Maximum in array which is at-least twice of other elements
Last Updated :
03 Mar, 2023
Given an array of integers of length n. Our task is to return the index of the max element if the it is at least twice as much as every other number in the array. If the max element does not satisfy the condition return -1.
Examples:
Input : arr = {3, 6, 1, 0}
Output : 1
Here, 6 is the largest integer, and for
every other number in the array x, 6 is
more than twice as big as x. The index of
value 6 is 1, so we return 1.
Input : arr = {1, 2, 3, 4}
Output : -1
4 isn't at least as big as twice the value
of 3, so we return -1.
Approach : Scan through the array to find the unique largest element m, keeping track of it’s index maxIndex. Scan through the array again. If we find some x != m with m < 2*x, we should return -1. Otherwise, we should return maxIndex.
Algorithm:
Step 1: Start
Step 2: Create a function of int return type name “findIndex” which takes an integer array as an input parameter and returns index of max element that satisfies the condition.
a. initialize an int variable called “maxIndex” with value 0.
b. start a for loop from i=0 to i < length of the array
1. check if the value of the current index is greater than the value at the maxIndex of the array if true then set maxIndex to the current index.
c. start a for loop from i=0 to i < length of the array
1. check if maxIndex is not equal to the current index and value of the array at maxIndex is less than twice of the value of the array at the current index if true then return -1
2. If there does not exist such a pair and you came out of the loop then return the maxIndex.
Step 3: End
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int findIndex( int arr[], int len)
{
int maxIndex = 0;
for ( int i = 0; i < len; ++i)
if (arr[i] > arr[maxIndex])
maxIndex = i;
for ( int i = 0; i < len; ++i)
if (maxIndex != i && arr[maxIndex] < 2 * arr[i])
return -1;
return maxIndex;
}
int main()
{
int arr[] = { 3, 6, 1, 0 };
int len = sizeof (arr) / sizeof (arr[0]);
printf ( "%d" , findIndex(arr, len));
}
|
C
#include <stdio.h>
int findIndex( int arr[], int len)
{
int maxIndex = 0;
for ( int i = 0; i < len; ++i)
if (arr[i] > arr[maxIndex])
maxIndex = i;
for ( int i = 0; i < len; ++i)
if (maxIndex != i && arr[maxIndex] < 2 * arr[i])
return -1;
return maxIndex;
}
int main()
{
int arr[] = { 3, 6, 1, 0 };
int len = sizeof (arr) / sizeof (arr[0]);
printf ( "%d" , findIndex(arr, len));
}
|
Java
import java.util.*;
import java.lang.*;
class GfG {
public static int findIndex( int [] arr) {
int maxIndex = 0 ;
for ( int i = 0 ; i < arr.length; ++i)
if (arr[i] > arr[maxIndex])
maxIndex = i;
for ( int i = 0 ; i < arr.length; ++i)
if (maxIndex != i && arr[maxIndex] < 2 * arr[i])
return - 1 ;
return maxIndex;
}
public static void main(String argc[]){
int [] arr = new int []{ 3 , 6 , 1 , 0 };
System.out.println(findIndex(arr));
}
}
|
Python3
def findIndex(arr):
maxIndex = 0
for i in range ( 0 , len (arr)):
if (arr[i] > arr[maxIndex]):
maxIndex = i
for i in range ( 0 , len (arr)):
if (maxIndex ! = i and
arr[maxIndex] < ( 2 * arr[i])):
return - 1
return maxIndex
arr = [ 3 , 6 , 1 , 0 ]
print (findIndex(arr))
|
C#
using System;
class GfG {
public static int findIndex( int [] arr) {
int maxIndex = 0;
for ( int i = 0; i < arr.Length; ++i)
if (arr[i] > arr[maxIndex])
maxIndex = i;
for ( int i = 0; i < arr.Length; ++i)
if (maxIndex != i && arr[maxIndex] < 2 * arr[i])
return -1;
return maxIndex;
}
public static void Main()
{
int [] arr = new int []{3, 6, 1, 0};
Console.WriteLine(findIndex(arr));
}
}
|
PHP
<?php
function findIndex( $arr , $len )
{
$maxIndex = 0;
for ( $i = 0; $i < $len ; ++ $i )
if ( $arr [ $i ] > $arr [ $maxIndex ])
$maxIndex = $i ;
for ( $i = 0; $i < $len ; ++ $i )
if ( $maxIndex != $i and
$arr [ $maxIndex ] < 2 * $arr [ $i ])
return -1;
return $maxIndex ;
}
$arr = array (3, 6, 1, 0);
$len = count ( $arr );
echo findIndex( $arr , $len );
?>
|
Javascript
<script>
function findIndex(arr, len) {
let maxIndex = 0;
for (let i = 0; i < len; ++i)
if (arr[i] > arr[maxIndex])
maxIndex = i;
for (let i = 0; i < len; ++i)
if (maxIndex != i &&
arr[maxIndex] < 2 * arr[i])
return -1;
return maxIndex;
}
let arr = [3, 6, 1, 0];
let len = arr.length;
document.write(findIndex(arr, len));
</script>
|
Time Complexity:
Auxiliary Space: O(1)
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