Una array ordenada arr[] de tamaño N se gira en algún punto desconocido, encuentre el elemento mínimo en ella.
Nota: Se supone que todos los elementos son distintos.
C++
// C++ program to find minimum
// element in a sorted and rotated array
#include <bits/stdc++.h>
using namespace std;
int findMin(int arr[], int low, int high)
{
// This condition is needed to
// handle the case when array is not
// rotated at all
if (high < low)
return arr[0];
// If there is only one element left
if (high == low)
return arr[low];
// Find mid
int mid = low + (high - low) / 2; /*(low + high)/2;*/
// Check if element (mid+1) is minimum element. Consider
// the cases like {3, 4, 5, 1, 2}
if (mid < high && arr[mid + 1] < arr[mid])
return arr[mid + 1];
// Check if mid itself is minimum element
if (mid > low && arr[mid] < arr[mid - 1])
return arr[mid];
// Decide whether we need to go to left half or right
// half
if (arr[high] > arr[mid])
return findMin(arr, low, mid - 1);
return findMin(arr, mid + 1, high);
}
// Driver program to test above functions
int main()
{
int arr1[] = { 5, 6, 1, 2, 3, 4 };
int n1 = sizeof(arr1) / sizeof(arr1[0]);
cout << "The minimum element is "
<< findMin(arr1, 0, n1 - 1) << endl;
int arr2[] = { 1, 2, 3, 4 };
int n2 = sizeof(arr2) / sizeof(arr2[0]);
cout << "The minimum element is "
<< findMin(arr2, 0, n2 - 1) << endl;
int arr3[] = { 1 };
int n3 = sizeof(arr3) / sizeof(arr3[0]);
cout << "The minimum element is "
<< findMin(arr3, 0, n3 - 1) << endl;
int arr4[] = { 1, 2 };
int n4 = sizeof(arr4) / sizeof(arr4[0]);
cout << "The minimum element is "
<< findMin(arr4, 0, n4 - 1) << endl;
int arr5[] = { 2, 1 };
int n5 = sizeof(arr5) / sizeof(arr5[0]);
cout << "The minimum element is "
<< findMin(arr5, 0, n5 - 1) << endl;
int arr6[] = { 5, 6, 7, 1, 2, 3, 4 };
int n6 = sizeof(arr6) / sizeof(arr6[0]);
cout << "The minimum element is "
<< findMin(arr6, 0, n6 - 1) << endl;
int arr7[] = { 1, 2, 3, 4, 5, 6, 7 };
int n7 = sizeof(arr7) / sizeof(arr7[0]);
cout << "The minimum element is "
<< findMin(arr7, 0, n7 - 1) << endl;
int arr8[] = { 2, 3, 4, 5, 6, 7, 8, 1 };
int n8 = sizeof(arr8) / sizeof(arr8[0]);
cout << "The minimum element is "
<< findMin(arr8, 0, n8 - 1) << endl;
int arr9[] = { 3, 4, 5, 1, 2 };
int n9 = sizeof(arr9) / sizeof(arr9[0]);
cout << "The minimum element is "
<< findMin(arr9, 0, n9 - 1) << endl;
return 0;
}
// This is code is contributed by rathbhupendra
C
// C program to find minimum element in a sorted and rotated
// array
#include <stdio.h>
int findMin(int arr[], int low, int high)
{
// This condition is needed to handle the case when
// array is not rotated at all
if (high < low)
return arr[0];
// If there is only one element left
if (high == low)
return arr[low];
// Find mid
int mid = low + (high - low) / 2; /*(low + high)/2;*/
// Check if element (mid+1) is minimum element. Consider
// the cases like {3, 4, 5, 1, 2}
if (mid < high && arr[mid + 1] < arr[mid])
return arr[mid + 1];
// Check if mid itself is minimum element
if (mid > low && arr[mid] < arr[mid - 1])
return arr[mid];
// Decide whether we need to go to left half or right
// half
if (arr[high] > arr[mid])
return findMin(arr, low, mid - 1);
return findMin(arr, mid + 1, high);
}
// Driver program to test above functions
int main()
{
int arr1[] = { 5, 6, 1, 2, 3, 4 };
int n1 = sizeof(arr1) / sizeof(arr1[0]);
printf("The minimum element is %d\n",
findMin(arr1, 0, n1 - 1));
int arr2[] = { 1, 2, 3, 4 };
int n2 = sizeof(arr2) / sizeof(arr2[0]);
printf("The minimum element is %d\n",
findMin(arr2, 0, n2 - 1));
int arr3[] = { 1 };
int n3 = sizeof(arr3) / sizeof(arr3[0]);
printf("The minimum element is %d\n",
findMin(arr3, 0, n3 - 1));
int arr4[] = { 1, 2 };
int n4 = sizeof(arr4) / sizeof(arr4[0]);
printf("The minimum element is %d\n",
findMin(arr4, 0, n4 - 1));
int arr5[] = { 2, 1 };
int n5 = sizeof(arr5) / sizeof(arr5[0]);
printf("The minimum element is %d\n",
findMin(arr5, 0, n5 - 1));
int arr6[] = { 5, 6, 7, 1, 2, 3, 4 };
int n6 = sizeof(arr6) / sizeof(arr6[0]);
printf("The minimum element is %d\n",
findMin(arr6, 0, n6 - 1));
int arr7[] = { 1, 2, 3, 4, 5, 6, 7 };
int n7 = sizeof(arr7) / sizeof(arr7[0]);
printf("The minimum element is %d\n",
findMin(arr7, 0, n7 - 1));
int arr8[] = { 2, 3, 4, 5, 6, 7, 8, 1 };
int n8 = sizeof(arr8) / sizeof(arr8[0]);
printf("The minimum element is %d\n",
findMin(arr8, 0, n8 - 1));
int arr9[] = { 3, 4, 5, 1, 2 };
int n9 = sizeof(arr9) / sizeof(arr9[0]);
printf("The minimum element is %d\n",
findMin(arr9, 0, n9 - 1));
return 0;
}
Java
// Java program to find minimum element in a sorted and
// rotated array
import java.io.*;
import java.lang.*;
import java.util.*;
class Minimum {
static int findMin(int arr[], int low, int high)
{
// This condition is needed to handle the case when
// array is not rotated at all
if (high < low)
return arr[0];
// If there is only one element left
if (high == low)
return arr[low];
// Find mid
int mid
= low + (high - low) / 2; /*(low + high)/2;*/
// Check if element (mid+1) is minimum element.
// Consider the cases like {3, 4, 5, 1, 2}
if (mid < high && arr[mid + 1] < arr[mid])
return arr[mid + 1];
// Check if mid itself is minimum element
if (mid > low && arr[mid] < arr[mid - 1])
return arr[mid];
// Decide whether we need to go to left half or
// right half
if (arr[high] > arr[mid])
return findMin(arr, low, mid - 1);
return findMin(arr, mid + 1, high);
}
// Driver Program
public static void main(String[] args)
{
int arr1[] = { 5, 6, 1, 2, 3, 4 };
int n1 = arr1.length;
System.out.println("The minimum element is "
+ findMin(arr1, 0, n1 - 1));
int arr2[] = { 1, 2, 3, 4 };
int n2 = arr2.length;
System.out.println("The minimum element is "
+ findMin(arr2, 0, n2 - 1));
int arr3[] = { 1 };
int n3 = arr3.length;
System.out.println("The minimum element is "
+ findMin(arr3, 0, n3 - 1));
int arr4[] = { 1, 2 };
int n4 = arr4.length;
System.out.println("The minimum element is "
+ findMin(arr4, 0, n4 - 1));
int arr5[] = { 2, 1 };
int n5 = arr5.length;
System.out.println("The minimum element is "
+ findMin(arr5, 0, n5 - 1));
int arr6[] = { 5, 6, 7, 1, 2, 3, 4 };
int n6 = arr6.length;
System.out.println("The minimum element is "
+ findMin(arr6, 0, n6 - 1));
int arr7[] = { 1, 2, 3, 4, 5, 6, 7 };
int n7 = arr7.length;
System.out.println("The minimum element is "
+ findMin(arr7, 0, n7 - 1));
int arr8[] = { 2, 3, 4, 5, 6, 7, 8, 1 };
int n8 = arr8.length;
System.out.println("The minimum element is "
+ findMin(arr8, 0, n8 - 1));
int arr9[] = { 3, 4, 5, 1, 2 };
int n9 = arr9.length;
System.out.println("The minimum element is "
+ findMin(arr9, 0, n9 - 1));
}
}
Python3
# Python program to find minimum element
# in a sorted and rotated array
def findMin(arr, low, high):
# This condition is needed to handle the case when array is not
# rotated at all
if high < low:
return arr[0]
# If there is only one element left
if high == low:
return arr[low]
# Find mid
mid = int((low + high)/2)
# Check if element (mid+1) is minimum element. Consider
# the cases like [3, 4, 5, 1, 2]
if mid < high and arr[mid+1] < arr[mid]:
return arr[mid+1]
# Check if mid itself is minimum element
if mid > low and arr[mid] < arr[mid - 1]:
return arr[mid]
# Decide whether we need to go to left half or right half
if arr[high] > arr[mid]:
return findMin(arr, low, mid-1)
return findMin(arr, mid+1, high)
# Driver program to test above functions
arr1 = [5, 6, 1, 2, 3, 4]
n1 = len(arr1)
print("The minimum element is " + str(findMin(arr1, 0, n1-1)))
arr2 = [1, 2, 3, 4]
n2 = len(arr2)
print("The minimum element is " + str(findMin(arr2, 0, n2-1)))
arr3 = [1]
n3 = len(arr3)
print("The minimum element is " + str(findMin(arr3, 0, n3-1)))
arr4 = [1, 2]
n4 = len(arr4)
print("The minimum element is " + str(findMin(arr4, 0, n4-1)))
arr5 = [2, 1]
n5 = len(arr5)
print("The minimum element is " + str(findMin(arr5, 0, n5-1)))
arr6 = [5, 6, 7, 1, 2, 3, 4]
n6 = len(arr6)
print("The minimum element is " + str(findMin(arr6, 0, n6-1)))
arr7 = [1, 2, 3, 4, 5, 6, 7]
n7 = len(arr7)
print("The minimum element is " + str(findMin(arr7, 0, n7-1)))
arr8 = [2, 3, 4, 5, 6, 7, 8, 1]
n8 = len(arr8)
print("The minimum element is " + str(findMin(arr8, 0, n8-1)))
arr9 = [3, 4, 5, 1, 2]
n9 = len(arr9)
print("The minimum element is " + str(findMin(arr9, 0, n9-1)))
# This code is contributed by Pratik Chhajer
C#
// C# program to find minimum element
// in a sorted and rotated array
using System;
class Minimum {
static int findMin(int[] arr, int low, int high)
{
// This condition is needed to handle
// the case when array
// is not rotated at all
if (high < low)
return arr[0];
// If there is only one element left
if (high == low)
return arr[low];
// Find mid
// (low + high)/2
int mid = low + (high - low) / 2;
// Check if element (mid+1) is minimum element.
// Consider the cases like {3, 4, 5, 1, 2}
if (mid < high && arr[mid + 1] < arr[mid])
return arr[mid + 1];
// Check if mid itself is minimum element
if (mid > low && arr[mid] < arr[mid - 1])
return arr[mid];
// Decide whether we need to go to
// left half or right half
if (arr[high] > arr[mid])
return findMin(arr, low, mid - 1);
return findMin(arr, mid + 1, high);
}
// Driver Program
public static void Main()
{
int[] arr1 = { 5, 6, 1, 2, 3, 4 };
int n1 = arr1.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr1, 0, n1 - 1));
int[] arr2 = { 1, 2, 3, 4 };
int n2 = arr2.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr2, 0, n2 - 1));
int[] arr3 = { 1 };
int n3 = arr3.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr3, 0, n3 - 1));
int[] arr4 = { 1, 2 };
int n4 = arr4.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr4, 0, n4 - 1));
int[] arr5 = { 2, 1 };
int n5 = arr5.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr5, 0, n5 - 1));
int[] arr6 = { 5, 6, 7, 1, 2, 3, 4 };
int n6 = arr6.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr6, 0, n1 - 1));
int[] arr7 = { 1, 2, 3, 4, 5, 6, 7 };
int n7 = arr7.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr7, 0, n7 - 1));
int[] arr8 = { 2, 3, 4, 5, 6, 7, 8, 1 };
int n8 = arr8.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr8, 0, n8 - 1));
int[] arr9 = { 3, 4, 5, 1, 2 };
int n9 = arr9.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr9, 0, n9 - 1));
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP program to find minimum
// element in a sorted and
// rotated array
function findMin($arr, $low,
$high)
{
// This condition is needed
// to handle the case when
// array is not rotated at all
if ($high < $low) return $arr[0];
// If there is only
// one element left
if ($high == $low) return $arr[$low];
// Find mid
$mid = $low + ($high - $low) / 2; /*($low + $high)/2;*/
// Check if element (mid+1)
// is minimum element.
// Consider the cases like
// (3, 4, 5, 1, 2)
if ($mid < $high &&
$arr[$mid + 1] < $arr[$mid])
return $arr[$mid + 1];
// Check if mid itself
// is minimum element
if ($mid > $low &&
$arr[$mid] < $arr[$mid - 1])
return $arr[$mid];
// Decide whether we need
// to go to left half or
// right half
if ($arr[$high] > $arr[$mid])
return findMin($arr, $low,
$mid - 1);
return findMin($arr,
$mid + 1, $high);
}
// Driver Code
$arr1 = array(5, 6, 1, 2, 3, 4);
$n1 = sizeof($arr1);
echo "The minimum element is " .
findMin($arr1, 0, $n1 - 1) . "\n";
$arr2 = array(1, 2, 3, 4);
$n2 = sizeof($arr2);
echo "The minimum element is " .
findMin($arr2, 0, $n2 - 1) . "\n";
$arr3 = array(1);
$n3 = sizeof($arr3);
echo "The minimum element is " .
findMin($arr3, 0, $n3 - 1) . "\n";
$arr4 = array(1, 2);
$n4 = sizeof($arr4);
echo "The minimum element is " .
findMin($arr4, 0, $n4 - 1) . "\n";
$arr5 = array(2, 1);
$n5 = sizeof($arr5);
echo "The minimum element is " .
findMin($arr5, 0, $n5 - 1) . "\n";
$arr6 = array(5, 6, 7, 1, 2, 3, 4);
$n6 = sizeof($arr6);
echo "The minimum element is " .
findMin($arr6, 0, $n6 - 1) . "\n";
$arr7 = array(1, 2, 3, 4, 5, 6, 7);
$n7 = sizeof($arr7);
echo "The minimum element is " .
findMin($arr7, 0, $n7 - 1) . "\n";
$arr8 = array(2, 3, 4, 5, 6, 7, 8, 1);
$n8 = sizeof($arr8);
echo "The minimum element is " .
findMin($arr8, 0, $n8 - 1) . "\n";
$arr9 = array(3, 4, 5, 1, 2);
$n9 = sizeof($arr9);
echo "The minimum element is " .
findMin($arr9, 0, $n9 - 1) . "\n";
// This code is contributed by ChitraNayal
?>
Javascript
<script>
// Javascript program to find minimum element in a sorted and rotated array
function findMin(arr,low,high)
{
// This condition is needed to handle the case when array
// is not rotated at all
if (high < low)
return arr[0];
// If there is only one element left
if (high == low)
return arr[low];
// Find mid
let mid =low + Math.floor((high - low)/2); /*(low + high)/2;*/
// Check if element (mid+1) is minimum element. Consider
// the cases like {3, 4, 5, 1, 2}
if (mid < high && arr[mid+1] < arr[mid])
return arr[mid+1];
// Check if mid itself is minimum element
if (mid > low && arr[mid] < arr[mid - 1])
return arr[mid];
// Decide whether we need to go to left half or right half
if (arr[high] > arr[mid])
return findMin(arr, low, mid-1);
return findMin(arr, mid+1, high);
}
// Driver Program
let arr1=[5, 6, 1, 2, 3, 4];
let n1 = arr1.length;
document.write("The minimum element is "+ findMin(arr1, 0, n1-1)+"<br>");
let arr2=[1, 2, 3, 4];
let n2 = arr2.length;
document.write("The minimum element is "+ findMin(arr2, 0, n2-1)+"<br>");
let arr3=[1];
let n3 = arr3.length;
document.write("The minimum element is "+ findMin(arr3, 0, n3-1)+"<br>");
let arr4=[1,2];
let n4 = arr4.length;
document.write("The minimum element is "+ findMin(arr4, 0, n4-1)+"<br>");
let arr5=[2,1];
let n5 = arr5.length;
document.write("The minimum element is "+ findMin(arr5, 0, n5-1)+"<br>");
let arr6=[5, 6, 7, 1, 2, 3, 4];
let n6 = arr6.length;
document.write("The minimum element is "+ findMin(arr6, 0, n6-1)+"<br>");
let arr7=[1, 2, 3, 4, 5, 6, 7];
let n7 = arr7.length;
document.write("The minimum element is "+ findMin(arr7, 0, n7-1)+"<br>");
let arr8=[2, 3, 4, 5, 6, 7, 8, 1];
let n8 = arr8.length;
document.write("The minimum element is "+ findMin(arr8, 0, n8-1)+"<br>");
let arr9=[3, 4, 5, 1, 2];
let n9 = arr9.length;
document.write("The minimum element is "+ findMin(arr9, 0, n9-1)+"<br>");
// This code is contributed by avanitrachhadiya2155
</script>
C++
// C++ program to find minimum element in a sorted
// and rotated array containing duplicate elements.
#include <bits/stdc++.h>
using namespace std;
// Function to find minimum element
int findMin(int arr[], int low, int high)
{
while (low < high) {
int mid = low + (high - low) / 2;
if (arr[mid] == arr[high])
high--;
else if (arr[mid] > arr[high])
low = mid + 1;
else
high = mid;
}
return arr[high];
}
// Driver code
int main()
{
int arr1[] = { 5, 6, 1, 2, 3, 4 };
int n1 = sizeof(arr1) / sizeof(arr1[0]);
cout << "The minimum element is "
<< findMin(arr1, 0, n1 - 1) << endl;
int arr2[] = { 1, 2, 3, 4 };
int n2 = sizeof(arr2) / sizeof(arr2[0]);
cout << "The minimum element is "
<< findMin(arr2, 0, n2 - 1) << endl;
int arr3[] = { 1 };
int n3 = sizeof(arr3) / sizeof(arr3[0]);
cout << "The minimum element is "
<< findMin(arr3, 0, n3 - 1) << endl;
int arr4[] = { 1, 2 };
int n4 = sizeof(arr4) / sizeof(arr4[0]);
cout << "The minimum element is "
<< findMin(arr4, 0, n4 - 1) << endl;
int arr5[] = { 2, 1 };
int n5 = sizeof(arr5) / sizeof(arr5[0]);
cout << "The minimum element is "
<< findMin(arr5, 0, n5 - 1) << endl;
int arr6[] = { 5, 6, 7, 1, 2, 3, 4 };
int n6 = sizeof(arr6) / sizeof(arr6[0]);
cout << "The minimum element is "
<< findMin(arr6, 0, n6 - 1) << endl;
int arr7[] = { 1, 2, 3, 4, 5, 6, 7 };
int n7 = sizeof(arr7) / sizeof(arr7[0]);
cout << "The minimum element is "
<< findMin(arr7, 0, n7 - 1) << endl;
int arr8[] = { 2, 3, 4, 5, 6, 7, 8, 1 };
int n8 = sizeof(arr8) / sizeof(arr8[0]);
cout << "The minimum element is "
<< findMin(arr8, 0, n8 - 1) << endl;
int arr9[] = { 3, 4, 5, 1, 2 };
int n9 = sizeof(arr9) / sizeof(arr9[0]);
cout << "The minimum element is "
<< findMin(arr9, 0, n9 - 1) << endl;
return 0;
}
// This code is contributed by Sania Kumari Gupta
// (kriSania804)
C
// C program to find minimum element in a sorted
// and rotated array containing duplicate elements.
#include <stdio.h>
// Function to find minimum element
int findMin(int arr[], int low, int high)
{
while (low < high) {
int mid = low + (high - low) / 2;
if (arr[mid] == arr[high])
high--;
else if (arr[mid] > arr[high])
low = mid + 1;
else
high = mid;
}
return arr[high];
}
// Driver code
int main()
{
int arr1[] = { 5, 6, 1, 2, 3, 4 };
int n1 = sizeof(arr1) / sizeof(arr1[0]);
printf("The minimum element is %d \n",
findMin(arr1, 0, n1 - 1));
int arr2[] = { 1, 2, 3, 4 };
int n2 = sizeof(arr2) / sizeof(arr2[0]);
printf("The minimum element is %d \n",
findMin(arr2, 0, n2 - 1));
int arr3[] = { 1 };
int n3 = sizeof(arr3) / sizeof(arr3[0]);
printf("The minimum element is %d \n",
findMin(arr3, 0, n3 - 1));
int arr4[] = { 1, 2 };
int n4 = sizeof(arr4) / sizeof(arr4[0]);
printf("The minimum element is %d \n",
findMin(arr4, 0, n4 - 1));
int arr5[] = { 2, 1 };
int n5 = sizeof(arr5) / sizeof(arr5[0]);
printf("The minimum element is %d \n",
findMin(arr5, 0, n5 - 1));
int arr6[] = { 5, 6, 7, 1, 2, 3, 4 };
int n6 = sizeof(arr6) / sizeof(arr6[0]);
printf("The minimum element is %d \n",
findMin(arr6, 0, n6 - 1));
int arr7[] = { 1, 2, 3, 4, 5, 6, 7 };
int n7 = sizeof(arr7) / sizeof(arr7[0]);
printf("The minimum element is %d \n",
findMin(arr7, 0, n7 - 1));
int arr8[] = { 2, 3, 4, 5, 6, 7, 8, 1 };
int n8 = sizeof(arr8) / sizeof(arr8[0]);
printf("The minimum element is %d \n",
findMin(arr8, 0, n8 - 1));
int arr9[] = { 3, 4, 5, 1, 2 };
int n9 = sizeof(arr9) / sizeof(arr9[0]);
printf("The minimum element is %d \n",
findMin(arr9, 0, n9 - 1));
return 0;
}
// This code is contributed by Sania Kumari Gupta
// (kriSania804)
Java
// Java program to find minimum element
// in a sorted and rotated array containing
// duplicate elements.
import java.lang.*;
import java.util.*;
class GFG {
// Function to find minimum element
public static int findMin(int arr[], int low, int high)
{
while (low < high) {
int mid = low + (high - low) / 2;
if (arr[mid] == arr[high])
high--;
else if (arr[mid] > arr[high])
low = mid + 1;
else
high = mid;
}
return arr[high];
}
// Driver code
public static void main(String args[])
{
int arr1[] = { 5, 6, 1, 2, 3, 4 };
int n1 = arr1.length;
System.out.println("The minimum element is "
+ findMin(arr1, 0, n1 - 1));
int arr2[] = { 1, 2, 3, 4 };
int n2 = arr2.length;
System.out.println("The minimum element is "
+ findMin(arr2, 0, n2 - 1));
int arr3[] = { 1 };
int n3 = arr3.length;
System.out.println("The minimum element is "
+ findMin(arr3, 0, n3 - 1));
int arr4[] = { 1, 2 };
int n4 = arr4.length;
System.out.println("The minimum element is "
+ findMin(arr4, 0, n4 - 1));
int arr5[] = { 2, 1 };
int n5 = arr5.length;
System.out.println("The minimum element is "
+ findMin(arr5, 0, n5 - 1));
int arr6[] = { 5, 6, 7, 1, 2, 3, 4 };
int n6 = arr6.length;
System.out.println("The minimum element is "
+ findMin(arr6, 0, n6 - 1));
int arr7[] = { 1, 2, 3, 4, 5, 6, 7 };
int n7 = arr7.length;
System.out.println("The minimum element is "
+ findMin(arr7, 0, n7 - 1));
int arr8[] = { 2, 3, 4, 5, 6, 7, 8, 1 };
int n8 = arr8.length;
System.out.println("The minimum element is "
+ findMin(arr8, 0, n8 - 1));
int arr9[] = { 3, 4, 5, 1, 2 };
int n9 = arr9.length;
System.out.println("The minimum element is "
+ findMin(arr9, 0, n9 - 1));
}
}
// This is code is contributed by SoumikMondal
Python3
# Python3 program to find
# minimum element in a sorted
# and rotated array containing
# duplicate elements.
# Function to find minimum element
def findMin(arr, low, high):
while (low < high):
mid = low + (high - low) // 2
if (arr[mid] == arr[high]):
high -= 1
elif (arr[mid] > arr[high]):
low = mid + 1
else:
high = mid
return arr[high]
# Driver code
if __name__ == '__main__':
arr1 = [5, 6, 1, 2, 3, 4]
n1 = len(arr1)
print("The minimum element is ",
findMin(arr1, 0, n1 - 1))
arr2 = [1, 2, 3, 4]
n2 = len(arr2)
print("The minimum element is ",
findMin(arr2, 0, n2 - 1))
arr3 = [1]
n3 = len(arr3)
print("The minimum element is ",
findMin(arr3, 0, n3 - 1))
arr4 = [1, 2]
n4 = len(arr4)
print("The minimum element is ",
findMin(arr4, 0, n4 - 1))
arr5 = [2, 1]
n5 = len(arr5)
print("The minimum element is ",
findMin(arr5, 0, n5 - 1))
arr6 = [5, 6, 7, 1, 2, 3, 4]
n6 = len(arr6)
print("The minimum element is ",
findMin(arr6, 0, n6 - 1))
arr7 = [1, 2, 3, 4, 5, 6, 7]
n7 = len(arr7)
print("The minimum element is ",
findMin(arr7, 0, n7 - 1))
arr8 = [2, 3, 4, 5, 6, 7, 8, 1]
n8 = len(arr8)
print("The minimum element is ",
findMin(arr8, 0, n8 - 1))
arr9 = [3, 4, 5, 1, 2]
n9 = len(arr9)
print("The minimum element is ",
findMin(arr9, 0, n9 - 1))
# This code is contributed by Princi Singh
C#
// C# program to find minimum element
// in a sorted and rotated array
// containing duplicate elements.
using System;
class GFG {
// Function to find minimum element
public static int findMin(int[] arr, int low, int high)
{
while (low < high) {
int mid = low + (high - low) / 2;
if (arr[mid] == arr[high])
high--;
else if (arr[mid] > arr[high])
low = mid + 1;
else
high = mid;
}
return arr[high];
}
// Driver code
public static void Main(String[] args)
{
int[] arr1 = { 5, 6, 1, 2, 3, 4 };
int n1 = arr1.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr1, 0, n1 - 1));
int[] arr2 = { 1, 2, 3, 4 };
int n2 = arr2.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr2, 0, n2 - 1));
int[] arr3 = { 1 };
int n3 = arr3.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr3, 0, n3 - 1));
int[] arr4 = { 1, 2 };
int n4 = arr4.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr4, 0, n4 - 1));
int[] arr5 = { 2, 1 };
int n5 = arr5.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr5, 0, n5 - 1));
int[] arr6 = { 5, 6, 7, 1, 2, 3, 4 };
int n6 = arr6.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr6, 0, n6 - 1));
int[] arr7 = { 1, 2, 3, 4, 5, 6, 7 };
int n7 = arr7.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr7, 0, n7 - 1));
int[] arr8 = { 2, 3, 4, 5, 6, 7, 8, 1 };
int n8 = arr8.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr8, 0, n8 - 1));
int[] arr9 = { 3, 4, 5, 1, 2 };
int n9 = arr9.Length;
Console.WriteLine("The minimum element is "
+ findMin(arr9, 0, n9 - 1));
}
}
// This code is contributed by Amit Katiyar
Javascript
<script>
// JavaScript program to find minimum element in a sorted
// and rotated array containing duplicate elements.
// Function to find minimum element
function findMin(arr, low, high)
{
while(low < high)
{
let mid = Math.floor(low + (high - low)/2);
if (arr[mid] == arr[high])
high--;
else if(arr[mid] > arr[high])
low = mid + 1;
else
high = mid;
}
return arr[high];
}
var arr1 = [5, 6, 1, 2, 3, 4];
var n1 = arr1.length;
document.write("The minimum element is " + findMin(arr1, 0, n1-1) + "<br>");
var arr2 = [1, 2, 3, 4];
var n2 = arr2.length;
document.write("The minimum element is " + findMin(arr2, 0, n2-1) + "<br>");
var arr3 = [1];
var n3 = arr3.length;
document.write("The minimum element is " + findMin(arr3, 0, n3-1) + "<br>");
var arr4 = [1, 2];
var n4 = arr4.length;
document.write("The minimum element is " + findMin(arr4, 0, n4-1) + "<br>");
var arr5 = [2, 1];
var n5 = arr5.length;
document.write("The minimum element is " + findMin(arr5, 0, n5-1) + "<br>");
var arr6 = [5, 6, 7, 1, 2, 3, 4];
var n6 = arr6.length;
document.write("The minimum element is " + findMin(arr6, 0, n6-1) + "<br>");
var arr7 = [1, 2, 3, 4, 5, 6, 7];
var n7 = arr7.length;
document.write("The minimum element is " + findMin(arr7, 0, n7-1) + "<br>");
var arr8 = [2, 3, 4, 5, 6, 7, 8, 1];
var n8 = arr8.length;
document.write("The minimum element is " + findMin(arr8, 0, n8-1) + "<br>");
var arr9 = [3, 4, 5, 1, 2];
var n9 = arr9.length;
document.write("The minimum element is " + findMin(arr9, 0, n9-1) + "<br>");
// This code is contributed by probinsah
</script>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA