Check if minimum element in array is less than or equals half of every other element
Last Updated :
05 Dec, 2022
Given an array arr[], the task is to check if the minimum element in the array is less than or equal to half of every other element. If it is then print “yes” otherwise print “no”.
Note: The minimum number in the given array is always unique.
Examples:
Input: arr = {2, 1, 4, 5}
Output: Yes
Explanation:
1 is the minimum element in the array arr[] and on dividing 2, 4, 5 by 2 we get 1, 2, 2.5 which is greater than or equal to the minimum number. Hence, print “yes”.
Input : arr = {2, 4, 5, 3}
Output : No
Explanation:
2 is the minimum element in the array arr[] and on dividing 4, 5, 3 by 2 we get 2, 2.5, 1.5 in which the integer 3 does not return a value which is greater than or equal to the minimum number ( 1.5 < 2). Hence, print “no”.
Method 1:
To solve the problem mentioned above we have to find the smallest element with the help of loops and then scan through the entire array again and check if twice the smallest element is smaller than or equal to every other element. But this solution takes O(N) time using two loops and can be optimized further where only one iteration is involved.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool checkMin( int arr[], int len)
{
int smallest = 0;
for ( int i = 0; i < len; i++) {
if (arr[i] < arr[smallest]) {
smallest = i;
}
}
for ( int i = 0; i < len; i++) {
if (i != smallest && arr[i] / 2 < arr[smallest]) {
return false ;
}
}
return true ;
}
int main()
{
int arr[] = { 2, 30, 4, 5 };
int len = sizeof (arr) / sizeof (arr[0]);
if (checkMin(arr, len)) {
cout << "Yes" ;
}
else {
cout << "No" ;
}
}
|
Java
import java.io.*;
class GFG {
public static boolean checkMin( int arr[], int len)
{
int smallest = 0 ;
for ( int i = 0 ; i < len; i++) {
if (arr[i] < arr[smallest]) {
smallest = i;
}
}
for ( int i = 0 ; i < len; i++) {
if (i != smallest && arr[i] / 2 < arr[smallest]) {
return false ;
}
}
return true ;
}
public static void main (String[] args) {
int arr[] = { 2 , 30 , 4 , 5 };
int len = arr.length;
if (checkMin(arr, len)) {
System.out.println( "Yes" );
}
else {
System.out.println( "No" );
}
}
}
|
Python3
class GFG :
@staticmethod
def checkMin( arr, len ) :
smallest = 0
i = 0
while (i < len ) :
if (arr[i] < arr[smallest]) :
smallest = i
i + = 1
i = 0
while (i < len ) :
if (i ! = smallest and int (arr[i] / 2 ) < arr[smallest]) :
return False
i + = 1
return True
@staticmethod
def main( args) :
arr = [ 2 , 30 , 4 , 5 ]
len1 = len (arr)
if (GFG.checkMin(arr, len1)) :
print ( "Yes" )
else :
print ( "No" )
if __name__ = = "__main__" :
GFG.main([])
|
C#
using System;
class GFG {
static bool checkMin( int [] arr, int len)
{
int smallest = 0;
for ( int i = 0; i < len; i++) {
if (arr[i] < arr[smallest]) {
smallest = i;
}
}
for ( int i = 0; i < len; i++) {
if (i != smallest
&& arr[i] / 2 < arr[smallest]) {
return false ;
}
}
return true ;
}
static void Main()
{
int [] arr = { 2, 30, 4, 5 };
int len = arr.Length;
if (checkMin(arr, len) == true ) {
Console.Write( "Yes" );
}
else {
Console.Write( "No" );
}
}
}
|
Javascript
function checkMin(arr, len)
{
let smallest = 0;
for (let i = 0; i < len; i++) {
if (arr[i] < arr[smallest]) {
smallest = i;
}
}
for (let i = 0; i < len; i++) {
if (i != smallest && arr[i] / 2 < arr[smallest]) {
return false ;
}
}
return true ;
}
let arr = [ 2, 30, 4, 5 ];
let len = arr.length;
if (checkMin(arr, len) == true ) {
console.log( "Yes" );
}
else {
console.log( "No" );
}
|
Time Complexity: O(N), where N is the length of the given array.
Auxiliary Space: O(1).
Method 2:
To optimize the above solution we can find the smallest as well as the second smallest element in a single iteration itself. Then simply check if twice of the smallest element is smaller than or equal to the second smallest element.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void checkMin( int arr[], int len)
{
int smallest = INT_MAX, secondSmallest = INT_MAX;
for ( int i = 0; i < len; i++) {
if (arr[i] < smallest) {
secondSmallest = smallest;
smallest = arr[i];
}
else if (arr[i] < secondSmallest) {
secondSmallest = arr[i];
}
}
if (2 * smallest <= secondSmallest)
cout << "Yes" ;
else
cout << "No" ;
}
int main()
{
int arr[] = { 2, 3, 4, 5 };
int len = sizeof (arr) / sizeof (arr[0]);
checkMin(arr, len);
}
|
Java
import java.util.*;
class GFG{
static void checkMin( int arr[], int len)
{
int smallest = Integer.MAX_VALUE;
int secondSmallest = Integer.MAX_VALUE;
for ( int i = 0 ; i < len; i++)
{
if (arr[i] < smallest)
{
secondSmallest = smallest;
smallest = arr[i];
}
else if (arr[i] < secondSmallest)
{
secondSmallest = arr[i];
}
}
if ( 2 * smallest <= secondSmallest)
System.out.print( "Yes" );
else
System.out.print( "No" );
}
public static void main(String[] args)
{
int arr[] = { 2 , 3 , 4 , 5 };
int len = arr.length;
checkMin(arr, len);
}
}
|
Python3
import math
def checkMin(arr, n):
smallest = math.inf
secondSmallest = math.inf
for i in range (n):
if (arr[i] < smallest):
secondSmallest = smallest
smallest = arr[i]
elif (arr[i] < secondSmallest):
secondSmallest = arr[i]
if ( 2 * smallest < = secondSmallest):
print ( "Yes" )
else :
print ( "No" )
if __name__ = = '__main__' :
arr = [ 2 , 3 , 4 , 5 ]
n = len (arr)
checkMin(arr, n)
|
C#
using System;
class GFG{
static void checkMin( int []arr, int len)
{
int smallest = int .MaxValue;
int secondSmallest = int .MaxValue;
for ( int i = 0; i < len; i++)
{
if (arr[i] < smallest)
{
secondSmallest = smallest;
smallest = arr[i];
}
else if (arr[i] < secondSmallest)
{
secondSmallest = arr[i];
}
}
if (2 * smallest <= secondSmallest)
Console.Write( "Yes" );
else
Console.Write( "No" );
}
public static void Main(String[] args)
{
int []arr = { 2, 3, 4, 5 };
int len = arr.Length;
checkMin(arr, len);
}
}
|
Javascript
<script>
function checkMin(arr, len)
{
var smallest = Number.INFINITY,
secondSmallest = Number.INFINITY;
for ( var i = 0; i < len; i++)
{
if (arr[i] < smallest)
{
secondSmallest = smallest;
smallest = arr[i];
}
else if (arr[i] < secondSmallest)
{
secondSmallest = arr[i];
}
}
if (2 * smallest <= secondSmallest)
document.write( "Yes" );
else
document.write( "No" );
}
var arr = [ 2, 3, 4, 5 ];
var len = 4;
checkMin(arr, len);
</script>
|
Time Complexity: O(N), where N is the length of the given array.
Auxiliary Space: O(1).
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