Patching Array in Java
"Patching an array" is a standard term for modifying or adding elements to an array in Java. The process can include adding new components to the array, deleting existing elements from the array, or doing any combination of these operations. Depending on the particular needs of your program, there are several situations in which the word "patching" might be used.
Example 1:
Input
nums = [1,3], n = 6
Output
1
Explanation
Combinations of nums are [1], [3], [1,3], which form possible sums of: 1, 3, 4. Now if we add/patch 2 to nums, the combinations are: [1], [2], [3], [1,3], [2,3], [1,2,3]. Possible sums are 1, 2, 3, 4, 5, 6, which now covers the range [1, 6]. So, we only need 1 patch.
Example 2:
Input
nums = [1, 5, 10], n = 20
Output
2
Explanation
The two patches that can be done are [2, 4]. With the help of them we can cover all the numbers in range [1, 20]. The combinations are: [1], [2], [1,2], [4], [5], [1,5], [2,5], [1,2,5], [4,5], [10], [1,10], [2,10], [1,2,10], [4,10], [5,10], [1,5,10], [2,5,10], [1,2,5,10], [4,5,10], [10,10].
Approach 1: Greedy Approach
The greedy approach is a popular algorithmic paradigm used to solve optimization problems. It involves making a series of locally optimal choices with the hope of finding a global optimum solution. In the context of patching arrays or other problems, the greedy approach selects the best possible option at each step without considering future consequences, aiming to achieve the overall best solution.
Algorithm
Step 1: Initialize patches variable to count the number of patches added. Initialize maxReach variable to store the maximum number reachable using current patches.
Step 2: Iterative Process: Iterate through each number in the range [1, n]. If the current number in nums is less than or equal to maxReach + 1, extend the range by adding nums[i] to maxReach.
Step 3: If the current number in nums is greater than maxReach + 1, add a new patch to extend the range.
Step 4: Return the value of patches, which represents the minimum number of patches needed to cover the range [1, n].
Filename: ArrayPatching.java
import java.util.*;
public class ArrayPatching {
public static int minPatches(int[] nums, int n) {
int patches = 0; // Variable to count the number of patches added
long maxReach = 0; // Variable to store the maximum number reachable using current patches
// Iterate through each number in the range [1, n]
for (int i = 0; maxReach < n; ) {
// If the current number in nums is less than or equal to maxReach + 1,
// we can extend the range by adding nums[i] to maxReach
if (i < nums.length && nums[i] <= maxReach + 1) {
maxReach += nums[i++];
} else {
// If the current number in nums is greater than maxReach + 1,
// we need to add a new patch to extend the range
maxReach += maxReach + 1;
patches++;
}
}
return patches;
}
public static void main(String[] args) {
int[] nums = {1, 3};
int n = 6;
int patches = minPatches(nums, n);
System.out.println("Minimum number of patches needed: " + patches);
}
}
Output:
Minimum number of patches needed: 1
Time Complexity
The algorithm iterates through each number in the range [1, n] once. Inside the loop, the operations involve comparing elements of the nums array and updating maxReach. Since each iteration of the loop takes constant time O(1), and there are at most n iterations, the overall time complexity is O(n).
Space Complexity
The space complexity of the algorithm is O(1) because it only uses a constant amount of extra space regardless of the input size. The additional space used does not depend on the size of the input array nums or the target number n. Therefore, the space complexity is constant O(1).
Approach 2: Dynamic Programming
Dynamic programming is a powerful algorithmic technique used to solve optimization problems by breaking them down into simpler subproblems and storing the solutions to these subproblems to avoid redundant computations. In the context of patching an array, dynamic programming can be applied to efficiently determine the minimum number of patches needed to cover a given range of numbers.
Algorithm
Step 1: Create a boolean array dp of size n + 1 to track whether each number from 0 to n can be formed using the elements of nums. Set dp[0] to true since the number 0 can always be formed.
Step 2: Dynamic Programming: Iterate through each number num in the array nums. For each num, update the dp array starting from num up to n. Update dp[i] to true if dp[i - num] is true, indicating that the number i can be formed using num and the previously patched numbers.
Step 3: Check Missing Numbers: Iterate through the range [1, n] to check for missing numbers. If dp[i] is false, it means that the number i cannot be formed using the elements of nums.
Step 4: Add a patch for the missing number i and extend the range by updating the dp array accordingly.
Step 5: Count the number of patches added to cover the missing numbers and return the count of patches as the minimum number of patches needed to cover the range [1, n].
Filename: ArrayPatchingDynamicProgramming.java
public class ArrayPatchingDynamicProgramming {
public static int minPatches(int[] nums, int n) {
// Initialize a boolean array to track whether each number from 0 to n can be formed using the elements of nums
boolean[] dp = new boolean[n + 1];
dp[0] = true; // Base case: Number 0 can always be formed
// Iterate through each number in nums
for (int num : nums) {
// Update dp array for each number in nums
for (int i = n; i >= num; i--) {
dp[i] |= dp[i - num]; // Update dp[i] if dp[i - num] is true
}
}
int patches = 0; // Variable to count the number of patches added
// Iterate through the range [1, n] to check for missing numbers
for (int i = 1; i <= n; i++) {
if (!dp[i]) {
patches++; // Increment patches if the number i cannot be formed
// Extend the range by adding the missing number i
for (int j = n; j >= i; j--) {
dp[j] |= dp[j - i]; // Update dp[j] if dp[j - i] is true
}
}
}
return patches; // Return the count of patches
}
public static void main(String[] args) {
int[] nums = {1, 5, 10};
int n = 20;
int patches = minPatches(nums, n);
System.out.println("Minimum number of patches needed: " + patches);
}
}
Output:
Minimum number of patches needed: 2
Time Complexity
The overall time complexity is O(n * m), where n is the target number and m is the number of elements in the nums array. It is because we iterate through each number from 1 to n, and for each number, we iterate through each element of nums.
Space Complexity
The overall space complexity is O(n), where n is the target number. It is due to the boolean array dp of size n + 1 used to store the solutions to the subproblems.
Approach 3: Binary Search
Binary search is a widely-used algorithmic technique for searching for a target value within a sorted array. It works by repeatedly dividing the search interval in half until the target value is found or the interval becomes empty. In the context of patching an array, binary search can be employed to efficiently locate the missing numbers in the given range.
Algorithm
Step 1: Sort the given array nums in non-decreasing order. Initialize a variable maxReach to track the maximum reachable number starting from 0. Initialize a variable patch to count the number of patches added.
Step 2: Iterate through the range [1, n]: Use binary search to find the first element in nums that is greater than or equal to the current number in the range.
Step 3: If the element is found, update maxReach to include the current number. If the element is not found, add the current number to nums as a patch, update maxReach accordingly, and increment patches.
Step 4: Repeat until the entire range is covered and return the count of patches as the minimum number of patches needed to cover the range [1, n].
Filename: ArrayPatchingBinarySearch.java
import java.util.Arrays;
public class ArrayPatchingBinarySearch {
public static int minPatches(int[] nums, int n) {
Arrays.sort(nums); // Sort the array
long maxReach = 0; // Initialize maximum reachable number
int patches = 0; // Initialize number of patches added
// Iterate through the range [1, n]
for (int i = 1; maxReach < n; i++) {
if (i <= nums.length && nums[i - 1] <= maxReach + 1) {
// If nums[i - 1] is within the reachable range, update maxReach
maxReach += nums[i - 1];
} else {
// If nums[i - 1] is not within the reachable range, add a patch
maxReach += maxReach + 1; // Add the next missing number as a patch
patches++; // Increment the patch count
i--; // Repeat the iteration for the same number
}
}
return patches; // Return the count of patches
}
public static void main(String[] args) {
int[] nums = {1, 5, 10};
int n = 20;
int patches = minPatches(nums, n);
System.out.println("Minimum number of patches needed: " + patches);
}
}
Output:
Minimum number of patches needed: 2
Time Complexity
Sorting the array takes O(m log m) time, where m is the number of elements in the array nums. The iterative binary search process takes O(n log m) time, where n is the target number. Therefore, the overall time complexity is O(m log m + n log m).
Space Complexity
The space complexity is O(1) because the algorithm uses only a constant amount of extra space, regardless of the size of the input array or the target number.