Given a sorted array arr[] (may be distinct or may contain duplicates) of size N that is rotated at some unknown point, the task is to find the minimum element in it.
Examples:
Input: arr[] = {5, 6, 1, 2, 3, 4}
Output: 1
Explanation: 1 is the minimum element present in the array.Input: arr[] = {1, 2, 3, 4}
Output: 1Input: arr[] = {2, 1}
Output: 1
Find the Minimum element in a Sorted and Rotated Array using Linear Search:
A simple solution is to use linear search to traverse the complete array and find a minimum.
Follow the steps mentioned below to implement the idea:
- Declare a variable (say min_ele) to store the minimum value and initialize it with arr[0].
-
Traverse the array from the start.
- Update the minimum value (min_ele) if the current element is less than it.
- Return the final value of min_ele as the required answer.
Below is the implementation of the above approach.
// C++ code to implement the approach #include <bits/stdc++.h> using namespace std;
// Function to find the minimum value int findMin( int arr[], int n)
{ int min_ele = arr[0];
// Traversing over array to
// find minimum element
for ( int i = 0; i < n; i++) {
if (arr[i] < min_ele) {
min_ele = arr[i];
}
}
return min_ele;
} // Driver code int main()
{ int arr[] = { 5, 6, 1, 2, 3, 4 };
int N = sizeof (arr) / sizeof (arr[0]);
// Function call
cout << findMin(arr, N) << endl;
return 0;
} |
/*package whatever //do not write package name here */ import java.io.*;
class GFG {
// Function to find the minimum value
static int findMin( int arr[], int n)
{
int min_ele = arr[ 0 ];
// Traversing over array to
// find minimum element
for ( int i = 0 ; i < n; i++) {
if (arr[i] < min_ele) {
min_ele = arr[i];
}
}
return min_ele;
}
public static void main (String[] args) {
int arr[] = { 5 , 6 , 1 , 2 , 3 , 4 };
int N = arr.length;
System.out.println(findMin(arr, N));
}
} // This code is contributed by aadityaburujwale. |
// C# code to implement above approach using System;
class Minimum {
static int findMin( int [] arr, int N)
{
int min_ele = arr[0];
// Traversing over array to
// find minimum element
for ( int i = 0; i < N; i++) {
if (arr[i] < min_ele) {
min_ele = arr[i];
}
}
return min_ele;
}
// Driver Program
public static void Main()
{
int [] arr = { 5, 6, 1, 2, 3, 4 };
int N = arr.Length;
Console.WriteLine(findMin(arr, N));
}
} // This code is contributed by aditya942003patil. |
// JS code to implement the approach // Function to find the minimum value function findMin(arr, n) {
let min_ele = arr[0];
// Traversing over array to
// find minimum element
for (let i = 0; i < n; i++) {
if (arr[i] < min_ele) {
min_ele = arr[i];
}
}
return min_ele;
} // Driver code let arr = [5, 6, 1, 2, 3, 4]; let N = arr.length; // Function call console.log(findMin(arr, N)); // This code is contributed by adityamaharshi21. |
# python3 code to implement the approach def findMin(arr, N):
min_ele = arr[ 0 ];
# Traversing over array to
# find minimum element
for i in range (N) :
if arr[i] < min_ele :
min_ele = arr[i]
return min_ele;
# Driver program arr = [ 5 , 6 , 1 , 2 , 3 , 4 ]
N = len (arr)
print (findMin(arr,N))
# This code is contributed by aditya942003patil |
1
Time Complexity: O(N)
Auxiliary Space: O(1)
Find the Minimum element in a Sorted and Rotated Array using Binary Search:
We start by checking if the array is not rotated. If it is not rotated, the minimum element is simply the first element of the array.
If the array is rotated, we use binary search to narrow down the search space. We compare the middle element with the left and right elements to determine which half of the array contains the minimum element.
However, we need to handle the case where the middle element is equal to the left and right elements. This indicates that the array is rotated and the minimum element could be either the middle element, the left element, or the right element.
To handle this case, we update the minimum element to the minimum of the current minimum and the middle element. We then increment the left pointer and decrement the right pointer to continue the search.
By repeatedly narrowing down the search space and handling duplicates, we can efficiently find the minimum element in a rotated sorted array.
Below is the implementation of the above approach:
#include <bits/stdc++.h> using namespace std;
int findMin(vector< int >& arr, int low, int high)
{ // If the array is not rotated
if (arr[low] < arr[high]) {
return arr[low];
}
int ans = 1e9;
// Binary search
while (low <= high) {
int mid = (low + high) / 2;
// if left most element is equal with right most and
// middle element then we can reduce the search
// space only by increasing the lower limit and
// decreasing the upper limit
if (arr[mid] == arr[low]
and arr[mid] == arr[high]) {
ans = min(ans,arr[mid]);
low++;
high--;
}
// If the left half is sorted, the minimum element
// must be in the right half
else if (arr[mid] > arr[high]) {
low = mid + 1;
}
// If the right half is sorted, the minimum element
// must be in the left half
else {
ans = min(ans,arr[mid]);
high = mid - 1;
}
}
// If no minimum element is found, return -1
return ans;
} // Driver program to test above functions int main()
{ vector< int > arr = {7, 8, 9, 1, 2, 3, 4, 5, 6};
int N = arr.size();
cout << "The minimum element is "
<< findMin(arr, 0, N - 1) << endl;
return 0;
} |
import java.util.*;
public class Main {
public static int findMin(List<Integer> arr, int low,
int high)
{
// If the array is not rotated
if (arr.get(low) < arr.get(high)) {
return arr.get(low);
}
int ans = 1000000000 ;
// Binary search
while (low <= high) {
int mid = (low + high) / 2 ;
if (arr.get(mid)== arr.get(low) && arr.get(mid) == arr.get(high)) {
ans = Math.min(ans,arr.get(mid));
low++;
high--;
}
// If the left half is sorted, the minimum
// element must be in the right half
else if (arr.get(mid) > arr.get(high)) {
low = mid + 1 ;
}
// If the right half is sorted, the minimum
// element must be in the left half
else {
ans = Math.min(ans,arr.get(mid));
high = mid - 1 ;
}
}
// If no minimum element is found, return -1
return ans;
}
// Driver program to test above functions
public static void main(String[] args)
{
List<Integer> arr = new ArrayList<>(
Arrays.asList( 5 , 6 , 1 , 2 , 3 , 4 ));
int N = arr.size();
System.out.println( "The minimum element is "
+ findMin(arr, 0 , N - 1 ));
}
} |
using System;
using System.Collections.Generic;
public class Program {
public static int findMin(List< int > arr, int low, int high) {
// If the array is not rotated
if (arr[low] <= arr[high]) {
return arr[low];
}
// Binary search
while (low <= high) {
int mid = (low + high) / 2;
// Check if mid is the minimum element
if (arr[mid] < arr[mid - 1]) {
return arr[mid];
}
// If the left half is sorted, the minimum element must be in the right half
if (arr[mid] > arr[high]) {
low = mid + 1;
}
// If the right half is sorted, the minimum element must be in the left half
else {
high = mid - 1;
}
}
// If no minimum element is found, return -1
return -1;
}
// Driver program to test above functions
public static void Main() {
List< int > arr = new List< int > {5, 6, 1, 2, 3, 4};
int N = arr.Count;
Console.WriteLine( "The minimum element is " + findMin(arr, 0, N - 1));
}
} // This code is contributed by Prajwal Kandekar |
function findMin(arr, low, high) {
// If the array is not rotated
if (arr[low] <= arr[high]) {
return arr[low];
}
// Binary search
while (low <= high) {
let mid = Math.floor((low + high) / 2);
// Check if mid is the minimum element
if (arr[mid] < arr[mid - 1]) {
return arr[mid];
}
// If the left half is sorted, the minimum element must be in the right half
if (arr[mid] > arr[high]) {
low = mid + 1;
}
// If the right half is sorted, the minimum element must be in the left half
else {
high = mid - 1;
}
}
// If no minimum element is found, return -1
return -1;
} // Driver program to test above functions let arr = [5, 6, 1, 2, 3, 4]; let N = arr.length; console.log( "The minimum element is " + findMin(arr, 0, N - 1));
|
def findMin(arr, low, high):
# If the array is not rotated
if arr[low] < = arr[high]:
return arr[low]
# Binary search
while low < = high:
mid = (low + high) / / 2
# Check if mid is the minimum element
if arr[mid] < arr[mid - 1 ]:
return arr[mid]
# If the left half is sorted, the minimum element must be in the right half
if arr[mid] > arr[high]:
low = mid + 1
# If the right half is sorted, the minimum element must be in the left half
else :
high = mid - 1
# If no minimum element is found, return None
return None
# Driver program to test above functions if __name__ = = '__main__' :
arr = [ 5 , 6 , 1 , 2 , 3 , 4 ]
N = len (arr)
print ( "The minimum element is " + \
str (findMin(arr, 0 , N - 1 )))
|
The minimum element is 1
Time complexity: Worst Case: O(N) Average Case: O(logn)
Where n is the number of elements in the array. This is because the algorithm uses binary search, which has a logarithmic time complexity.
Auxiliary Space: O(1), the algorithm uses a constant amount of extra space to store variables such as low, high, and mid, regardless of the size of the input array.