Find the minimum element in a sorted and rotated array
A sorted array is rotated at some unknown point, find the minimum element in it.
The following solution assumes that all elements are distinct.
Examples:
Input: {5, 6, 1, 2, 3, 4}
Output: 1
Input: {1, 2, 3, 4}
Output: 1
Input: {2, 1}
Output: 1
A simple solution is to traverse the complete array and find a minimum. This solution requires O(n) time.
We can do it in O(Logn) using Binary Search. If we take a closer look at the above examples, we can easily figure out the following pattern:
- The minimum element is the only element whose previous is greater than it. If there is no previous element, then there is no rotation (the first element is minimum). We check this condition for the middle element by comparing it with (mid-1)’th and (mid+1)’th elements.
- If the minimum element is not at the middle (neither mid nor mid + 1), then the minimum element lies in either the left half or right half.
- If the middle element is smaller than the last element, then the minimum element lies in the left half
- Else minimum element lies in the right half.
We strongly recommend you to try it yourself before seeing the following implementation.
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; } |
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 |
Java
// Java program to find minimum element in a sorted and rotated array import java.util.*; import java.lang.*; import java.io.*; 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)); } } |
Python
# 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 ?> |
Output:
The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1
How to handle duplicates?
The above approach in the worst case(If all the elements are the same) takes O(N).
Below is the code to handle duplicates in O(log n) time.
C++
// C++ program to find minimum element in a sorted // and rotated array contating 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 is code is contributed by Saptakatha Adak. |
Java
// Java program to find minimum element // in a sorted and rotated array contating // duplicate elements. import java.util.*; import java.lang.*; 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 contating # 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 // contating 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 |
Output:
The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1
This article is contributed by Abhay Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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