Find the Minimum element in a Sorted and Rotated Array
Last Updated :
07 Mar, 2024
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: 1
Input: 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++
#include <bits/stdc++.h>
using namespace std;
int findMin(int arr[], int n)
{
int min_ele = arr[0];
for (int i = 0; i < n; i++) {
if (arr[i] < min_ele) {
min_ele = arr[i];
}
}
return min_ele;
}
int main()
{
int arr[] = { 5, 6, 1, 2, 3, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << findMin(arr, N) << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
static int findMin(int arr[], int n)
{
int min_ele = arr[0];
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));
}
}
|
C#
using System;
class Minimum {
static int findMin(int[] arr, int N)
{
int min_ele = arr[0];
for (int i = 0; i < N; i++) {
if (arr[i] < min_ele) {
min_ele = arr[i];
}
}
return min_ele;
}
public static void Main()
{
int[] arr = { 5, 6, 1, 2, 3, 4 };
int N = arr.Length;
Console.WriteLine(findMin(arr, N));
}
}
|
Javascript
function findMin(arr, n) {
let min_ele = arr[0];
for (let i = 0; i < n; i++) {
if (arr[i] < min_ele) {
min_ele = arr[i];
}
}
return min_ele;
}
let arr = [5, 6, 1, 2, 3, 4];
let N = arr.length;
console.log(findMin(arr, N));
|
Python3
def findMin(arr, N):
min_ele = arr[0];
for i in range(N) :
if arr[i] < min_ele :
min_ele = arr[i]
return min_ele;
arr = [5, 6, 1, 2, 3, 4]
N = len(arr)
print(findMin(arr,N))
|
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:
C++
#include <bits/stdc++.h>
using namespace std;
int findMin(vector<int>& arr, int low, int high)
{
if (arr[low] < arr[high]) {
return arr[low];
}
int ans = 1e9;
while (low <= high) {
int mid = (low + high) / 2;
if (arr[mid] == arr[low]
and arr[mid] == arr[high]) {
ans = min(ans,arr[mid]);
low++;
high--;
}
else if (arr[mid] > arr[high]) {
low = mid + 1;
}
else {
ans = min(ans,arr[mid]);
high = mid - 1;
}
}
return ans;
}
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;
}
|
Java
import java.util.*;
public class Main {
public static int findMin(List<Integer> arr, int low,
int high)
{
if (arr.get(low) < arr.get(high)) {
return arr.get(low);
}
int ans = 1000000000;
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--;
}
else if (arr.get(mid) > arr.get(high)) {
low = mid + 1;
}
else {
ans = Math.min(ans,arr.get(mid));
high = mid - 1;
}
}
return ans;
}
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));
}
}
|
C#
using System;
using System.Collections.Generic;
public class Program {
public static int findMin(List<int> arr, int low, int high) {
if (arr[low] <= arr[high]) {
return arr[low];
}
while (low <= high) {
int mid = (low + high) / 2;
if (arr[mid] < arr[mid - 1]) {
return arr[mid];
}
if (arr[mid] > arr[high]) {
low = mid + 1;
}
else {
high = mid - 1;
}
}
return -1;
}
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));
}
}
|
Javascript
function findMin(arr, low, high) {
if (arr[low] <= arr[high]) {
return arr[low];
}
while (low <= high) {
let mid = Math.floor((low + high) / 2);
if (arr[mid] < arr[mid - 1]) {
return arr[mid];
}
if (arr[mid] > arr[high]) {
low = mid + 1;
}
else {
high = mid - 1;
}
}
return -1;
}
let arr = [5, 6, 1, 2, 3, 4];
let N = arr.length;
console.log("The minimum element is " + findMin(arr, 0, N - 1));
|
Python3
def findMin(arr, low, high):
if arr[low] <= arr[high]:
return arr[low]
while low <= high:
mid = (low + high) // 2
if arr[mid] < arr[mid-1]:
return arr[mid]
if arr[mid] > arr[high]:
low = mid + 1
else:
high = mid - 1
return None
if __name__ == '__main__':
arr = [5, 6, 1, 2, 3, 4]
N = len(arr)
print("The minimum element is " + \
str(findMin(arr, 0, N-1)))
|
Output
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.
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...