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completed comeptitive coding - 2#1190

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completed comeptitive coding - 2#1190
yashhh-23 wants to merge 1 commit into
super30admin:masterfrom
yashhh-23:master

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Copilot AI review requested due to automatic review settings July 8, 2026 12:52

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Pull request overview

Adds a Java solution for the classic “Two Sum” problem as part of the competitive coding exercises in this repository.

Changes:

  • Added twoSum implementation using a HashMap lookup strategy.
  • Added header comments describing complexity and approach.

Reviewed changes

Copilot reviewed 1 out of 2 changed files in this pull request and generated 3 comments.

File Description
twosum.java Introduces a Two Sum solution; needs minor adjustments to be self-contained Java and to simplify redundant logic.
Problem1.java Listed in the PR prompt, but the file wasn’t present in the repository snapshot provided to the reviewer, so changes could not be verified here.

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Comment thread twosum.java
// Did this code successfully run on Leetcode : yes
// Any problem you faced while coding this : difficult to come up with the approach of using hashmap to store the values and their indices.
// Your code here along with comments explaining your approach: hashmap is used to store the values and their indices. In the first loop, we store all the values and their indices in the hashmap. In the second loop, we check if the complement of the current value (target - current value) exists in the hashmap and if it does, we return the indices of the two values.
class Solution {
Comment thread twosum.java
Comment on lines +8 to +22
HashMap<Integer, Integer> map = new HashMap<>();

for(int i = 0; i < nums.length; i++) {
map.put(nums[i], i);
}

for (int i = 0; i < nums.length; i++) {
int cmp = target - nums[i];
if (map.containsKey(cmp) && map.get(cmp) != i) {
return new int[]{map.get(cmp),i};
}
map.put(nums[i], i);
}

return new int[]{};
Comment thread twosum.java
// Space Complexity : O(n)
// Did this code successfully run on Leetcode : yes
// Any problem you faced while coding this : difficult to come up with the approach of using hashmap to store the values and their indices.
// Your code here along with comments explaining your approach: hashmap is used to store the values and their indices. In the first loop, we store all the values and their indices in the hashmap. In the second loop, we check if the complement of the current value (target - current value) exists in the hashmap and if it does, we return the indices of the two values.
@super30admin

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Two Sum (twosum.java)

Strengths:

  • Excellent achievement of O(n) time complexity, meeting the follow-up challenge
  • Clear and informative comments explaining the approach
  • Proper use of HashMap for O(1) lookup operations
  • Good variable naming (cmp for complement is clear)
  • Handles the constraint of not using the same element twice with the != i check

Areas for Improvement:

  • The second loop's map.put(nums[i], i) is redundant since all elements are already in the map from the first loop
  • Could simplify by combining both operations in a single pass: check if complement exists, then add current element
  • The empty return new int[]{} could be more descriptive, though the problem guarantees a solution exists

Minor Optimization:

for (int i = 0; i < nums.length; i++) {
    int cmp = target - nums[i];
    if (map.containsKey(cmp)) {
        return new int[]{map.get(cmp), i};
    }
    map.put(nums[i], i);
}

This single-pass approach is more efficient as it avoids the redundant map population.

VERDICT: PASS


Interview Problem: 0-1 Knapsack Problem

The student appears to have submitted code for the wrong problem entirely. This could indicate:

  • Confusion about which problem to solve
  • Copy-paste error from a different assignment
  • Misunderstanding of the problem requirements

If this was intentional for a different assignment, the Two Sum solution itself is well-written with clear explanations. However, for the 0-1 Knapsack Problem specifically, the student needs to implement a dynamic programming approach using a 2D array where dp[i][w] represents the maximum value achievable with the first i items and a capacity of w.

For the Two Sum solution, the redundant HashMap population in the second loop could be removed to slightly improve efficiency.

VERDICT: NEEDS_IMPROVEMENT

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3 participants