Two pointer

[YogeshPardeshi]

No description provided.

[super30admin]

Arrange Colors (Problem1.java)

Strengths:

• Excellent implementation of the Dutch National Flag algorithm
• Clean, readable code with appropriate variable names
• Proper separation of concerns with a dedicated swap method
• Correctly handles edge cases (single element, all same color, etc.)

Areas for Improvement:

• Consider adding comments to explain the algorithm logic, especially for the three-pointer approach
• The solution is already optimal; no algorithmic improvements needed

This is a textbook-perfect implementation of the classic solution to this problem.

VERDICT: PASS

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3 sum (Problem2.java)

Strengths:

1. Good understanding of the sorting-based approach to handle duplicates
2. Clean code structure with separation of concerns (binary search in a separate method)
3. Proper use of early continue statements to skip duplicate elements
4. Better time complexity than the reference O(n³) solution

Areas for Improvement:

1. Critical Bug: The solution doesn't skip duplicates for the third element found via binary search. After finding a valid triplet, you need to move the left/right pointers past all duplicates to avoid adding duplicate triplets.
2. Suboptimal Algorithm: Using binary search gives O(n² log n) complexity. The classic two-pointer technique (moving left and right pointers toward each other) achieves O(n²) which is more efficient.
3. Edge Case Consideration: When binary search finds an element, you should verify that the index is valid and handle edge cases properly.

Recommended Fix:
Replace the binary search with a two-pointer approach:

int left = j + 1;
int right = n - 1;
while (left < right) {
    int sum = nums[i] + nums[j] + nums[left];
    if (sum == target) {
        result.add(Arrays.asList(nums[i], nums[j], nums[left]));
        while (left < right && nums[left] == nums[left+1]) left++;
        while (left < right && nums[right] == nums[right-1]) right--;
        left++;
        right--;
    } else if (sum < target) {
        left++;
    } else {
        right--;
    }
}

VERDICT: NEEDS_IMPROVEMENT

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Container With Most Water (Problem3.java)

Strengths:

1. Excellent implementation of the two-pointer technique - a classic and optimal approach for this problem
2. Clean, readable code with meaningful variable names
3. Correctly handles the pointer movement logic (moving the pointer with the shorter height)
4. Achieves O(n) time complexity, which is optimal for this problem
5. Proper handling of edge cases (minimum array length is 2)

Areas for Improvement:

1. Consider adding comments to explain the two-pointer logic for future maintainability
2. The variable h could be declared as final since it's recalculated each iteration, but this is a minor point
3. Could add input validation for null or empty arrays as a defensive programming practice

Overall, this is a solid, efficient solution that demonstrates good algorithmic thinking.

VERDICT: PASS