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The task is to find all unique triplets in an array that add up to zero. A triplet consists of three numbers, and the sum of these three numbers should be zero. For example:
Input: [-1, 0, 1, 2, -1, -4]
Output: [[-1, -1, 2], [-1, 0, 1]]
Sort the Array:
First, sort the array in ascending order. Sorting helps to efficiently find the triplets using two pointers.
Fix One Element:
Iterate through the array and fix one element. For each fixed element, find two other elements (using two pointers) such that their sum equals the negative of the fixed element.
Use Two Pointers:
After fixing one element, use two pointers:
Avoid Duplicates:
Skip duplicate elements to ensure the triplets are unique.
Add Valid Triplets:
If the sum of the triplet is zero, add it to the result list.
import java.util.*;
public class ZeroSumTriplets {
public static List<List<Integer>> findTriplets(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
Arrays.sort(nums); // Step 1: Sort the array
for (int i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue; // Skip duplicates for the first element
}
int left = i + 1; // Pointer 1
int right = nums.length - 1; // Pointer 2
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if (sum == 0) {
result.add(Arrays.asList(nums[i], nums[left], nums[right]));
// Move both pointers and skip duplicates
while (left < right && nums[left] == nums[left + 1]) left++;
while (left < right && nums[right] == nums[right - 1]) right--;
left++;
right--;
} else if (sum < 0) {
left++; // Move the left pointer to increase the sum
} else {
right--; // Move the right pointer to decrease the sum
}
}
}
return result;
}
public static void main(String[] args) {
int[] nums = {-1, 0, 1, 2, -1, -4};
List<List<Integer>> triplets = findTriplets(nums);
System.out.println("Triplets with zero sum: " + triplets);
}
}
This solution is efficient and easy to implement for finding triplets with zero sum.
| operation | REG | DCFL | CFL | CSL | RC | RE |
|---|---|---|---|---|---|---|
| union | Y | N | Y | Y | Y | Y |
| intersection | Y | N | N | Y | Y | Y |
| set difference | Y | N | N | Y | Y | N |
| complementation | Y | Y | N | Y | Y | N |
| intersection with a regular language | Y | Y | Y | Y | Y | Y |
| concatenation | Y | N | Y | Y | Y | Y |
| Kleene star | Y | N | Y | Y | Y | Y |
| Kleene plus | Y | N | Y | Y | Y | Y |
| reversal | Y | Y | Y | Y | Y | Y |
| lambda-free homomorphism | Y | N | Y | Y | Y | Y |
| homomorphism | Y | N | Y | N | N | Y |
| inverse homomorphism | Y | Y | Y | Y | Y | Y |
| lambda-free substitution | Y | N | Y | Y | Y | Y |
| substitution | Y | N | Y | N | N | Y |
| lambda-free GSM mapping | Y | N | Y | Y | Y | Y |
| GSM mapping | Y | N | Y | N | N | Y |
| inverse GSM mapping | Y | Y | Y | Y | Y | Y |
| lambda- limited erasing | Y | Y | Y | Y | ||
| rational transduction | Y | N | Y | N | N | Y |
| right quotient with a regular language | Y | Y | Y | N | Y | |
| left quotient with a regular language | Y | Y | Y | N | Y |
| Abbreviation | Name |
|---|---|
| REG | regular |
| DCFL | deterministic context-free |
| CFL | context-free |
| CSL | context-sensitive |
| RC | recursive |
| RE | recursively enumerable |
A tri-state buffer can be thought of as a switch. If B is on, the switch is closed. If B is off, the switch is open.
|
| INPUT | OUTPUT | |
| A | B | C |
| 0 | 1 | 0 |
| 1 | 1 | |
| X | 0 | Z (high impedance) |
| FLIP-FLOP NAME | FLIP-FLOP SYMBOL | CHARACTERISTIC TABLE | CHARACTERISTIC EQUATION | EXCITATION TABLE | |||||||||||||||||||||||||||||||||||
| SR | ![]() |
| Q(next) = S + R'QSR = 0 |
| |||||||||||||||||||||||||||||||||||
| JK | ![]() |
| Q(next) = JQ' + K'Q |
| |||||||||||||||||||||||||||||||||||
| D | ![]() |
| Q(next) = D |
| |||||||||||||||||||||||||||||||||||
| T | ![]() |
| Q(next) = TQ' + T'Q |
|
class Functor
{
public:
int operator()(int a, int b)
{
return a < b;
}
};
int main()
{
Functor f;
int a = 5;
int b = 7;
int ans = f(a, b);
}
| A run of the heapsort algorithm sorting an array of randomly permuted values. In the first stage of the algorithm the array elements are reordered to satisfy the heap property. Before the actual sorting takes place, the heap tree structure is shown briefly for illustration. | |
| Class | Sorting algorithm |
|---|---|
| Data structure | Array |
| Worst case performance | ![]() |
| Best case performance | [1] |
| Average case performance | ![]() |
| Worst case space complexity | auxiliary |
JMP LATER
...
...
LATER:
This is known as a forward reference. If the assembler is processing the file one line at a time, then it doesn't know where LATER is when it first encounters the jump instruction. So, it doesn't know if the jump is a short jump, a near jump or a far jump. There is a large difference amongst these instructions. They are 2, 3, and 5 bytes long respectively. The assembler would have to guess how far away the instruction is in order to generate the correct instruction. If the assembler guesses wrong, then the addresses for all other labels later in the program woulds be wrong, and the code would have to be regenerated. Or, the assembler could alway choose the worst case. But this would mean generating inefficiency in the program, since all jumps would be considered far jumps and would be 5 bytes long, where actually most jumps are short jumps, which are only 2 bytes long.Note: If an error is found in the first pass, the assembly process terminates and does not continue to the second pass. If this occurs, the assembler listing only contains errors and warnings generated during the first pass of the assembler.

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