
Problem Statement
Imagine we have an array of integers of length 'n' that is initially sorted in ascending order. This array is then subjected to multiple rotations, between 1
and n
times. For instance, the array nums = [0, 1, 4, 4, 5, 6, 7]
could be transformed into [4, 5, 6, 7, 0, 1, 4]
after being rotated four times. With each rotation, the last element of the array moves to the front, shifting all other elements one position to the right. The challenge presented is to determine and return the smallest element in such a rotated and possibly duplicate-laden array efficiently, minimizing the number of operational steps needed to achieve this.
Examples
Example 1
Input:
nums = [1,3,5]
Output:
1
Example 2
Input:
nums = [2,2,2,0,1]
Output:
0
Constraints
n == nums.length
1 <= n <= 5000
-5000 <= nums[i] <= 5000
nums
is sorted and rotated between1
andn
times.
Approach and Intuition
To efficiently find the minimum element in a rotated sorted array that may include duplicates, we can leverage a modified binary search. Given the constraints and nature of the problem, here’s a systematic step-by-step approach:
Initiate pointers: start (
low
) at0
and end (high
) atn-1
of thenums
array.Enter a while loop that continues as long as
low < high
:- Calculate the midpoint
mid
usingmid = low + (high - low) / 2
. - If the element at
mid
is greater than the element athigh
, it implies the smallest value is in the right half of the array. Thus, setlow = mid + 1
. - If the element at
mid
is less than the element athigh
, it indicates the smallest value is in the left half or at mid. Hence, sethigh = mid
. - If
nums[mid]
equalsnums[high]
, there is an ambiguity due to duplicate values. Reduce the search space conservatively by decrementinghigh
by one.
- Calculate the midpoint
After exiting the loop, the pointer
low
will point to the smallest element since the loop narrows down the search window to the point wherelow
equalshigh
.
Considering the array may contain duplicates, this approach is particularly beneficial as straightforward binary search would not suffice without modifications. The slight adjustment where we decrement high
addresses cases where the middle element is the same as the end element, a scenario common in arrays with duplicates. This approach ensures the algorithm remains efficient and runs in logarithmic time complexity in the best and average case, making it suitable for the input size constraints provided.
Solutions
- C++
- Java
- C
- JavaScript
- Python
class Solution {
public:
int getMinimum(vector<int>& numbers) {
int start = 0, end = numbers.size() - 1;
while (start < end) {
int mid = start + (end - start) / 2;
if (numbers[mid] < numbers[end])
end = mid;
else if (numbers[mid] > numbers[end])
start = mid + 1;
else
end -= 1;
}
return numbers[start];
}
};
This solution tackles the problem of finding the minimum element in a potentially duplicated, rotated sorted array using C++. The approach leverages a binary search technique but with modifications to handle duplicate values and the rotation of the array.
The function getMinimum
operates as follows:
- Initialize two pointers:
start
at index 0 andend
at the last index of thenumbers
vector. - Execute a while loop that continues as long as
start
is less thanend
.- Calculate the midpoint (
mid
) of the currentstart
andend
. - Compare the element at the midpoint (
numbers[mid]
) with the element atend
(numbers[end]
):- If
numbers[mid]
is less thannumbers[end]
, narrow the search to the left half by settingend
tomid
. - If
numbers[mid]
is greater thannumbers[end]
, narrow the search to the right half by settingstart
tomid + 1
. - If they are equal, decrement
end
by one to reduce the search space slowly, handling potential duplicates effectively.
- If
- Calculate the midpoint (
- Once the loop exits,
start
should be at the smallest element's position. Returnnumbers[start]
.
This strategy efficiently finds the minimum value in the array, ensuring optimum performance even in cases where duplicates exist and the typical binary search might falter.
class Solution {
public int getMinimum(int[] array) {
int start = 0, end = array.length - 1;
while (start < end) {
int mid = start + (end - start) / 2;
if (array[mid] < array[end]) end = mid;
else if (array[mid] > array[end]) start = mid + 1;
else end -= 1;
}
return array[start];
}
}
This article explains how to solve the problem of finding the minimum element in a rotated sorted array that may contain duplicates using Java.
The provided Java solution implements a binary search strategy to efficiently find the minimum element. Here's an overview of the approach:
- Initialize
start
to0
andend
to the last index of the array. - Use a
while
loop to continue searching as long asstart
is less thanend
. - Calculate the middle index
mid
betweenstart
andend
. - Compare the middle element with the end element to decide if you need to search the left half or the right half of the array:
- If the middle element is less than the end element, set
end
tomid
. - If the middle element is greater than the end element, set
start
tomid + 1
. - If they are equal, decrement
end
by one to narrow the search range without skipping the minimum element.
- If the middle element is less than the end element, set
- Once the loop terminates, the minimum element is at the index
start
.
By employing this method, you ensure that you deal efficiently with duplicates and potentially reduce the search space more strategically.
int getMinimum(int* array, int size) {
int start = 0, end = size - 1;
while (start < end) {
int middle = start + (end - start) / 2;
if (array[middle] < array[end])
end = middle;
else if (array[middle] > array[end])
start = middle + 1;
else
end -= 1;
}
return array[start];
}
This summary addresses the problem of finding the minimum element in an array that has been rotated and might contain duplicates. You will implement this using the C programming language with the provided function getMinimum
.
The function
getMinimum
takes two parameters:array
- a pointer to the first element of the array.size
- the number of elements in the array.
The function utilizes a binary search mechanism to efficiently find the minimum value in the rotated sorted array. The approach deals effectively with duplicates by adjusting the search space.
Here’s a breakdown of the logic:
- Initialize two pointers,
start
andend
, to reference the beginning and end of the array respectively. - Use a while loop to continue searching while
start
is less thanend
. - Calculate the middle index of the current search interval.
- Compare the element at the middle index with the element at the end index:
- If the middle element is less than the end element, adjust the
end
pointer tomiddle
. - If the middle element is greater than the end element, adjust the
start
pointer tomiddle + 1
. - If they are equal, decrement the
end
pointer by 1 to narrow the search interval.
- If the middle element is less than the end element, adjust the
- Once the loop concludes, the
start
index will point to the minimum element in the rotated array.
- Initialize two pointers,
Implement this in your C code to effectively find the smallest element, even when the input array has duplicates and is rotated.
var minimum = function (array) {
let start = 0,
end = array.length - 1;
while (start < end) {
let mid = start + Math.floor((end - start) / 2);
if (array[mid] < array[end]) end = mid;
else if (array[mid] > array[end]) start = mid + 1;
else end -= 1;
}
return array[start];
};
This Javascript implementation effectively solves the problem of finding the minimum element in a rotated sorted array which may contain duplicates. The code utilizes a binary search approach, optimizing the search process even when the array is rotated and duplicates are present.
- In the function
minimum
, two pointers are initialized,start
andend
, representing the boundaries of the current search index within the array. - The while loop continues as long as
start
is less thanend
.- Calculate the middle index
mid
using thestart
andend
. - Three conditions help to refine the search area:
- If the element at the middle (
array[mid]
) is less than the element atend
, modifyend
to bemid
, narrowing the search to the left segment. - If the middle element is greater than the end element, move
start
tomid + 1
, focusing on the right segment. - If the middle element equals the end element, decrement the
end
index to reduce redundancy without missing the minimum.
- If the element at the middle (
- Calculate the middle index
- Once the loop finishes,
array[start]
holds the minimum value, which is returned as the result.
This code ensures efficiency even in scenarios with large arrays and maintains a low time complexity, ideal for cases where performance is critical.
class Solution:
def findMinimum(self, arr: List[int]) -> int:
left = 0
right = len(arr) - 1
while right > left:
mid = left + (right - left) // 2
if arr[mid] < arr[right]:
right = mid
elif arr[mid] > arr[right]:
left = mid + 1
else:
right -= 1
return arr[left]
Explore how to find the minimum element in a sorted, potentially rotated, array where the array may contain duplicates using Python. The provided method uses a binary search approach to efficiently locate the minimum element even when duplicates are present.
- Initialize two pointers,
left
andright
, at the start and end of the array respectively. - Continue to narrow down the search space by adjusting the
left
andright
pointers until they converge:- Calculate the midpoint of the current search space.
- Compare the middle element with the element at the
right
pointer:- If the middle element is less than the element at
right
, adjust theright
pointer tomid
. - If the middle element is greater than the element at
right
, adjust theleft
pointer to one position right ofmid
. - If they're equal, decrement the
right
pointer by one to skip over the duplicate.
- If the middle element is less than the element at
- Eventually, the
left
pointer points to the smallest element whenright
converges toleft
.
This technique makes it possible to handle scenarios where the smallest number might be a duplicate and ensures the solution remains efficient even in those cases.
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