Find frequency of smallest value in an array
Given an array A of N elements. Find the frequency of the smallest value in the array.
Examples:
Input : N = 5, arr[] = {3, 2, 3, 4, 4}
Output : 1
The smallest element in the array is 2
and it occurs only once.
Input : N = 6, arr[] = {4, 3, 5, 3, 3, 6}
Output : 3
The smallest element in the array is 3
and it occurs 3 times.
Simple Approach: A simple method is to first find the minimum element in the array in first traversal. Then traverse the array again and find the number of occurrences of the minimum element.
Efficient Approach: The efficient approach is to do this in a single traversal.
Let us assume the first element to be the current minimum, so the frequency of the current minimum would be 1. Now let’s iterate through the array (1 to N), we come across 2 cases:
- Current element is smaller than our current minimum: We change the current minimum to be equal to current element and since this is the first time we are coming across this element, we make its frequency 1.
- Current element is equal to the current minimum: We increment the frequency of current minimum.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int frequencyOfSmallest( int n, int arr[])
{
int mn = arr[0], freq = 1;
for ( int i = 1; i < n; i++) {
if (arr[i] < mn) {
mn = arr[i];
freq = 1;
}
else if (arr[i] == mn)
freq++;
}
return freq;
}
int main()
{
int N = 5;
int arr[] = { 3, 2, 3, 4, 4 };
cout << frequencyOfSmallest(N, arr);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static int frequencyOfSmallest( int n, int arr[])
{
int mn = arr[ 0 ], freq = 1 ;
for ( int i = 1 ; i < n; i++)
{
if (arr[i] < mn)
{
mn = arr[i];
freq = 1 ;
}
else if (arr[i] == mn)
freq++;
}
return freq;
}
public static void main (String[] args)
{
int N = 5 ;
int arr[] = { 3 , 2 , 3 , 4 , 4 };
System.out.println (frequencyOfSmallest(N, arr));
}
}
|
Python3
def frequencyOfSmallest(n,arr):
mn = arr[ 0 ]
freq = 1
for i in range ( 1 ,n):
if (arr[i] < mn):
mn = arr[i]
freq = 1
elif (arr[i] = = mn):
freq + = 1
return freq
if __name__ = = '__main__' :
N = 5
arr = [ 3 , 2 , 3 , 4 , 4 ]
print (frequencyOfSmallest(N, arr))
|
C#
using System;
class GFG
{
static int frequencyOfSmallest( int n, int []arr)
{
int mn = arr[0], freq = 1;
for ( int i = 1; i < n; i++)
{
if (arr[i] < mn)
{
mn = arr[i];
freq = 1;
}
else if (arr[i] == mn)
freq++;
}
return freq;
}
public static void Main()
{
int N = 5;
int []arr = { 3, 2, 3, 4, 4 };
Console.WriteLine(frequencyOfSmallest(N, arr));
}
}
|
PHP
<?php
function frequencyOfSmallest( $n , $arr )
{
$mn = $arr [0];
$freq = 1;
for ( $i = 1; $i < $n ; $i ++)
{
if ( $arr [ $i ] < $mn )
{
$mn = $arr [ $i ];
$freq = 1;
}
else if ( $arr [ $i ] == $mn )
$freq ++;
}
return $freq ;
}
$N = 5;
$arr = array ( 3, 2, 3, 4, 4 );
print (frequencyOfSmallest( $N , $arr ));
?>
|
Javascript
<script>
function frequencyOfSmallest(n, arr)
{
let mn = arr[0], freq = 1;
for (let i = 1; i < n; i++)
{
if (arr[i] < mn)
{
mn = arr[i];
freq = 1;
}
else if (arr[i] == mn)
freq++;
}
return freq;
}
let N = 5;
let arr = [ 3, 2, 3, 4, 4 ];
document.write(frequencyOfSmallest(N, arr));
</script>
|
Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Last Updated :
15 Nov, 2022
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