the new keyword is used to create new objects or instances of a class. It is essential for memory allocation in Java, as it initializes objects dynamically during runtime.

what is the output of the ff.

String ace = "Ace Coding";
System.out.println(ace.startsWith('A');
System.out.println(ace.endsWith('g');

💻 A2SV prep: Two pointers

🟢 Two sum
🟡 Two sum ll input array is sorted

👆The above two are easy; warm up

🟡 3sum Click here

Solution:

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        res = []
        nums.sort()
        n = len(nums)
        for i in range(n-2):
            if nums[i] > 0:
                break
            if i > 0 and nums[i] == nums[i-1]:
                continue
            
            L, R = i + 1, n - 1
            while L < R:
                sum = nums[i] + nums[L] + nums[R]
                if sum == 0:
                    res.append([nums[i], nums[L], nums[R]])
                    L +=1 
                    R -=1
                    while L < R and nums[L] == nums[L-1]:
                        L += 1
                    while L < R and nums[R] == nums[R+1]:
                        R -= 1
                    
                elif sum > 0:
                    R -= 1
                else:
                    L += 1
    
        return res

🟡 3sum closest :- click here

Solution:

python 
class Solution:
    def threeSumClosest(self, nums: List[int], target: int) -> int:
        nums.sort()
        res = sum(nums[:3])
        for i in range(len(nums)):
            L, R = i+1, len(nums)-1

            while L < R:
                closest = nums[i] + nums[L] + nums[R]
                if abs(target - closest) < abs(target - res):
                    res = closest
                
                if closest > target:
                    R -= 1
                elif closest < target:
                    L += 1
                else:
                    return closest
        
        return res

🟡 4Sum :- click here

Solution:

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        nums.sort()
        n = len(nums)
        res = []
        
        for i in range(n-3):
            if i > 0 and nums[i] == nums[i-1]:
                continue

            for j in range(i+1, n-2):
                if j > i+1 and nums[j] == nums[j-1]:
                    continue

                L, R = j + 1, n - 1
                while L < R:
                    four_sum = nums[i] + nums[j] + nums[L] + nums[R]
                    if four_sum == target:
                        res.append([nums[i], nums[j], nums[L], nums[R]])
                    
                        L += 1
                        R -= 1
                        while L < R and nums[L] == nums[L-1]:
                            L += 1
                        while L < R and nums[R] == nums[R+1]:
                            R -= 1

                    elif four_sum > target:
                        R -= 1
                    else:
                        L += 1

        return res

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✅ 100% Yesified 😂

Theory vs Practice

@AceCoding

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💻 A2SV prep: Two pointers

🟢 1995. Count Special Quadruplets

Solution: This is a good example of questions with a small amount of input, or constraints that can be solved using a simple brute force.

class Solution:
    def countQuadruplets(self, nums: List[int]) -> List[List[int]]:
        res = 0 
        n = len(nums)
        for i in range(n-3):
            for j in range(i+1, n-2):
                for k in range(j +1, n-1):
                    for l in range(k +1,n):
                        if nums[i] + nums[k] + nums[j] - nums[l] == 0:
                            res += 1
        return res

🟡 454. 4Sum II

Intuition : from the first two arrays find the count of sum of paris in nums1 and nums2 next get the compliment from the last two arrays nums3 and nums4

class Solution:
    def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
        sum_count = Counter(a + b for a in nums1 for b in nums2)
        count = 0
        for c in nums3:
            for d in nums4:
                complement = -(c + d)
                count += sum_count.get(complement, 0)
        
        return count

🌟🚀 @AceCoding Presents! 🚀🌟