Introduction
In programming, dealing with arrays is a common task. One interesting problem that often arises is finding the longest subarray of 1’s after deleting one element.
This problem has various implementations in different programming languages such as C, C++, Java, and Python. In this article, we will explore different approaches to solving this problem in each of these languages.
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By the end, you will have a clear understanding of how to find the longest subarray of 1’s after deleting one element in C, C++, Java, and Python.
Understanding the Problem
Before diving into the implementations, let’s first establish a clear understanding of the problem at hand. We are given an array consisting of 0’s and 1’s.
Our task is to find the length of the longest subarray containing only 1’s that can be obtained by deleting one element from the given array.
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To tackle this problem efficiently, we will explore different strategies and code implementations in C, C++, Java, and Python. Let’s start with C.
Longest Subarray of 1’s After Deleting One Element in C
To find the longest subarray of 1’s after deleting one element in C, we can use a two-pointer approach. Here’s an example code snippet:
#include <stdio.h>
int longestSubarray(int arr[], int n) {
int left = 0, right = 0, maxLen = 0, count = 0;
while (right < n) {
if (arr[right] == 1)
count++;
if (right - left + 1 - count > 1) {
if (arr[left] == 1)
count--;
left++;
}
maxLen = (right - left + 1 > maxLen) ? right - left + 1 : maxLen;
right++;
}
return maxLen;
}
int main() {
int arr[] = {1, 0, 1, 1, 0, 1, 1, 1};
int n = sizeof(arr) / sizeof(arr[0]);
int longest = longestSubarray(arr, n);
printf("The longest subarray of 1's after deleting one element is: %d\n", longest);
return 0;
}
In the above C code, we initialize two pointers, left
and right
, to track the subarray boundaries. We also maintain a count
variable to keep track of the number of 1’s in the current subarray.
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By moving the pointers appropriately and updating the count
, we can find the length of the longest subarray of 1’s after deleting one element.
Longest Subarray of 1’s After Deleting One Element in C++
Moving on to C++, we can leverage the power of the Standard Template Library (STL) to solve this problem. Here’s an example code snippet:
#include <iostream>
#include <vector>
using namespace std;
int longestSubarray(vector<int>& arr) {
int left = 0, right = 0, maxLen = 0, count = 0;
while (right < arr.size()) {
if (arr[right] == 1)
count++;
if (right - left + 1 - count > 1) {
if (arr[left] == 1)
count--;
left++;
}
maxLen = max(maxLen, right - left + 1);
right++;
}
return maxLen;
}
int main() {
vector<int> arr = {1, 0, 1, 1, 0, 1, 1, 1};
int longest = longestSubarray(arr);
cout << "The longest subarray of 1's after deleting one element is: " << longest << endl;
return 0;
}
In the above C++ code, we utilize the std::vector
container from the STL to represent the array. The rest of the logic remains the same as the C implementation.
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Longest Subarray of 1’s After Deleting One Element in Java
Moving further, let’s explore the Java implementation. Here’s an example code snippet:
public class Main {
public static int longestSubarray(int[] arr) {
int left = 0, right = 0, maxLen = 0, count = 0;
while (right < arr.length) {
if (arr[right] == 1)
count++;
if (right - left + 1 - count > 1) {
if (arr[left] == 1)
count--;
left++;
}
maxLen = Math.max(maxLen, right - left + 1);
right++;
}
return maxLen;
}
public static void main(String[] args) {
int[] arr = {1, 0, 1, 1, 0, 1, 1, 1};
int longest = longestSubarray(arr);
System.out.println("The longest subarray of 1's after deleting one element is: " + longest);
}
}
The Java implementation follows a similar approach using two pointers and a count variable. By traversing the array and updating the pointers, we can find the length of the longest subarray of 1’s after deleting one element.
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Longest Subarray of 1’s After Deleting One Element in Python
Lastly, let’s explore the Python implementation. Here’s an example code snippet:
def longest_subarray(arr):
left = right = max_len = count = 0
while right < len(arr):
if arr[right] == 1:
count += 1
if right - left + 1 - count > 1:
if arr[left] == 1:
count -= 1
left += 1
max_len = max(max_len, right - left + 1)
right += 1
return max_len
arr = [1, 0, 1, 1, 0, 1, 1, 1]
longest = longest_subarray(arr)
print("The longest subarray of 1's after deleting one element is:", longest)
The Python implementation is similar to the previous ones, utilizing two pointers and a count variable.
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By iterating through the array and updating the pointers, we can find the length of the longest subarray of 1’s after deleting one element.
Frequently Asked Questions (FAQs)
The time complexity of the solution is O(n), where n is the size of the given array. The solution performs a single pass over the array, making it efficient.
No, the problem statement specifies that the array consists only of 0’s and 1’s. If the array contains other elements, the solution may not produce the correct result.
In the case of multiple longest subarrays, the solution will return the length of the first longest subarray encountered during the traversal.
Yes, the solution uses a constant amount of additional space. It only requires a few variables to track the pointers, count, and maximum length.
Yes, there can be multiple approaches to solve this problem. One alternative approach is to use dynamic programming to keep track of the longest subarray length at each index.
Yes, the solution can handle large arrays efficiently. It performs a single pass over the array, making it scalable for larger input sizes.
Conclusion
In this article, we explored different approaches to finding the longest subarray of 1’s after deleting one element in C, C++, Java, and Python.
By using a two-pointer approach, we can efficiently solve this problem in each of these programming languages. Remember to adapt the code to your specific use case and consider any edge cases that may arise.