - 1, x) << " times";
x = 4;
cout << "\n " << x << " Occurring "
<< task->countFrequency(arr1, 0, n - 1, x) << " times";
n = sizeof(arr2) / sizeof(arr2[0]);
/*
arr2 = [8 , 20 , 16 , 10 , 6]
*/
x = 16;
cout << "\n " << x << " Occurring "
<< task->countFrequency(arr2, 0, n - 1, x) << " times";
x = 9;
cout << "\n " << x << " Occurring "
<< task->countFrequency(arr2, 0, n - 1, x) << " times";
return 0;
}
Output
3 Occurring 2 times
4 Occurring 3 times
16 Occurring 1 times
9 Occurring 0 times
// Include namespace system
using System;
// Csharp Program
// Find the frequency of a number in an array using divide and conquer
public class Counting
{
public int countFrequency(int[] arr,
int low,
int high,
int x)
{
if (low > high)
{
return 0;
}
if (low == high)
{
if (x == arr[low])
{
// We get element
return 1;
}
return 0;
}
// Find middle element
int mid = low + ((high - low) / 2);
// Count element in left and right subtree using recursion
return this.countFrequency(arr, low, mid, x) +
this.countFrequency(arr, mid + 1, high, x);
}
public static void Main(String[] args)
{
Counting task = new Counting();
int[] arr1 = {
4 , 4 , 10 , 3 , 12 , 4 , 6 , 3
};
int[] arr2 = {
8 , 20 , 16 , 10 , 6
};
/*
arr1 = [ 4, 4, 10 , 3 , 12 ,4 , 6 , 3]
*/
int n = arr1.Length;
int x = 3;
Console.Write("\n " + x + " Occurring " +
task.countFrequency(arr1, 0, n - 1, x) + " times");
x = 4;
Console.Write("\n " + x + " Occurring " +
task.countFrequency(arr1, 0, n - 1, x) + " times");
n = arr2.Length;
/*
arr2 = [8 , 20 , 16 , 10 , 6]
*/
x = 16;
Console.Write("\n " + x + " Occurring " +
task.countFrequency(arr2, 0, n - 1, x) + " times");
x = 9;
Console.Write("\n " + x + " Occurring " +
task.countFrequency(arr2, 0, n - 1, x) + " times");
}
}
Output
3 Occurring 2 times
4 Occurring 3 times
16 Occurring 1 times
9 Occurring 0 times
package main
import "fmt"
// Go Program
// Find the frequency of a number in an array using divide and conquer
type Counting struct {}
func getCounting() * Counting {
var me *Counting = &Counting {}
return me
}
func(this Counting) countFrequency(arr[] int,
low int, high int, x int) int {
if low > high {
return 0
}
if low == high {
if x == arr[low] {
// We get element
return 1
}
return 0
}
// Find middle element
var mid int = low + ((high - low) / 2)
// Count element in left and right subtree using recursion
return this.countFrequency(arr, low, mid, x) +
this.countFrequency(arr, mid + 1, high, x)
}
func main() {
var task * Counting = getCounting()
var arr1 = [] int {
4,
4,
10,
3,
12,
4,
6,
3,
}
var arr2 = [] int {
8,
20,
16,
10,
6,
}
/*
arr1 = [ 4, 4, 10 , 3 , 12 ,4 , 6 , 3]
*/
var n int = len(arr1)
var x int = 3
fmt.Print("\n ", x, " Occurring ",
task.countFrequency(arr1, 0, n - 1, x), " times")
x = 4
fmt.Print("\n ", x, " Occurring ",
task.countFrequency(arr1, 0, n - 1, x), " times")
n = len(arr2)
/*
arr2 = [8 , 20 , 16 , 10 , 6]
*/
x = 16
fmt.Print("\n ", x, " Occurring ",
task.countFrequency(arr2, 0, n - 1, x), " times")
x = 9
fmt.Print("\n ", x, " Occurring ",
task.countFrequency(arr2, 0, n - 1, x), " times")
}
Output
3 Occurring 2 times
4 Occurring 3 times
16 Occurring 1 times
9 Occurring 0 times
<?php
// Php Program
// Find the frequency of a number in an array using divide and conquer
class Counting
{
public function countFrequency($arr, $low, $high, $x)
{
if ($low > $high)
{
return 0;
}
if ($low == $high)
{
if ($x == $arr[$low])
{
// We get element
return 1;
}
return 0;
}
// Find middle element
$mid = $low + ((int)(($high - $low) / 2));
// Count element in left and right subtree using recursion
return $this->countFrequency($arr, $low, $mid, $x) +
$this->countFrequency($arr, $mid + 1, $high, $x);
}
}
function main()
{
$task = new Counting();
$arr1 = array(4, 4, 10, 3, 12, 4, 6, 3);
$arr2 = array(8, 20, 16, 10, 6);
/*
arr1 = [ 4, 4, 10 , 3 , 12 ,4 , 6 , 3]
*/
$n = count($arr1);
$x = 3;
echo("\n ".$x.
" Occurring ".$task->countFrequency($arr1, 0, $n - 1, $x).
" times");
$x = 4;
echo("\n ".$x.
" Occurring ".$task->countFrequency($arr1, 0, $n - 1, $x).
" times");
$n = count($arr2);
/*
arr2 = [8 , 20 , 16 , 10 , 6]
*/
$x = 16;
echo("\n ".$x.
" Occurring ".$task->countFrequency($arr2, 0, $n - 1, $x).
" times");
$x = 9;
echo("\n ".$x.
" Occurring ".$task->countFrequency($arr2, 0, $n - 1, $x).
" times");
}
main();
Output
3 Occurring 2 times
4 Occurring 3 times
16 Occurring 1 times
9 Occurring 0 times
// Node JS Program
// Find the frequency of a number in an array using divide and conquer
class Counting
{
countFrequency(arr, low, high, x)
{
if (low > high)
{
return 0;
}
if (low == high)
{
if (x == arr[low])
{
// We get element
return 1;
}
return 0;
}
// Find middle element
var mid = low + (parseInt((high - low) / 2));
// Count element in left and right subtree using recursion
return this.countFrequency(arr, low, mid, x) +
this.countFrequency(arr, mid + 1, high, x);
}
}
function main()
{
var task = new Counting();
var arr1 = [4, 4, 10, 3, 12, 4, 6, 3];
var arr2 = [8, 20, 16, 10, 6];
/*
arr1 = [ 4, 4, 10 , 3 , 12 ,4 , 6 , 3]
*/
var n = arr1.length;
var x = 3;
process.stdout.write("\n " + x + " Occurring " +
task.countFrequency(arr1, 0, n - 1, x) + " times");
x = 4;
process.stdout.write("\n " + x + " Occurring " +
task.countFrequency(arr1, 0, n - 1, x) + " times");
n = arr2.length;
/*
arr2 = [8 , 20 , 16 , 10 , 6]
*/
x = 16;
process.stdout.write("\n " + x + " Occurring " +
task.countFrequency(arr2, 0, n - 1, x) + " times");
x = 9;
process.stdout.write("\n " + x + " Occurring " +
task.countFrequency(arr2, 0, n - 1, x) + " times");
}
main();
Output
3 Occurring 2 times
4 Occurring 3 times
16 Occurring 1 times
9 Occurring 0 times
# Python 3 Program
# Find the frequency of a number in an array using divide and conquer
class Counting :
def countFrequency(self, arr, low, high, x) :
if (low > high) :
return 0
if (low == high) :
if (x == arr[low]) :
# We get element
return 1
return 0
# Find middle element
mid = low + (int((high - low) / 2))
# Count element in left and right subtree using recursion
return self.countFrequency(
arr, low, mid, x
) + self.countFrequency(
arr, mid + 1, high, x
)
def main() :
task = Counting()
arr1 = [4, 4, 10, 3, 12, 4, 6, 3]
arr2 = [8, 20, 16, 10, 6]
# arr1 = [ 4, 4, 10 , 3 , 12 ,4 , 6 , 3]
n = len(arr1)
x = 3
print("\n ", x ," Occurring ",
task.countFrequency(arr1, 0, n - 1, x) ," times", end = "")
x = 4
print("\n ", x ," Occurring ",
task.countFrequency(arr1, 0, n - 1, x) ," times", end = "")
n = len(arr2)
# arr2 = [8 , 20 , 16 , 10 , 6]
x = 16
print("\n ", x ," Occurring ",
task.countFrequency(arr2, 0, n - 1, x) ," times", end = "")
x = 9
print("\n ", x ," Occurring ",
task.countFrequency(arr2, 0, n - 1, x) ," times", end = "")
if __name__ == "__main__": main()
Output
3 Occurring 2 times
4 Occurring 3 times
16 Occurring 1 times
9 Occurring 0 times
# Ruby Program
# Find the frequency of a number in an array using divide and conquer
class Counting
def countFrequency(arr, low, high, x)
if (low > high)
return 0
end
if (low == high)
if (x == arr[low])
# We get element
return 1
end
return 0
end
# Find middle element
mid = low + ((high - low) / 2)
# Count element in left and right subtree using recursion
return self.countFrequency(arr, low, mid, x) +
self.countFrequency(arr, mid + 1, high, x)
end
end
def main()
task = Counting.new()
arr1 = [4, 4, 10, 3, 12, 4, 6, 3]
arr2 = [8, 20, 16, 10, 6]
# arr1 = [ 4, 4, 10 , 3 , 12 ,4 , 6 , 3]
n = arr1.length
x = 3
print("\n ", x ," Occurring ",
task.countFrequency(arr1, 0, n - 1, x) ," times")
x = 4
print("\n ", x ," Occurring ",
task.countFrequency(arr1, 0, n - 1, x) ," times")
n = arr2.length
# arr2 = [8 , 20 , 16 , 10 , 6]
x = 16
print("\n ", x ," Occurring ",
task.countFrequency(arr2, 0, n - 1, x) ," times")
x = 9
print("\n ", x ," Occurring ",
task.countFrequency(arr2, 0, n - 1, x) ," times")
end
main()
Output
3 Occurring 2 times
4 Occurring 3 times
16 Occurring 1 times
9 Occurring 0 times
// Scala Program
// Find the frequency of a number in an array using divide and conquer
class Counting()
{
def countFrequency(arr: Array[Int],
low: Int, high: Int, x: Int): Int = {
if (low > high)
{
return 0;
}
if (low == high)
{
if (x == arr(low))
{
// We get element
return 1;
}
return 0;
}
// Find middle element
var mid: Int = low + ((high - low) / 2);
// Count element in left and right subtree using recursion
return countFrequency(arr, low, mid, x) +
countFrequency(arr, mid + 1, high, x);
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: Counting = new Counting();
var arr1: Array[Int] = Array(4, 4, 10, 3, 12, 4, 6, 3);
var arr2: Array[Int] = Array(8, 20, 16, 10, 6);
/*
arr1 = [ 4, 4, 10 , 3 , 12 ,4 , 6 , 3]
*/
var n: Int = arr1.length;
var x: Int = 3;
print("\n " + x + " Occurring " +
task.countFrequency(arr1, 0, n - 1, x) + " times");
x = 4;
print("\n " + x + " Occurring " +
task.countFrequency(arr1, 0, n - 1, x) + " times");
n = arr2.length;
/*
arr2 = [8 , 20 , 16 , 10 , 6]
*/
x = 16;
print("\n " + x + " Occurring " +
task.countFrequency(arr2, 0, n - 1, x) + " times");
x = 9;
print("\n " + x + " Occurring " +
task.countFrequency(arr2, 0, n - 1, x) + " times");
}
}
Output
3 Occurring 2 times
4 Occurring 3 times
16 Occurring 1 times
9 Occurring 0 times
import Foundation;
// Swift 4 Program
// Find the frequency of a number in an array using divide and conquer
class Counting
{
func countFrequency(_ arr: [Int],
_ low: Int,
_ high: Int,
_ x: Int) -> Int
{
if (low > high)
{
return 0;
}
if (low == high)
{
if (x == arr[low])
{
// We get element
return 1;
}
return 0;
}
// Find middle element
let mid: Int = low + ((high - low) / 2);
// Count element in left and right subtree using recursion
return self.countFrequency(arr, low, mid, x) +
self.countFrequency(arr, mid + 1, high, x);
}
}
func main()
{
let task: Counting = Counting();
let arr1: [Int] = [4, 4, 10, 3, 12, 4, 6, 3];
let arr2: [Int] = [8, 20, 16, 10, 6];
/*
arr1 = [ 4, 4, 10 , 3 , 12 ,4 , 6 , 3]
*/
var n: Int = arr1.count;
var x: Int = 3;
print("\n ", x ," Occurring ",
task.countFrequency(arr1, 0, n - 1, x) ," times", terminator: "");
x = 4;
print("\n ", x ," Occurring ",
task.countFrequency(arr1, 0, n - 1, x) ," times", terminator: "");
n = arr2.count;
/*
arr2 = [8 , 20 , 16 , 10 , 6]
*/
x = 16;
print("\n ", x ," Occurring ",
task.countFrequency(arr2, 0, n - 1, x) ," times", terminator: "");
x = 9;
print("\n ", x ," Occurring ",
task.countFrequency(arr2, 0, n - 1, x) ," times", terminator: "");
}
main();
Output
3 Occurring 2 times
4 Occurring 3 times
16 Occurring 1 times
9 Occurring 0 times
// Kotlin Program
// Find the frequency of a number in an array using divide and conquer
class Counting
{
fun countFrequency(arr: Array < Int > ,
low: Int,
high: Int,
x: Int): Int
{
if (low > high)
{
return 0;
}
if (low == high)
{
if (x == arr[low])
{
// We get element
return 1;
}
return 0;
}
// Find middle element
val mid: Int = low + ((high - low) / 2);
// Count element in left and right subtree using recursion
return this.countFrequency(arr, low, mid, x) +
this.countFrequency(arr, mid + 1, high, x);
}
}
fun main(args: Array < String > ): Unit
{
val task: Counting = Counting();
val arr1: Array < Int > = arrayOf(4, 4, 10, 3, 12, 4, 6, 3);
val arr2: Array < Int > = arrayOf(8, 20, 16, 10, 6);
/*
arr1 = [ 4, 4, 10 , 3 , 12 ,4 , 6 , 3]
*/
var n: Int = arr1.count();
var x: Int = 3;
print("\n " + x + " Occurring " +
task.countFrequency(arr1, 0, n - 1, x) + " times");
x = 4;
print("\n " + x + " Occurring " +
task.countFrequency(arr1, 0, n - 1, x) + " times");
n = arr2.count();
/*
arr2 = [8 , 20 , 16 , 10 , 6]
*/
x = 16;
print("\n " + x + " Occurring " +
task.countFrequency(arr2, 0, n - 1, x) + " times");
x = 9;
print("\n " + x + " Occurring " +
task.countFrequency(arr2, 0, n - 1, x) + " times");
}
Output
3 Occurring 2 times
4 Occurring 3 times
16 Occurring 1 times
9 Occurring 0 times
Time Complexity
The time complexity of this algorithm can be analyzed based on the number of recursive calls. In each recursive
call, we split the array into two halves. Since each call only considers one half, the maximum number of recursive
calls is O(log n)
, where n
is the number of elements in the array. Within each recursive
call, we perform constant time operations. Hence, the overall time complexity is O(log n)
.
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